1 I J. A. HENDERSON'S JNTELLECTUAL AND ^PRACTICAL CORRECTED, REVISED. AND SUPPLEMENTED BY IMPORTANT RULES IN FRENCH, SPMISH AND GERMAN SECOND EDITION. IllFUj KANCISCO. f873. OF THE UNIVERSITY HENDERSON'S Intellectual and Practical LIGHTNING CALCULATOR. BY J, A. HENDERSON, GRADUATE OF UNION COLLEGE. .Author of Calculator, Book of Blocks Illus? trating Roots, and the New Decirnal Method of Computing Interest and Imparting the same. ADDRESSED TO. J.-.A. HENDERSON^ SAN FRANCISCO. ENTERED according to Act of Congress, in tne year of our Lord 1872, By J. A. HENDERSON, In the Office of Librarian of Congress, Washington.. PREFACE. It is better to know everything about something, than something about everything. Early ideas are not usually true ideas, but need to be revised and re-revised. Right means straight, and wrong means crooked. And knowing that thought kindles at the fire of thought, we do not hesitate to offer any apology for presenting to the public some new seed- thoughts, and right methods of operation in business calculations. The practical utility of this book consists in the brevity, conciseness, and general application of its rules. Par- ticular attention is invited to the grand improvements in computing time, all possible cases in interest, squaring and multiplying numbers, dividing and ! multiplying fractions, an infinite number of ways and the absolute right method, of extracting roots. TABLE OF CONTENTS. Lithograph of the Author Title Page 1 Preface The Arithmetical Alphabet 5 Numeration , 6 Method of Acquiring Multiplication Table 7 Method of Addition 8 The Lightning Process by Combination 9 Multiplication, Useful Contractions 10 Bapid Process of Marking Goods . . : 12 To Multiply and Divide by the Aliquot parts of 100 anc 1,000 14 Lightning Process of Calculating Interest 16 Problems in Interest 20 Method of Squaring Numbers by their Complement and Supplement 27 Method of Multiplying Numbers 29 Greatest Common Factor or Divisor 30 Least Common Multiple 31 Method of Adding and Substracting Fractions 32 General Principles of Fractions 33 Division of Fractions 35 To find the value of Currency when Gold is at a Stated Price 36 Interest Table and Form for making Tables 37 To find the difference of time between two dates, and tell the day of the week from the day of the month. 39 Powers and Hoots 42 Method of Extracting Square Root 44 Infinite number of ways of finding the Square Boot of any number 45 Absolute right method of extracting Cube Boot 47 Bule for Extracting Cube Boot 49 Examples in Cube Boot . . ^ 50 Square and Cube Boot of Fractions 52 To Find the Surface of Plane Figures 54 Method of Measuring Land 55 Method of Measuring Grain 57 Some of the Miscellaneous Weights to the Bushel 58 Short Methods in Division and Multiplication 58 Mental Exercise or Mental Calisthenics 62 Squaring and Multiplying Numbers 64 Miscellaneous Problems 66 General Information 72 THE LIGHTNING CALCULATOR. The arithmetical alphabet, as written and read, g g M rf The second 1 g 1 | I g is two times S B g I 1 the first; the -5 3 jy 5} is 30,1 , by 61 is 42* RULE, The integer increase by one; multiply by the integer, and the product of the fractional parts annex; increase the integer by one since the mm of the fractional parts make a Knit, RAPID PROCESS OF MARKING GOODS, To tell what an article should retail for to make a profit of 20 per cent., is done by re- moving the decimal point one place to the left. For instance, if hate cost $17.50 per dozen* remove the decimal point one place to the left, LIGHTNING CALCULATOR. 13 making $1.75 what they should be sold for apiece to gain 20 per cent, on the cost. If they cost $13 per dozen, they should be sold for $1.30 apiece, etc. RULE. Remove the point one place to the left, on the cost per dozen, to gain 20 per cent.; increase or diminish to suit the required rate. NOTE. Remove the point one place to the left, for 12 tens make 120. TABLE For marking all articles bought by the aozen. N. B. Most of these are used in business. To make 20 per cent., remove the point one place to the left. To make 80 per cent, remove tho point and add one half itself. 60 " " " " " third 50 fourth 44 i * fifth " 40 " 37H * 35 4 * 33>i < 30 28 " 25 " < 12 ^ 18^ seventh eighth ninth tenth twelfth " fifteenth itself. twentieth " twenty -fourth 4 * subtract ona sixteeeuth. tweniy-pixth If you buy one dozen shirts for $28, what shall I retail them for to make 50 per cent.? Ana. $3.50. 14 HENDERSON'S TABLE Of the Aliquot parts o/lOO and 1000. N. B. Most of these are used in business. 12* is X part of 100 8* is 1-12 part of 100 25 is 2-8 or X of 100 16^ is 2-12 or 1-6 of 100 37* is 3-8 part of 100 33 H is 4-12 or ^ of 100 50 i*4-8or*of 100 66- is 8-12 or % of 100 f2* is & part of 100 83}$ is 10-12 or 5-6 of 100 75 is 6-8 or % of 100 125 is ^ part of 1000 87* is 7-8 part of 100 250 is 2-8 or ^ of.... .... 1000 6U is 1-16 part of 100 375 is % part of 1000 18?4 is 3-16 part of 100 C25 is % part of 1000 3Ui is 5-16 part of 100 875 is % part of 1000 To multiply by an aliquot part of 100. KULE. Take such a part of the multiplicand as the multiplier is part of 100, and call it hundreds. To multiply by an aliquot part of 1000 : Tako such part of the multiplicand as the multiplier if part of 1000, and call it thousands. To divide by the aliquot parts of 100. To divide any number by 12 J: remove the pohit two places to the left, and multiply the quotient by 8. Multiply the quotient by 8, be- cause 12| is % of 100. To divide any number by 25: remove the point two places to the left, and multiply by 4. 345-^-25 3.45 4 Ans. 13.80 LIGHTNING CALCULATOR. 15 . To divide any number by 50: remove the point two places to the left and multiply by 2. .75-50 2 1.50 To divide any number by 75: remove tho point two places to the left, multiply by 4 and divide by 3; because 75 is f of 100 . To divide by the aliquot parts of 1000. To divide any number by 125: remove the point three places to the left and multiply by 8. Kemove the point fchree places to the loft to divide by a thousand and multiply by 8; because 125 is of a thousand. Thus: 34G7+425 9,712.29+425 8 8 Ans. 27.736 Ans. 77.69832 Etc., for all other examples. To divide any number by 250 : remove the point three places to the left and multiply by 4. 4.357^-250 4 Ans. 17.428 357.25-250 4 Ans. 1.42900 16 HENDEESON'S HENDERSON'S LIGHTNING PROCESS OF CALCU- LATING INTEREST. The base of our system of notation being 10, numbers increase and diminish in a tenfold ratio; increasing from the decimal point to the left, and decreasing from the decimal point to- wards the right. Hence, to divide any number by 10, remove the point one place to the left. To divide any number by 100, remove the point two places to the left. To divide any number by 1000, remove the point three places to the left. To multiply any number by 10, remove the point one place to the right. To multiply any number by 100, remove the point two places to the right. To multiply any number by 1000, remove the point three places to the right. INTEREST. Since the interest is generally a part of the principal, the method of calculating it, will come under the method of dividing. The rule establishes the time when a dollar makes a cent, and we remove the point two places to the left; for one hundredth of the principal equals the interest. In ten times that time, a dollar makes ten cents, and we remove the point LIGHTNING CALCULATOB. 17 one place to the left, because a tenth of the principal is the interest; in one tenth of the same time a dollar makes a mill, and we remove the point three places to the left, because one thousandth of the principal equals the interest. RULE. The reciprocal of the rate, or the rate in- verted, indicates the time when the decimal point can be removed, two places to the left in all cases; ten times that time one place to the left, and one tenth of the same time three places to the left. Increase or dimmish the results to suit the time given. TKe arithmetical alphabet is \, f , |, \ , -f-> f > J, \, f, and 0; -J- inverted is 1; f inverted is J; f- inverted is -J-, etc. If the rate is 1 per cent, per month, one inverted gives the time when a dollar makes a cent, and the point removed two places to left, shows the interest in every case. Ten times one inverted, or ten months, a dol- lar makes ten cents, and the point being removed one place to the left, all examples for that rate and time are calculated. One tenth of one month, or three days, a dollar makes a mill, and the point removed three places to the left, shows the interest in all examples for that rate and time. We remove the point one place to tha left, be- cause a tenth of the principal is the interest. We remove the point two places, because a hun- dredth of the principal is the interest. We remove the point three places, because a thou- 18 HENDERSON'S sandth of the principal is the interest. To reach all other time, simply increase or dimmish the results to suit the time given. $600.00 @ 1 per cent, per mo. for two months, remove the point two places to the left = $6. 00 the interest for one month. Twice $6.00, the interest for one month, is $12.00, the interest for two months. The interest on the same amount for three days is .60 cts., simply remov- ing the point three places to the left. The in- terest for ten months on the same amount, would be $60.00. Simply removing the point one place to the left, $60.00, the interest for ten months, plus $12,00, the interest for two months is $72.00, the interest for one year. By this method we can calculate an infinite number of examples in a moment when working from the base. At one per cent, per month: Thus. $2 9, 11, 22, 5J67.35 861.50 746.75 463.25 5J38.40 0100.50 LIGHTNING CALCULATOR, 19 By this method a world of work is done in the twinkling of an eye, and the way opened to the answer of every example in interest. The rate is 2 per cent, per month, 2 inverted is -|, or 15 clays, the point removed two places to the left, all examples are calculated for that rate and date, 10 times half a month or five months^ the point removed one place to the left, all ex- amples are calculated. One tenth of 15 days, or a day and a half, the point removed three places to the left, all examples are performed for that time *" 3 $10, 25, 2, 9, 0:00.50 650.00 475.30 460.50 2150.31 Simply increasing or diminishing the results ; we find the answer for any other time. 20 * HENDERSON'S PROBLEMS IN INTEREST. PROBLEM 1. What is the interest of $50 for 4 years at 6 per cent. SOLUTION. Bemoving the point one place to the left, we have $5.00 the interest for 20 months. J?or 40 months, it is $10.00; 8 months, being the fifth of 40 months, the interest would be $2.00; $10.00 plus $2.00 is $12.00 the interest for 48 months, or 4 years. PROBLEM 2. What is the interest of $10.00 for 2 years, at 5 per cent ? Simply remove the point one place to left, and you have the in- terest. PROBLEM 3. What is the interest of $48.00 for 6 years, at 5 per cent.? PROBLEM 4. What is the interest of $70.00 for 7 years, at 5 per cent.? PROBLEM 5. What is the interest of $68.00 for 5 years, at 6 per cent.? PROBLEM 6. What is the interest of $70.00 for 2 years, at 5 per cent.? PROBLEM 7. What is the interest of $75.00 for 5 years, at 3 per cent.? PROBi.Ejyf 8, What is the interest of $120.00 for 3 years, at 5 per cent.? PROBLEM 9. What is the interest of $100.00 for 10 years, at 6 per c T ent.? PROBLEM 10. What i$ tlie interest of $140.00 for 12 years, at 5 per p LIGHTNING CALCULATOB. 21; PROBLEM 11. What is the interest of $150.00 for 5 years, at 3 per cent.? PROBLEM 12. What is the interest of $145.00 for 6 years, at 5 per cent.? PROBLEM 13. What is the interest of $200.00 for 10 years, at 8 per cent.? PROBLEM 14. What is the interest of $250.00 for 3 years, at 8 per cent.? PROBLEM 15. What is the interest of $500.00 for 9 years, at 8 per cent.? PROBLEM 16. What is the interest of $50.00 for 2 years and 2 months, at 2 per cent.? PROBLEM 17. What is the interest of $80.00 for 8 years and 6 months, at 6 per cent.? PROBLEM 18. What is the interest of $90.00 for 5 years and 6 months, at 6 per cent.? SOLUTION. Remove the point one place to the left, we have $9.00 the interest for 20 months. The interest would be 3 times $9.00, which is $27.00 for 5 years'. The interest for 6 months would be one tenth of $27.00, which is $2.70, which added to $27.00, makes $29.70, Ans. PROBLEM 19. What is the interest of $90.00 for 12 years and 10 months, at 6 per cent.? PROBLEM 20. What is the interest of $200.00 for 4 years and 8 months, at 3 per cent.? PROBLEM 21. What is the interest of $70.00 for 8 years and 4 months, at 2 per cent.? PROBLEM 22. What is the interest of $225.00 for 52 days, at 7 per cent.? Ans. $2.25., 22 HENDERSON S PROBLEM 23. What is the interest of $500.00 for 26 days, at 7 per cent.? Ans. $2.50. PROBLEM 24. What is the interest of $500.00 for 2 years, 6 months, and 15 days, at 4 per cent.? SOLUTION. Remove the point one place to the left we have 50.00, the interest for 2 years and 6 months. Removing the point two places to the left, we have $5.00 the interest for 3 months; 15 days being one sixth of three months, we have 83J cts. the interest for 15 days, which added to $50.00 makes $50.83J, Ans. PROBLEM 25. What is the interest of $200.00 for 5 years, 9 months and 18 days, at 5 percent.? SOLUTION. Removing the point one place to the left we have $20.00, the interest for 2 years. The interest for 5 years would be 2J times $20.00, or $50.00. The interest for 1 year is $10.00; for 9 months it would be | of $10.00, which is $7.50. Removing the point 2 places to the left, we have $2.00, the interest for 72 days, the interest for 18 days would I e the fourth of $2.00, which is 50 cents, added to $57.50, would be $58.00, Ans, PROBLEM 26. What is the ii terest of $700.00 for 1 year, 7 months, and 18 days, at 6 per cent.? PROBLEM 27. What is the interest of $250.00 for 3 months, at 1 per cent, per month? SOLUTION. Remove the point two places to left, we have $2.50, the interest for one month. LIGHTNING CALCULATOB. 23 The interest for 3 months would be three times $2.50, which is $7.50, Ans. PROBLEM 28. "What is the interest of $60.00 for 6 years. 4 months, and 24 days, at 5 per cent.? PROBLEM 29. What is the interest of $40.00 for 1 year, at 1 per cent, per month ? PROBLEM 30. What is the interest of $950.25 for 9 months, at 1 per cent, per month? PROBLEM 31. What is the interest of $55.00 for 11 months, at 1 per cent, per month? PROBLEM 32. What is the interest of $200.00 for 10 months, at 1 per cent, per month? PROBLEM 33. What is the interest of $144.50 for 15 months, at 1 per cent, per month? PROBLEM 34. What is the interest of $60.00 for 22 months, al 1 per cent, per month ? PROBLEM 35. What is Ihe interest of $600.00 for 18 days, at 10 per cent, per ennum? SOLUTION. Eeinove the point two places to the left, we have $6.00, the interest for 36 days. The interest for 18 days is one-half of $6.00, which is $3.00, Ans. PROBLEM 36. WTiat is the interest of $250.25 for 35 days, at 10 per cent, per annum ? PROBLEM 37. What is the interest of $360.50 for 1 year, at 10 per cent, per annum ? $36.05, Ans. PROBLEM 38. What is the interest of $200.00 for 72 days, at 10 per cent, per annum? 24 HENDERSON'S PROBLEM 39. What is the interest of $80.00 for one year, at f per cent, per month ? Ans. $8.00. PROBLEM 40. What is the interest of $500.00 for 2 years, at f per cent, per month ? PROBLEM 41. What is the interest of $250.00 for 3 years, at -| per cent, per month? NOTE. Remove the point one place to the left, because a tenth of the principal is the in- terest. Two places, because a hundredth of the principal is the interest, etc. PROBLEM 52. What is the interest of $250.00 for one month, at 1 per cent, per month? SOLUTION. At 1 per cent, per month, one one hundredth of the principal is the interest, we therefore remove the point two places to the left. Ans. $2.50. Removing the point two places to the left, we have the answer. PROBLEM 43. What is the interest of $250.50 for 2 months, at 1 per cent, per month ? SOLUTION. Removing the point two places, we get the interest $2.505 for 1 month; for 2 months, the interest would be twice $2.505, which would be $5.01. PROBLEM 44. What is the interest of $100.00 for 15 days, at 1 per cent, per month? SOLUTION. Removing the point two places to the left, we get the interest $1.00, for 1 month. The interest for 15 days would be one half of $1.00, or 50 cents. LIGHTNING CALCULATOR. 25 PROBLEM 45. What is the interest of $145.00 for 3 days, at 1 per cent, per month ? SOLUTION. Eemove the point three places to the left, and we have $1.45, Ans. PROBLEM 46. What is She interest of $2000.00 for 9 days, at 1 per cent, per month ? SOLUTION. Remove the point three places, we have the interest $2.00 for 3 days; for 9 days, $6.00, Ans. PROBLEM 47. What is the interest of $250.25 for 12 days, at 1 per cent, per month ? PROBLEM 48. What is the interest % of $270.00 for 10 months, at 1 per cent, per month ? Ee- move the point 1 place to the left. Ans. $27.00. PROBLEM 49. What is the interest of $350.00 for 1 year, at 1 per cent, per month ? Kemove the point one place, we have the interest $35.00 for 10 months, for one year, one fifth more, $42.00, Ans. PROBLEM 50. What is the interest of $250.00 for 11 months and 3 days, at 1 per cent, per month? Interest 10 months, $25.00; interest 1 month, $2.50; interest 3 days, 25 cents, equals $27.75, Ans. PROBLEM 51. What is the interest of $2,500.00 for 1 year, at 1 per cent, per month? PROBLEM 52. What is the interest of $125.00 for 33 days, at 1 per cent, per month ? PROBLEM 53. What is the interest of $260.00 for 24 days, at 1J per cent, per month? 28 HENDERSON'S SOLUTION. Remove the point two places to the left, we have the interest $2.60, Ans. PROBLEM 54. What is the interest of $360.00 for 1 month, at 1J per cent, per month ? SOLUTION. Remove the point two places to lie left, we have $3.60, the interest for 24 days; add i, .90, we have $4.50, Ans. PROBLEM 55. What is the interest of $800.50 for 8 months, at 1J per cent, per month ? SOLUTION. Remove the point one place to the left. Ans. $80.05. PROBLEM 56. What is the interest of $500.00 for 1 year, at 1 J per cent, per month ? SOLUTION. Remove the point one place to the left, we have the interest $50.00 for 8 months. For 4 months, the interest would be $25.00, added to $50.00, equals $75.00, Ans. PROBLEM 57. What is the interest of $900.00 for 4 months, at 1J per cent, per month? SOLUTION. Remove the point one place to the left, we have $90.00, the interest for 8 months; for 4 months, the interest would be one half of $90.00, or $45.00, Ans. Removing the point one place to the left, gives the interest of any sum for 8 months, at 1J per cent., increase or diminish the result to suit the time given LIGHTNING CALCULATOR. 27 METHOD OF SQUARING NUMBERS BY THEIR COM- PLEMENT AND SUPPLEMENT, The complement of a number is the difference between the number and some particular num- ber above it. The supplement of a number is the difference of a number and some number below it. (99) 2 = 9801. Take the complement of 99 from it, call it hundreds, and add the square of the complement. EXPLANATION. Let N equal 99, and C equals 1. Then N plus = 100. N C = 98. Mul- tiplying the two equations together, we have N 2 C 2 = 9800. Add C 2 to both menbers o the equation, and we have N 2 9801, the square of 99. (98) 2 = 9604. Now 2, the complement of 98 from 98 = 96; call it hundreds, and add the square of 2, and we have 9604, the square of 98. (97) 2 = 9409. The complement 3 from 97 = 94; call it hundreds, and add the square of 3, and we have the square of 97. (96) 2 = 9216. The complement of 96 is 4; 4 from 96 = 92, call it hundreds, and add the square of 4, and we have the square of 96. (95) 2 = 9025. The completent of 95 is 5; 95 5 90, call it hundreds, and add the square of 5, and we have the square of 95. (101) 2 = 10201. The supplement of 101 is 1 ; 28 HENDEKSON'S 1 added to 101 is 102, call it hundreds, and add the square of 1, and we have 10201 the isquare of 101. (102) 2 104U4. The supplement is 2, added to 102 is 104 T call it hundreds, and add the square of 2, and we have 10404 the square of 103. RULE. WJien ubove the base, add the supple- ment, call it hundreds, and add the square of the supplement, call it hundreds, because the number ivhen increased by the supplement, is multiplied by one hundred in this case, when below t substract the complement. (103) 2 = 10609. The supplement 3 added, call it hundreds, and add T!he square of 3. (104) 2 = 10816. (1001) 2 =1002001. The supplement is 1 added to 1001 = 1002, call it thousands, and add the square of 1, and it equals 1002001. (1002) 2 = 1004004. (1003)'= 1006009. (1004) 2 = 1008016. (999) 2 = 998001. The complement is 1 from 999 equals 998, call it thousands, and add the square of 1, and we have the square of the num- ber. (998) 2 =996004. (997) 2 = 994009. (996) 2 = 992016. (995) : = 99002d. (994) 2 = 988036, etc. Take any number that is easy to multiply by for the base 10, 20, 30, 50, 80, 100, 1000, etc. 9 2 = 81. The complement of 9 is 1, 1 from LIGHTNING CALCULATOR. 2t> 9 leaves 8, call it tens and add the square of 1, and we have the square of 9. 8 2 =64. The complement of 8 is 2, 2 from 8 leaves 6, call it tens, and add the square of 2, and we have the square of 8. (II) 2 = 121. The supplement of 11 is 1, 1 added to 11 is 12, call it tens and add the square of 1, and we have the square of 11. (12)2 = 144. The supplement is 2, 2 added to 12 is 14, call it tens and add the square of 2, and we have the square of the number. (13) 3 = 169. The supplement of 13 is 3, 3 added is 16, call it tens and add the square of 3, and we have the square of the number. (14) 3 = 196. (15) 3 = 225. (19) 2 = 361. The complement is 1, 1 from 19 leaves 18, 18 multiplied by 20, equals 360, add the square of 1, and we have the square of the number. (18)2 = 324. (17)2=289. (16) s =256. (21)2 =441. (22)2484. (49)2=2401. The com- plement is 1, 1 from 49 is 48, call it fifties, and add the square of 1, and we have 2401, Ans. (51)2=2601. (52)2=2704. (53)2=2809. To multiply numbers. KULE. T/ie product of any two numbers is the square of their mean, diminished by the square of half their difference. 19x21=399. The mean is 20, the square of 20 is 400; 400 I 2 is 399, the product of 19 X 30 HENDEESON'S 21, 18 x 22. The mean is 20, the square of 20 is 400, 2 2 is 4; 4 from 400 leaves 396, the product, 17x23=391. The square of 3 is 9; 9 from 400 leaves 391 the product. 16 x 24=384. The square of 4 is 16. 16 from 400 leaves 384 the product. 15 x 25 =375. The square of 5 is 25. 25 from 400 leaves 375 the product. 29x31=899. The mean is 30. The square is 900, minus the square of 1 is 899 their pro- duct. 28 x 32 = 896. The square of the mean is 900, minus the square of 2 is 896 the product. 27x33=891. 26x34=884. 25x35=875. 39x41=1599. 38x42=1596. 37x43=1591. 36x44=1584. 35x45=1575. 34x46=1564. 49x51=2499 48x52=2496. 47x53=2491. GREATEST COMMON FACTOR OR DIVISOR What is the greatest common divisor of 21 and 77. Separating the numbers into their prime factors we have 21=7x3, 77=7 Xll, hence 7 is the greatest common factor or the greater com- mon divisor of the two numbers. RULE. Separate the numbers into their prime factors. Tlie product of all the factors that are common will be the greatest common divisor. LIGHTNING CALCULATOB. 31 What is the greatest divisor of 25 and 60. 25 5 X 5, 60=5 X 3 X 2 X 2 ? Hence 5 is the greatest common divisor. What is the greatest common divisor of 5, 15 and 20? What is the greatest common divisor of 36, 18, 24 and 12. 36=6x6, 18=6x3, 24=6x4 12 =6 X 2 ? Hence 6 is the greatest common factor or divisor. What is tne greatest common divisor of 135 and 225? What i* the greatest common divisor of 4, 8, 12, 16? What is the greatest common divisor of 25 and 75? What is the greatest common divisor of 13 ^nd 65? What is the greatest common divisor of 14 and 42? LEAST COMMON MULTIPLE. A multiple of a number is any number which contains it as a factor. A common multiple of two or more numbers is any number which contains them all as factors. The least common multiple of two or mo*e numbers is the least number which contains them all as factors. Hence it follows a multiple 32 HENDERSON'S of a number must contain all the prime factors of that number. A common multiple of two or more numbers must contain all the prime factors of those num- bers. The least common multiple of two or more numbers must be the least number that contains sill the prime factors of those numbers. KULE. Tlie product of all the prime factors of that number having the greatest number of prime factors, and those prime factors of the other num- bers not found in the factors of the number taken, will be the least common multiple. What is the least common multiple of 12 and 18? 12=2X2X3, 18=2X3X3. The least com- mon multiple is 2X2X3X3 or 36. What is the least common multiple of 4 and 6? What is the least common multiple of 18 and 36? What is the least common multiple of 4, 6, 8 and 10? WTiat is the least common multiple of 2, 4, 6, 9 and 18? What is the least common multiple of 2, 3, 4, 5 and 6 ? RULE FOE ADDING AND SUBTRACTING FRACTIONS. First make the fractions similar by reducing them to the same denominator. Add the numer- LIGHTNING CALCULATOR. 33 ators and place the sum over the common denomin- ator. In subtraction ivrite the difference of the numerators over the common denominator. What is the sum of -\ and |, | ^V, -J=*V> A +&= , Ans. |+l=l, 4+f=l*. What is the sum of T 5 and i=|. What is the sum of -fa and 4 = lf What is the sum of f and 1=1 What is the sum of f and J-=l What is the sum of f and -1=1 From f subtract ^= T V From | subtract |. |=Jf , |=^, Ji~A= 8 V From A take f |=^, A &=& What is the sum of 3, 2J, 4J, 5J= 15f . Add the fractions and whole numbers sepa- rately. What is the sum of 9J, 6J, 7|=23J. From 8t take 3, J=f , f i=J. 83=5; 5 +i=5i. From 23f take 9J. f=f , J=f, f-*=fc 25 GENERAL PRINCIPLES OF FRACTIONS. Multiplying the numerator multiplies the frac- tion. Dividing the numerator divides the fraction. Multiplying the denominator divides the frac- tion. 34 HENDERSON'S Dividing the denominator multiplies the frac- 1 tion. Multiplying both terms of the fraction by the ' same number does not change its value. Fractions are called similar when they have a common denominator, as f , |, f , ^. Dissimilar fractions are fractions which are not alike, as f , f , |, 5 , f . The numerators of similar fractions only can be added, The common denominator is written under the sum or difference . Multiply T * T by 8=ff=an. Multiply A by 14=j=5. Multiply 40 by f =5x5=25. Multiply 3J by 6. Multiply the whole num- ber and fraction separately. 6xJ=3, 6 x 3=18 x 3=21. Multiply 4J by 8. 8xJ 2f, 8x4=32+2f= 34|. Multiply 7|- by 9. 9xi=4J, 9x7=63-)-4i= Multiply 8J by 12. 12xi=6, 12x8=96+6= 102. Multiply 7J by 7. 7| x 7J=52. T 9 6 Multiply 1\ by 7J=56J. Multiply 8| by 8J=72|. Multiply 9f by 9f=9 f OF TBEB f TTNIVERSIT^ LIGHTNING CALCULATOB. '^ElllfM -- DIVISION OF FRACTIONS. RULE. Reduce mixed numbers to improper frac- tions, and whole numbers to the form of fractions; multiply the dividend by the divisor inverted, or multiply both numerator and denominator by the least common midtiple of the denominators of the fractional parts. $3i=sJ=2. Simply multiplying numerator and denominator by 2. Divide 5J by 2J-. Multiply both numerator and denominator by 6, the least common mul- tiple of 2 and 3. Divide 25 by =50. Divide 21 by 3J=fJ=6 T V To divide any number by 3| , remove the point one place to the left and multiply by 3. Divide 20 by 3J . Remove the point one place we have 2, 2x3=6 Ans. Divide 27 by 3J=8 T V- To divide any number by 2 J, remove the point one place to the left and multiply by 4. Divide 20^ by 2J. Remove the point one place to the left and multiply by 4. Removing the point one place to the left makes 2-^, 2^X^=84- Ans. To divide any number by 1-J-, remove the point one place to the left and multiply by 9. Divide 11 by 1-^9^. Divide any number by 5. Remove the point 36 HENDERSON'S one place and multiply by 2. Removing the point one place to the left divides the number by 10. In dividing by 10 we divide by a num- ber twice too large; therefore we multiply by 2 for the correct result. To divide any number by 12J, remove the point two places to the left and multiply by 8. Divide 125 by 12J. Divide 47 5 by 12J. Divide 96 by 12J. Divide 99 by 12J. To divide any number by 25, remove the point two places to the left and multiply by 4. To divide any number by 33J, remove the point two places to the left aud multiply by 3. To divide any number by 50, remove the point two places to the left and multiply by 2. To divide by 66f , remove the point two places to the left, divide bv 2 end multiply by 3. TO FIND THE VALUE OF CURRENCY WHEN GOLD IS AT A STATED PRICE, When gold is 111|, what is the value of $1.00 currency? "We take the 100, the number of cents in a dollar, as the -numerator, and the value of the gold as the denominator. Simplify the fraction by multiplying the numerator and de- LIGHTNING CALCULATOR. 37 nominator by 9 and we have T F of a dollar or 90 cents; the value of the currency. When gold ts 109^, what is the value of $1.00 currency? 100 900 450 ~ 319 109 : 982 : 491 When currency is worth 75 cents, what is the value of gold? inn 4.4 ^L =; |, | of 100 cents equals $1.33}. 75 o o When gold is worth 105J, what is the value of $1.00 currency? 100 _ 200 _ * 94 166 105! " Z 2lT 211 RULE. We take 100, the number of cents in a dollar, for the numerator, and the value of gold or currency, as the case may be, for the denominator. Simplify the fraction l>y annexing ciphers to the numerator and dividing by the denominator. INTEREST TABLE AND FORM FOR MAKING TABLES. Ehe following Table gives the Interest on any amount at 7 per cent., by simply removing the Doint to right or left, as the case ma,v require: 38 HENDERSON'S Number of Days. $100 $90 $80 $70 1 .0192 .01726 .01534 .01342 2 .0384 .03452 .03058 .02685 3 .0575 .05178 .04603 .04027 4 .0767 .06904 .06137 .05370 5 .0959 08630 .07671 .06712 6 .1151 .10356 .09205 .08055 7 .1342 .12082 .10740 .09897 8 1532 13808 12274 .10740 9 .1726 .15534 .13808 .12089 90 1 7260 1 5342 1 38082 1 . 20822 93 1.7836 1.60521 1.42685 1.24849 100 1.9178 1 82603 1.53425 1.24247 $60 $50 $40 $30 $20 .01151 .00950 .00767 .00575 .00384 .02301 .01918 . 01534 .01151 .00767 .03452 .02877 .02301 .01726 .01151 .04603 .02836 .03068 .02301 .01536 .05753 .04795 .03836 .02877 .01918 .06904 .05753 .04603 .03452 .02313 .08055 .06712 .05370 .04027 .02685 .09205 .07671 .06137 .04603 .03068 1.0356 .08630 .06904 .05178 .03452 1.03562 .86301 .69041 .51781 .34521 1.07014 .89178 .71342 .53508 .35671 1.15065 .95890 .76712 .57534 .48356 LIGHTNING CALCULATOR. 39 To FIND THE DIFFERENCE OF TIME BETWEEN Two DATES BY THE FOLLOWING TABLE : RULE. Opposite the day of the month is written the number of days of the year ivhich have expired. Subtract this number from the whole number of days that have expired at the last date. Thus : What is the time from the first day of March to the 27th day of September? The 1st day of March we find by the table that 60 days of the year are gone. The 27th day of Septem- ber we find that 270 days are gone. Hence 270 days minus 60 days equals 210 days, the time between the two dates. To FIND THE DAY OF THE WEEK FKOM THE DAY OF THE MONTH BY THE SAME TABLE : Cast the sevens out of the day of the month, the ratio of the month, the ratio of the year which is 3, and the year. One of a remainder will be the first day of the week, two, second, etc. the last day of the week. The ratio of the month is found above its name. The ratio of every month except January and February is one more in Leap Years. 40 HENDEKSONS 3 January 6 I ebruary 6 March 2 April 4 May June 1 1 1 32 1 60 1 91 1 121 1 152 2 2 2 33 2 61 2 92 2 122 2 153 3 3 3 34 3 62 3 93 3 123 3 154 4 4 4 35 4 63 4 94 4 124 4 155 5 5 5 36 5 64 5 95 5 125 5 156 6 6 6 37 6 65 6 96 6 126 6 157 7 7 7 38 7 66 7 97 7 127 7 158 8 8 8 39 8 67 8 98 8 128 8 159 9 9 9 40 9 68 9 09 9 129 9 160 10 10 10 41 10 69 10 100 10 130 10 161 11 11 11 42 11 70 11 101 11 131 11 162 12 12 12 43 12 71 12 102 12 132 12 163 13 13 13 44 13 72 13 103 13 132 13 164 14 14 14 45 14 73 14 104 14 134 14 165 15 15 15 46 15 74 15 105 15 135 15 166 16 16 16 47 16 75 16 106 16 136 16 167 17 17 17 48 17 76 17 107 17 137 17 168 18 18 18 49 18 77 18 108 18 138 18 169 19 19 19 50 19 78 19 109 19 139 19 170 20 20 20 51 20 79 20 110 20 140 20 171 21 21 21 52 21 80 21 111 21 141 21 172 22 22 22 53 22 81 22 112 22 142 22 173 23 23 23 54 23 82 23 113 22 143 23 174 24 24 24 55 24 83 24 114 24 144 24 175 25 25 25 56 25 84 25 115 25 145 25 176 26 26 26 57 26 85 26 116 26 146 26 177 27 27 27 58 27 86 27 117 27 147 27 178 28 28 28 59 28 87 28 118 28 148 28 179 29 29 V 29 88 29 119 29 149 29 180 30 30 30 89 30 120 30 150 30 181 31 31 31 90 31 151 LIGHTNING CALCULATOB. 2 July 5 August 1 September 3 October 6 November 1 December 1 182 1 213 1 244 1 274 1 305 1 335 2 183 2 214 2 245 2 275 2 306 2 336 3 184 3 215 3 246 3 276 3 307 3 337 4 185 4 216 4 247 4 277 4 308 4 338 5 186 5 217 5 248 5 278 5 309 5 339 6 187 6 218 6 249 6 279 6 310 6 340 7 188 7 219 7 250 7 280 7 311 7 341 8 189 8 220 8 251 8 281 8 312 8 342 9 190 9 221 9 252 9 282 9 313 9 343 10 191 10 222 10 253 10 283 10 314 10 344 11 192 11 223 11 254 11 284 11 315 11 345 12 193 12 224 12 255 12 285 12 316 12 346 13 194 13 225 13 256 13 286 13 317 13 347 14 195 14 226 14 257 14 287 14 318 14 348 15 196 15 227 15 258 15 288 15 319 15 349 16 197 16 228 36 259 16 289 16 320 16 350 17 198 17 229 17 260 17 290 17 321 17 351 18 199 18 230 18 261 18 291 18 322 18 352 19 200 19 231 . 19 262 19 292 19 323 19 353 20 201 20 232 20 263 20 293 20 324 20 354 21 202 21 233 21 264 21 294 21 325 21 355 22 203 22 234 22 265 22 295 22 326 22 356 23 204 23 235 23 266 23 296 23 327 23 357 24 205 24 236 24 267 24 297 24 328 24 358 25 206 25 237 25 268 25 298 25 329 25 359 26 207 26 238 26 269 26 299 26 330 26 360 27 208 27 239 27 270 27 300 27 331 27 361 28 209 28 240 28 271 28 301 28 332 28 362 29 210 29 241 29 272 29 302 29 333 29 363 30 21 30 242 30 273 30 303 30 334 30 364 31 21 31 243 31 304 31 365 42 HENDEKSON'S POWERS AND ROOTS. The product of a number taken any number of times as a factor, is called a power of the number. A root of a number is such a number as taken some number of times as factor will produce a given number. If the root is taken twice as a factor to pro- duce the number, it is the square root. If three times, the cube root. If four times, the fourth root, etc. ILLUSTRATION. 5 is the square root of 25. The cube root of 125. The fourth root of 625, because (5) 2 =25, (5) 3 =125, (5) 4 =625. (1)=1 (1)3=1 (2) 2 =4 (2) 3 =8 (3) 2 =9 (3) 3 =27 (4) 2 =16 (4) 3 =64 (5) 2 =25 (5) 3 =125 (6) 2 =r36 (6) 3 =216 (7) 2 =49 (7) 3 =343 (8) 2 =64 (8) 3 =512 (9) 2 =81 (9) 3 =729 (10) 2 =100 (10) 3 =10QO We observe that the square of any one of the digits is less than 100. And the cube of any one of the digits is less than 1000. Hence the square root of two figures cannot give more than one figure. LIGHTNING CALCULATOR. 43 Hence if we begin at the right of any number and separate it into periods of two figures each, the number of periods would be the same as the number of figures in its square root. In order to understand the method of extract- ing square root, it is necessary to consider how the square of a number consisting of two parts is formed from those parts. To do this let a represent any number what- ever, ~b represent any other number, then will a -f- & represent the sum and (a + 6) 2 the square of the sum of any two numbers, but since the square of any two terms is the square of the first, plus two times the first into the second, plus the square of the second : we have (a -f 6) 2 =:a 2 -f 2 a 6 + & 2 . ILLUSTRATIONS. 23 here a =20 and & 3. HencO (a -j- Vf will equal (20 -j- 3) 2 . In applying the above formally, commence at the units in- stead of the tens to find the square of the num- ber. Thus 3 2 is 9, two times 3 into 2 is 12. Write down the 2 and carry the 1 to the square of the first term 2, and we have 529, the square of 23 and 23 is the square root of 529. The square of any number of terms is the square of the first, plus two times the first into the second, plus the square of the second, plus two times the sum of the first two into the third, plus the square of the third, plus two tjie sum of the first three into the fourth, plus the square of 44 HENDERSON'S the fourth, etc. Note In applying the above formula commence at the units to square num- bers. METHOD OF EXTRACTING SQUARE BOOT. 2 1 625. This number 'contains two periods; hence there are two figures in the roots. The greater square below 6, the first or left hand period is 4, the root of which is 2; and since there are two figures in the root, 2 will stand in the tens place and equal 20. Hence, we sub- tract the square of 20, which is 400, from 625, and we have 225 remaining. We have found a square 20 feet on a side. Now, in order to pre- serve the square, we make the addition on two adjacent sides. Hence, we double 20, the length of one side, and get 40, the trial divisor; dividing 225 by 40, we get the width of the ad- dition, 5 feet; adding 5 feet to 40 feet, the width of the little square in the corner, we get 45, the true divisor. Multiplying 45 by 5, we get 225, the surface of the addition. Hence, 25 is the length of one side of a square that contains 625 square feet. LIGHTNIN A />LCULATOB. 45 ICO J5625(100+20+5- 1st trial divisor 200 10000 1st true " 220 5625 2d trial " 240 4400 1225 2dtrue " 245 1225 We may have an infinite number of ways for finding the square root of any number. Thus : Presume the root of the number to be divided into a certain number of equal parts* Let 5 a equal the square root of 15625. Since the square of the root is equal to the number (5a) 2 or, 25a 2 =15625 and a 2 =625, and a=25. 5a is the root =125. Presume the root of 15625 to be 75a, then (25a) 2 =625(a) 2 =15625, (a) 2 = 25, a=5, 25a=125. In the same way, we may presume the root to be divided into 2, 3, 4, 5, or any number of equal parts. Hence the rule : Divide any number by the square of two, extract the square root of the quotient, "end we have one half of the root of the number. Divide any number by the square of three, ex- tract the square root of the quotient, and we have one third of the root of the number. Divide any number by the square of four, ex~ tract the square root of the quotient, and we have a fourth of the square root of the number, etc. 45 HENDERSON'? What is the square root of 9604? What is the square root of 2401 ? What is the square root of 225? What is the square root of 64? CUBE HOOT. RELATION OF CUBE TO BOOT. I 3 = 1 03 _ Q Bv observation we see that the zj O ^ 3 3 27 entire part of the cube root of any 4 3 = 64 number below 1000 will be less than 5 3 =125 jo and will, therefore, contain but /3 Q-| / 73 Hqlq one figure- The entire part of the gs 512 cu ^ e ro t f a number containing 9 3 =739 four, five or six figures, will contain 10 3 =1000 two figures, and so on with the larger numbers. Hence: If we begin at the right of any num- ber, and separate it into periods of three figures each, the number of periods will equal the number of figures in the entire part of the cube root. The cube of the highest denomination will be found in the left hand period. The cube of the two highest will be found in the two left hand periods, etc. A cube of any number of terms, is the cube of the first term, plus three times the square of the first into the second, plus three times the first into the square of the second, plus the cube LIGHTNING CALCULATOE. 47 of the second, plus three times the square of the sum of the first two into the third, plus three times the sum of the first two into the square of the third, plus the cube of the third, etc. METHOD OF EXTRACTING CUBE BOOT. 10000 3 1 2 953 1 125(100+20+5 1 000 000 30000 Trial divisor. 6000 953 125 4001 728000 36400 1st true divisor 225 125 6000 225 125 800 43200 2d trial divisor. 1800 25 45025 2d true divisor. EXPLANATION. We separate the number into periods of three figures each, by placing small digits over the periods. "We find the greatest cube in the first or left hand period, which is 1, the cube root of which is 1; and since there are three periods, fhere will be three figures in the root, and this 1 will stand in hundreds place and equal 100. We will presume the linear edge of a cubical block to be 100 feet. The surface of one side will be 100 times 100, or 48 HENDERSON'S 10000 square feet, and the solid contents will be 100 times 10000, or 1000000 solid feet; sub- tracting this number from the given number, we have 953125 feet remaining. To increase this cube and preserve the cubical form, we must make the addition on three ad- jacent sides; and since 10000 is the surface of one side, three times 10000, or 30000 will be the surface of three sides, which forms the trial divisor; dividing the dividend by this number, we find 20, the thickness of the addition; but besides these three large square pieces, there are three parallelopipedons, the length of each 100 feet, the width 20 feet. Hence, these sur- faces would be 20 times 300, or 6000. The little cube is 20 feet each way, the surface of one side of it would be 20 times 20, or 400, adding 30000, 6000 and 400, we have 36400, the sum of the surfaces of one side of each of the pieces making the addition. Multiply 36400 by 20, the thick- ness of the addition, we have 728000 the solid content of the addition. Subtracting from the last dividend, we have 225125 feet to be still added. The next trial divisor is three sides of the complete cube; by observation we see that 36400 lacks of being three sides of the complete ' cube. One side of each of the parallelopipedons and two sides of the little cube. Hence, by bringing down 6000 and doubling 400, adding the 36400, we obtain three sides of the com- LIGHTNING CALCULATOR. 49 plete cube, or tlie trial divisor; dividing, we find the thickness to be 5 feet. The three de- ficiencies, the length of one is 120 feet, the length of three would be 360 feet, the width 5 feet, the sum of the surfaces of one side of each would be five times 360 feet, which is 1800; the surface of one side of the small cube would be 5 times 5, or 25, adding to the 43200, we have 45025 feet; multiply by 5 to get; the solid contents of the last addition; if there were another figure in the root, we would simply bring down the 1800, double 25, and add to the last true divisor for the next trial divisor. This method of finding trial divisors is- of universal application, and the rule may be stated thus: Add to each true divisor, as they occur, twice the surface of one side of the small tube, and one each of the three parallelopipedons for the trial divisor 9 for that will make three sides of flie complete cube. We may have an infinite number of ways of finding the cube root of any number. Since the cube root of a number raised to the third power is alwafs equal to the number, we may presume the cube root of 1953125 to be divided into five equal parts represented by 5 a. The cube of (5a) 3 or 125a 3 =1953125, and a 3 wil equal 1953125-^125=15625. The cube root of 15625 is 25, a =25; 5 a=125 the cube root of the number. 50 HENDERSON'? In the same way we might presume the root to be divided into 25 equal parts represented by 25a. 25a 3 or 15625a 3 =1953125, and a 3 ^125 and a =5 and 25a the root of the number equals 5 times 25 or 125. EULE. Divide any number by the cube of 2, ex- tract tJie cube root of the quotient and we have half tliecube root of the number. Divide any num- ber by the cube of 3 or 27, extract the cube root of the quotient and we Jiave one third of the root of the number. Divide any number by the cube of 4 or 64, extract the cube root of the quotient and we have one fourth of the root of the number. Di- vide any number by the cube of 5 or 125, extract the cube root of the quotient and we have one fifth of the root of the number, etc. EXAMPLES IN CUBE BOOT, 1728)110592(64 10368 6912 6912 Presume the root to be divided into twelve equal parts. Hence the cube root of the quotient of the number divided by the cube of twelve, is T V of the root of the number. The cube of 12 is 1728, 110592 -f- 1728 =64, and the cube root of 64 is 4, 4 is T V of the cube root, the cube root would be 12 times 4 or 48. LIGHTNING CALCULATOR. 61 What is tlie cube root of 5931 J ? Presume the root to be divided into 13 equal parts. What is the cube root of 117649 ? Presume the root to be divided in 7 equal parts. What is the cube root of 97336 ? Presume the root to be divided in 23 equal parts. What is the cube root of 95112 ? Let the root be divided into 29 equal parts. The number divided by the cube of 29 equals 8, and the cube root of 8 is 2. Hence 29 times 2 is the cube root of the number or 58. What is the cube root of 91125 ? Let the root be divided into 9 equal parts, the number divided by the cube of 9 equals 125, the cube root of 125 is 5. Hence 9 times 5 or 45 is the cube root of the number. What is the cube root of 216x343? The cube root of 216 is 6. The root of 343 is 7. The cube root of the product is 6 times 7 or 42. What is the cube root of 64x125?; The cube root of 64 is 4. The cube root of 125 is 5, 5x4=20. The cube root of the pro- duct. What is the cube root of 125x125=5x5 ? What is the cube root of 125x15625=5x25 52 HENDERSON'S What is the cube root of 512x729 ? The cube root of 512 is 8. The cube root of 729 is 9. The cube root of the product is 8x9 or 72. What is the cube root of 216x729 ? The cube root of 216 is 6. The cube root of 729 is 9. The cube root of the product is 6x9 or 54. The methods which we have presented are of universal application, and are fully and clearly illustrated by Henderson's book of blocks illus- trating roots, copyrighted the llth clay of May, A. D. 1872. SQUARE AND CUBE ROOT OF FRACTIONS. ; To square a fraction, we square its numerator for the numerator^ and its denominator for the denominator. Hence,- to find the square root of a fraction, we must extract the square root of its numerator, for the numerator of the answer, and the square root of its denominator for the denominator otihe answer. ILLUSTEATIQNS. Find the square root of |. The squ'are root of 4, the numerator is 2. The square root of 9, the denominator, is 3 . Hence the answer, f . What is the square root of T 2 ^- = T V- LIGHTNING CALCULATOR. 53 What is the square root of .0081 = .09. When both terms of the fraction are not per- fect squares, only an approximate Value of the root can be obtained. In order that the denominator of a decimal fraction may be a perfect square, its numerator must contain an even number of decimal places. Hence, to extract the square root of a decimal fraction, make its number of decimal placqs even, by annexing a zero, if necessary; extract the root, as in whole numbers, observing that there will be one decimal place in the root for 'every two hi the given fraction, the root may be found to any number of decimal places by an- nexing two zeros for every additional figure. To extract the cube root of a fraction, we ex- tract the cube root of the numerator for the numerator of tlie answer, and the cube root , of the denominator for the denominator of the an- swer. If its numerator and denominator are not perfect cubes, the approximate value of the cube root can only be obtained. If the denom- inator is not a perfect cube, both terms should be multiplied by the square of the Denominator. Hence, to extract the square root of a decimal fraction, annex zeros, if necessary, to make. its number of decimal places some multiple of three; extract its root, as in whole numbers, observing that there will be one decimal place for everythree in the given fraction. 54 HENDERSON'S TO FIND THE SURFACE OF PLANE FIQUKES. A triangle is a figure having three sides and three angles. The altitude of a triangle is the perpendicular distance from the side assumed as its base to to the vertex of the opposite angle. B c is the perpendicular, and the A D the base. KULE. To find the surface of any triangle, multiply the base by half the altitude. A right-angle triangle is a triangle having a right angle. Lines are parallel when they lie in the same direction. A parallelogram is a fotir-sided figure having its opposite sides parallel. A trepizoid is a four-sided figure, having two of its sides parallel. A polygon is a figure bounded on all sides by straight lines. Similar figures are those which have the same shape. The corresponding sides are proportional. The base of a figure is the side on which it is supposed to stand. The altitude of a rectangle, a parallelogram or a trepizoid, is the perpendicular distance between its parallel basis. The area of a rectangle is 4he length multi- plied by the width. UGHTNING CALCULATOR. 55 METHOD OF MEASUBING LAND. Find the number of rods by multiplying the length by the width. Kemove the point two places to the left, divide by eight and multiply the quotient by five; or remove the point two places, take f of the result, and we have the number of acres. Thus: 3280 rods, the point removed two places leaves 32.80-5-8 = 4.1. 4,1 X 5 = 20.5 acres. What is the number of acres in 2440 rods ? Remove the point two places we have 24.40; f X 24.40 is 15J, the number of acres. This method is of universal application, and may be stated in the following words : Remove the deci- mal point two places to tlie left, and f of the quotient are tlie number of acres. We remove the point two places to reduce the number to units of a hundred, and since there are f of a hundred rods in one acre, five times J of the number of hundred rods must equal the number of acres; or simply ftie point removed two places and the quotient divided by f equals the number of acres. What are the number of acres in a field 160 rods wide and 480 rods long? Kemove the point two places on 160, and take f of the quo- tient, we find one acre multiplied by 480, the length, we get 480 acres, Ans. What is the number of acres in a field 2200 rods long and 640 wide? 56 HENDERSON'S What is the number of acres in a field of tri- angular shape ? The base of the triangle is 800 rods and the altitude 300; since the area is the base multiplied by half the altitude. Half the altitude is 150; remove the point two places on 800, and we have 8, and f X8=5, and 5x150= 750, the number of acres in the field. The area of a circle also equals the square of its radius multiplied by 3.1416, the ratio of the circumference to the diameter. If the radius is two feet the area of the circle is 3.1416x2 2 = 12.5664. Find the area of a circle 12 feet in diameter. Find the area of a circle of 8 feet radius; of a circle of 100 feet radius. The suface of a sphere equals the square of its diameter multiplied by 3.1416. ILLUSTRATION. The surface of a sphere 5 feet in diainetei 3. 1416x25. The surfaces of spheres are to each other as the squares of their diameters. The solidity of a sphere equals the product of the surface multiplied by ^ of the diameter, or it equals ^ of the cube of the diameter mul- tiplied by 3.1416. The solidities of spheres are to each other as the cubes of their diameters^ The solidities of similar solids arc to each, other as the cubes of their like dimensions. The solidity of a cylinder equals the product of the area of its base by its altitude. OF THB UNIVERSITY LIGHTNING CALCULATOR. The convex surface of a cylinder equals tlic product of the circumference of its base by its altitude. What is the solidity of a cylinder 8 feet higb with a base 4 feet in diameter? A cylinder 15 feet high, with a base 1 foot diameter? What is the diameter of a sphere containing 100 cubic feet? One bushel is about f of a cubic foot. Hence -f of the number of cubic feet equals the num- ber of bushels nearly. The dimensions of a box are 12 feet long, 6 feet in width, and^ 5 feet high. How many bush ols does it contain? The product of 12, 6 and 5 is 360; Number of cubic feet=288 bushels. Eemove the point one place to the left and multiply by 8. Hence the rule to find the number of bushels from the number of cubic feet. ,. Remove the decimal point one place to the left, and multiply the quotient by 8. ILLUSTRATION; In a bin of 800.9 cubic feet, remove the point one place to the left, we have 80.09; multiply by 8 and we have. 640. 72; the number of bushels. To find the number of cubic feet from the bushels, simply increase tho number by. olio quarter of itself. What is the number of bushels that a bin will contain,. 20 feet long, 8 wide, 5J deep? What is the number of cubic feet in 2150 bushels 2 68 HENDERSON'S SOME OF THE MISCELLANEOUS WEIGHTS TO THE BUSHEL. 60 !bs make 1 bushel of Wheat. 56 " 1 " Corn. 33 " 1 " Oats. 8 " 1 Barley. . 66 " 1 ' Bye. 60 " 1 * Beans. 52 " 1 ' Buckwheat. 70 1 * Corn in ear. 50 1 ' Corn meal. 60 i * Potatoes. 50 1 " Salt. 33 ' 1 " Peaches, dried. 25 1 " Apples, dried. 62 ' 1 " Clover seed. 45 * 1 " Timothy. 56 1 " Flax. SHORT METHODS IN DIVIS- ION AND MULTIPLICATION. Remove the point one place to the right to multiply by 10; two places to multiply by 100; three places 1000, etc. To divide, remove it to the left. To multiply by 25, divide by 4 and call the quotient hundreds. Thus: 25x48012000. 480-^4=120 call it hundreds, makes 12000. Divide by 4, be- cause 25 is one quarter of a hundred. To multiply by 2| divide by 4 and call it tens; call it tens, because 2J is the quarter of ten. LIGHTNING CALCULATOR. 69 To multiply by 125, divide by 8 and call it thousands. Call it thousands, because 125 is J of a thousand. To multiply by 12 J divide by 8; call it hun- dreds. To multiply by 1 divide by 8; call it tens. To multiply by 62| divide by 16 and call it thousands. To multiply by 6J divide by 16 and call it hundreds. To multiply by 31 divide by 32 and call it thousands. To multiply by 333, divide by 3 and call it thousands. To multiply by 33 J, divide by 3 and call it hundreds. To multiply by 3 J, divide by 3 and call it tens. To multiply by 50, divide by 2 and call it hun- dreds. To multiply by 66f , divide by 15 and call it thousands. To multiply oy 6f , divide by 15 and call it hundreds. To multiply by 833J, divide by 12 and call it ten thousands, by annexing four ciphers. To multiply by 83 J, divide by 12 and call it thousands. To multiply by 8J, divide by 12 and call it hundreds. Divide by 12 and -call it hundreds. 60 EENDEKSON'S because 8J- is -^ of a hundred. The reason is similar in. .each case. The primitive meaning of reason is hook some- thing to hold on by. Please get the reason in each case. To multiply by 166f, divide by 6 and call it thousands; because 166| is -f of 1000. To multiply by 16| , divide by 6 and call it hundreds. To multiply by If, divide by 6 and call it tens. To multiply by 37 J, take f of the number and call it hundreds; 87 J, ^ of the number, and call it hundreds, etc. We simply reverse these methods to divide. j To divide by 10, 100, 1000, etc., we remove tte point one, two, and three places to the left. To divide by 25, remove the decimal point two places to-the left and multiply by 4. Eemoving the point two places divides by one hundred; hence the (|uotient is 4 times to small; kence we remove the point two places $nd multiply by 4, : r To divide by 2J, remove the point one place to the left and ^multiply by 4. To divid^ ? by 125, remove the point three places to the left a : hd multiply by 8. To divide by 12 J, remove the point two places to the left and multiply fey 8. - < L To divide by 1| -, remove the point one place to the left and multiply by 8. There are about LIGHTNING CALCULATOB. 61 1J cubic feet in one bushel. Hence divide the number of cubic feet by 1J gives the number of bushels nearly. To divide by 625, remove the point four places to the leffc and multiply by 16. . To divide by 62J, remove the point three places to the left and multiply by 16. To divide by 6 J, remove the point two places to the left and multiply by 16. To divide by 3125, remove the point five places to the left and multiply by 32. To divide by 3J, remove the point two places to the left and multiply by 32. To divide 333 J, remove the point three places to the left and multiply by three. To divide by 666|, remove the point four places to the left and multiply by 15. To divide by 66|, remove the point three places to the left and multiply by 15. To divide by 833 J, remove, khe point four places to the left and multiply by 12. - To divide by 83^, remove .fhe point three places to the left and multiply by 12. To divide by 8^, remove the point two. places to the left and multiply by 12. To divide by 166f , remove the point three places to the left and multiply by 6. Eemoving the point three places divides by 1000; hence the quotient is 6 times too small. 166| is -J- of 1000. 62 HENDERSON'S MENTAL EXERCISE. PROBLEM 1. Take 1, multiply by 49, extract the square root, multiply by 4, subtract 1, and extract the cube root; what is the result? PROBLEM 2. Take 9, divide by 2, multiply by 6, extract the cube root, multiply by 27, and extract the fourth root; what is the result? PROBLEM 3. T&ke 48, divide by 2, multiply by 4, add 4, extract the square root, multiply by 5, subtract 1, divide by seven, and what is the result? PROB'.^^ 4. Take 8|, multiply by 8J, sub- tract f, divide by 8, extract the square root, multiply by 40 a&d divide by 10; what is the result? PROBLEM 5. Take 1J, multiply by 1J, 2| by 2J, 3| by 3f , run it up to 12J, in concert. PROBLEM 6. Take 1, multiply by If , 2 J by 2|, etc., up to 12. PROBLEM 7. Take If, multiply by If, 2f by 2|, etc., up to 15. PROBLEM 8. Take If, multiply by l, 2f by 2, etc., up to 20. PROBLEM 9. Take If, multiply by If, 2f by 2f, etc., up to 17. PROBLEM 10. Take If, multiply by If, 2f by 2|, etc. PROBLEM 11. Take 1 \, multiply by 1 T V, 2 T 5 ? by 2&, etc. LIGHTNING CALCULATOB. 68 PROBLEM 12. Take 1^, multiply by l T 4 f , 2 T V by 2 t V etc. PROBLEM 13. Take 12J, multiply by 12J, 11* by 11 J, etc., down to 1. PROBLEM 14. Take 11J, multiply by llf, 10J by 10|, etc., down to 1. PROBLEM 15. Take 12, multiply by 12J, 11 by 11, etc., down to 1. PROBLEM 16. Take 13f , multiply by 13f , 12| by 12|, etc., down to 1. PROBLEM 17. Take 12 T V, multiply by 12 T V, 11 T V by H^V e * c - down to 1. PROBLEM 18. Take 10 T \, multiply by 10^, etc., down to 1. PROBLEM 19. Take 12 T \, multiply by 12&, etc., down to 1. PROBLEM 20. Take 8^, multiply by 8 T \, 7^ by 7 T \, etc., down to 1. PROBLEM 21. Take 10 T V, multiply by 9ir ^J ^TT* e * c -> down to 1. PROBLEM 22. Take 12-&, multiply by etc., down to 1. PROBLEM 23. Take !!&, multiply etc., down to 1. PROBLEM 24. Take 12|, multiply by 12|, etc., down to 1. PROBLEM 25. Take 8 T \, multiply by 8j|, etc., down to 1. PROBLEM 26. Take 13|, multiply by 13^; etc., down to 1. 64 HENDERSON'S V The mean of two numbers, is half their sum, or the number equally distant from the two numbers. The product of two numbers is the square of their mean diminished by the square of half of their dif- ference. PROBLEM 27. 19 times 21, 18 times 22, etc., down to 15. Tims : The mean is 20, the square of 20 is 400, 400 the square of 1 is 399 ; the product, 18 times 22 is the square of 20, 400 the square of 2, 4, 396. 17 times 23 is 391, 16 times 24, 384 ; 15 times 25, 375. PROBLEM 28. Take 29 by 31, 28 by 32, etc., down to 20 and up to 40. PROBLEM 29. Take 39 by 41, 38 by 42, etc., down to 30 and up to 50. PROBLEM 30. Take 49 by 51, 48 by 52, etc., down to 40 and up to 60. PROBLEM 31. Take 59 by 61, 58 by 62, etc., dowm to 50 and up to 70. PROBLEM 32. Take 69 by 71, 68 by 72, etc., down to 60 and up to 80. PROBLEM 33. Take 79 by 81, 78 by 82, etc., down to 70 and up to 90. * PROBLEM 34. Take 89 by 91, 88 by 92, etc., down to 90 and up to 100. The complement of a number is the difference of that number and some particular number above it. The supplement of a number is the difference of that number and tome particular number be- LIGHTNING- CALCULATOR 65 Thus, the complement of 99 is the difference of 99 and 100, which is 1. The supplement of 101 is the difference of 101 and 100, which is 1. PEOBLEM 35. Commence at 99 and square num- bers down to 90. Thus : 99 times 99 is 9801, 98 times 98 is 9604, 97 times 97 is 9409, 96 times 96 is 9216, etc. Simply diminish the number by its complement, call it hundreds and add the square of the complement. When we use the supplement, w r e add it to the number, give it its proper name and add the square of the supplement. Thus : 101 times 101, the supplement 1 added to 101 makes 102, call it hundreds, is 10200, plus the square of tho supplement is 10201. PROBLEM 36. Commence at 101, square all the numbers up to 110 and down to 90. PEOBLEM 37. Commence at 51, square all the numbers up to 60 and clown to 40. PROBLEM 38. Commence at 21, square all the numbers up to 25 and down to 15, PEOBLEM 39. Commence at 11, square all the numbers up to 15 and down to 5. PEOBLEM 40. Commence at 999, square all the numbers down to 990 and up to 1010, etc.. etc., etc., etc. 66 HENDERSON'S MISCELLANEOUS PROBLEMS. PROBLEM 1. How many bushels in a bin 10 feet long, 4 feet wide and 4 feet deep? SOLUTION. Since there are ^ of a cubic foot in one bushel, the bin will contain 8 times T V of the number of cubic feet, in bushels. T V of 10 is 1, 8 times 1 are 8, 4 times 8, 32, and 4" times 32, 128, Ans. Or find the number of cubic feet in the bin, remove the decimal point one place to the left, and multiply by 8 in all cases. Thus : the product of 4, 4 and 10 is 160; remove the point one place to the left and we have 16, 16 multiplied by 8 is 128, Ans. PROBLEM 2. --How many bushels in a bin 32 feet long, 16 feet wide and 5J feet high? PROBLEM 3. How many bushels in a bin 24 f*et long, 12 feet wide, 4J feet high? PROBLEM 4. A cubic foot of water weighs 62 Ibs. 8 oz. what is the pressure 011 5 acres at the bottom of the sea, where the water is 1 mile deep ? PROBLEM 5. What would be the weight of this planet if one cubic foot weighs 62-| pounds? PROBLEM 6. If 21| bushels of oats are re- quired to seed 9f acres, how many bushels will be required to seed a field of 100 acres? PROBLEM 7. If 33 pounds of tea cost $27 J, Low much will 300 pounds cost? LIGHTNING CALCULATOR. 67 PROBLEM 8. A field 3J times as long as it is wide contains 30 acres what are its dimensions? PROBLEM 9. If each one of 20 pupils breathe 30 cubic feet of air per hour, in how long a time will they breathe as much air as a room 20 by 30 and 8 feet high contains? PROBLEM 10. If gold is 1.12J, what is cur- rency worth? SOLUTION. The value of currency would be TIK simply multiplying the numerator and denominator lyy 2 and we have ||f = f ; hence one dollar in currency is worth f X 1 y cents, or 88 1 cents. PROBLEM 11. If currency is worth 88f cents on the dollar, what is gold worth? Simply in- vert the preceding operation. PROBLEM 12. If gold is 1.10J, what is cur- rency ? PROBLEM 13. If currency is 95 cents on the dollar, what is gold? PROBLEM 14. If a wolf can eat a sheep in -J of an hour, and a bear in f of an hour, how long will it take them together to eat what re- mains of a sheep after the wolf has been eating half an hour? SOLUTION. In one hour the wolf eats f of a sheep, after eating half an hour f of the sheep would remain, since in one hour they eat - (- f or |f; to eat -| or r of a sheep it would take them as long as |f is contained in -^y, which is -% of an hour, Ans. 68 HENDERSON'S PROBLEM 15. John cuts a cord of wood in f of a day, James in f of a day, how long will it take them to cut a cord when they work to- gether? PROBLEM 16. A can do a piece of work in 8 days and A and B can do the same in 5 days ; after A did ^ of the work, B did the remainder how long did it take him? PROBLEM 17. Divide the number 108 into' t\ro such parts, that f of the first-j-8 shall equal the second. PROBLEM 18. A ship mast 63 feet in length, in a storm, was broken off; f of what was broken off equaled f of what remained; how much was broken off, and how much remained? PROBLEM 19. A farmer has 2290 sheep in two fields, f of the number in the first field equals f of the number in the second; how many are there in each field? PROBLEM 20. A market woman was requested to buy 99 fowls, consisting of two different kinds; J of the number of the first kind was to equal | of the second kind; how many of each kind must she buy? PROBLEM 21. A farmer, after selling f of 1-J times as much grain as he had, had 100 bushels remaining: how much had he at first? PROBLEM 22. Divide the number 170 into two parts, that shall be to each other as to f . PROBLEM 23. | of A's number of sheep plus LIGHTNING CALCULATOB. 09 f of B's number equals 900; how many sheep has each, providing f of B's number is 4- of A's number: PEOBLEM 24. A gold and silver watch were bought for $320; the silver watch cost T as much as the gold one; what was the cost of each ? PROBLEM 25 J of A's money -f~ of B's; equals 6800; and f of B's is 4 times -J- of A's; how much money has each? PROBLEM 26. Divide the number 60 into two parts, that shall be to each other as -J to f PEOBLEM 27. The sum of two numbers is 140, and the larger is to the smaller as 1 to |-; what are the numbers ?\ PEOBLEM 28. A and B together owe $207; B owes -i^- as much as A; how much does each owe ? PEOBLEM 29. I sold a horse for J more than he cost me, receiving $270 for him; how much did he cost me? PEOBLEM 30. What will f of a barrel of flour cost at $11.28 per barrel? PEOBLEM 31. What will | of a bag of coffee weigh if a bag weighs 147 BbB? PEOBLEM 32. What will | of a pound of tea cost at $1.25 per pound? PEOBLEM 33. What will T of a cord of wood cost at $6.25 per cord? PEOBLEM 34. What will T V of a hogshead of wine cost at $138.75 per hogshead? 70 HENDERSON'** PROBLEM 35. How much is | and J of -J- of 15? PROBLEM 36. A and B traded in company; A piji in as much as B; they gained $750; what was each man's share? PROBLEM 37. James says to John, give me 87.00 and I will have as much money as you. John says to James, give me $7.00 and I will have twice as much as you, Ans. 35 and 49. Simply multiply the $7.00 by the numbers 5 and 7 ; and for all similar problems simply mul- tiply the sum of money given, by the numbers 5 and 7. PROBLEM 38. A says t& B, give me $3J and I will have as much money as you. B says to A, give me $3J- and 1 will have twice as much as you. How much money has each? PROBLEM 39. Haight says to Booth, give me 1000 sheep and I will have as many as you. Booth says to Haight, give me 1000 and I will have twice as many as you. How much has each? PROBLEM 40. Friedlander says to Eeese, give me $500,000 and I will have as much as you. Eeese says to Friedlander, give me $500,000 and I will have twice as much as you. How much has each? PROBLEM 41. G says to D, give me $13.33J and I will have as much money as you. D says to G, give me 13.33J and I will have twice as much money as you. How much has each? PROBLEM 42. Greeley says to Grant, give mo LIGHTNING CALCtJLATOB. 71 50,000 votes and I will Lave as many as yon. Grant says to Greeley, give me 50,000 votes and I will have twice as many as you; how many has each? A PROBLEM 43. Two Hoodlums go imko a saloon; one says to the other, give me as much money as I have, and I will spend two bits with you. They go into another saloon, and he sajs, give me as much money as I now have, and I vail spend two bits with you. They went into the third saloon, and he made the same statement, and when they came out of the third saloon he had nothing left. How much had he when he went into the first saloon? Ans., If bits. Simply - of the sum borrowed in the first saloon is the answer. PROBLEM 44. A and B step into a hotel; A says to B, give me as much money as I have, and I will spend five dollars with you. They go into a second and third, A making the same statement; and when they came out of the third, he had nothing left. How much had he when they went into the first hotel ? PROBLEM 45. If 3 be the third of 6. what will the fourth of 20 be? Ans. 3J. SOLUTION. The third of 6 is 2, if 3 be 2, 1 is J of 2 or f, and 20 is 20 times or 4 ^, the -J of 20 is the J of V or V, 3J- Ans. PROBLEM 46. If the third of 6 be 3 what will the fourth of 20 be? Ans. 7i. 72 HENDERSON'S SOLUTION. If 2 be 3, 1 is of 3, 1J and 2Q is 20 times 1J or 30, i of 20 is the of 30 or 7* Ans. * GMENTERAL, IXFORY1ATION. The circumference of a circle equals the diam- eter multiplied by 3.1416, the ratio of the cir- cumference to tho diameter. The radius of a circle equals the circumfer- ence multiplied by 6.283185. The area of a circle equals the square of the radius multiplied by 3.1416. The area of a circle equals tiie square of the diameter multiplied by 7854. The area of a circle equals one quarter of the diameter multiplied by the circumference. The radius of a circle equals the circumfer- ence multiplied by 0.159155. The radius of a circle equals the square root of the area multiplied by 0.56419. The diameter of a circle equals the circum- ference multiplied by 0.31831. The diameter of a circle equals the square root of the area multiplied by 1.12838. The side of an inscribed equilateral triangle equals the diameter of the circle multiplied by 0.86. The side of an inscribed square equals the diameter multiplied by 0.7071. LIGHTNING CALCULATOR. 73 The side of an inscribed square equals the diameter of the circle multiplied by 0.225. The circumference of a circle multiplied by 0.282 equals one side of a square of the same area. The side of a square equals the diameter of a circle of the same area multiplied by 0.8862. The area of a triangle equals the base multi- tiplied by one half of its altitude. The area of an ellipse equals the product of both diameters and .7854. The solidity of a sphere equals its surface multiplied by one-sixth of its diameter. The surface equals the product of the diam- eter and circumference. The surface of a sphere equals the square of the diameter multiplied by 3.1416. The surface equals the square of the circum- ference multiplied by 0.3183. The solidity of a sphere equals the cube of the diameter multiplied by 0.5236. The diameter of a sphere equals the square root of the surface multiplied by 0.56419. The square root of the surface of a sphere multiplied by 1.772454 equals the circumference. The diameter of a sphere equals the cube root of its solidity multiplied by 1.2407. The circumference of a sphere equals the cube root of its solidity multiplied by 3.8978. The side of an inscribed cube equals the radius multiplied by 1.1547. 74 HENDEESON'S The solidity of a cone or pyramid equals the area of its base multiplied by one third of its altitude. 75 LIGHTNING CALCULATOR. COLLEGE DE L'UNION. DIPLOME DE BACHELIER DBS ARTS. Nous Directeurs du College de TUnion a Sche- nectady, Etat de New York, vu le Certificat cV apti- tude au grade de Bachelier es Arts, accorde par la Faculte du College au Sieur Jean Alexandre Hen- derson, ratinant le susdit Certincat. Donnons par ces presentes au dit Sieur, le Diplonie de Bachelier es Arts, pour en jouir avec les droits et prerogatives qui y sont attaches. En teinoignage de quoi nous avons muni ce Diplome de notre sceau et des sig- natures du President et des Professeurs de ce Col- lege. Fait & Schenectady le vingt huitieme Juillet 1864, E. NOTT, Pres< L. P. HICKOK, Acting Pres. J. H. JACKSON, Prof, de Math. JOHN FOSTER, Prof, de Physique. GUILL. M. GTLLESPIE, Prof, de Fonts et Chaussees. C. F. CHANDLER, Ghem. Prof. W. LAMOREUX, Acting Prof. Lang. Mod. N. G. CLARK, Prof, des Belles Lettres. JONATHAN PEARSON, Prof. Hist. Nat. N. B. J. A. Henderson a regu le A. M, degre de Maitre, 1867. LE CALCULATEUR INSTANTANE. PAR J. JL.- HENDERSON. L'alphabet arithmetique est~e*crit et lu: Le se- cond est deux a fois le premier, a 3 i 9 3 j .g ^ 6 le troisieme & g .2 H 2* ^ & ^S e^a troisfoislepre- r|&.S.y&Vg ' mier,etainside 1234567890 suite iusqu'a la & - fin. 1111111111 INTfiEET. Comme Tinte're^t est gen^ralement une portion du principal, la me'thode de le calculer viendra sur la methode de le diviser. La regie e*tablira le temps quand la piastre fait un cent, et nous place- rons le chiffre decimal deux places a la gauche, parce- que un centieme du principal egal Tinteret. Dans dix fois le temps la piastre fait dix cents, et nous placerons le- chiffre decimal une place a la gauche, parceque un dixieme du principal en est Tinteret; LIGHTNING CALCULATOR. 77 dansundixieme du meme temps une piastre fait un mille, et nous placerons le chiffre decimal trois places a la gauche, parceque un millieme du prin- cipal en est Tinte're^t. REGLE Le reciproque du prix, ou le prix ren- verse indique le temps quand nous pourrons chan- ger le chiffre decimal deux places a la gauche; en tout cas, dix fois le temps une place a la gauche, et un dixieme du meme temps trois place a la gauche, augmente ou diminue le resultat afin de retrouver le temps. L'alphabet numerical est \, $, ?, 1, f , ?, I, !, ?, etO; I renverse est 1; ? renverse' est ^; ^enverse J, etc. Si le prix est 1 pour cent par mois, 1 renverse indique le temps quand la piastre fait un cent, et le chiffre change deux places a la gauche montre Tintaret dans tout cas. Ainsi: $ 2 5 6 7.35 8 6 1.50 9, 7 4 6.75 11, 4 6 3.25 22, 5 3 8.40 1, 0.50 78 HENDERSON'S METHODE D'EGALISEB DES NOMBKES PAR LEUK COMPLEMENT ET SUPPLEMENT. Le complement d'un nombre est la difference cTun nombre et d'un autre nombre particulier avaiit lui. Le supplement d'un DS'inbre est la dif- f^rence d'un nombre et d'un autre nombre apres lui. (99)-=9801. Prenons le complement de 99, nous 1'appellerons centieme, et nous ajouterons le complement pour le rendre egale, EXPLICATION. Laissons que N e'gale 99, et C egal 1. Alors N plus 0=100. N 0=98. Multiplions les deux equivalents ensemble, nous avons N 2 2 =9800. Ajoutons C 2 aux uns et aux autres nombres, et nous aurons N 2 9801, le nombre egal de 99. IRE BEGLE. Quand le nombre est plus haut que la base, nous ajoutons le supplement, nous 1'appellerons centiemes, et nous ajouterons Tegal du supplement, nous 1'appellerons centieme, parce- que le nombre est augmente par le supplement quand il est multiplie par 100; en ce cas-ci, quand le nombre est moindre que la base, nous sous- trairons le complement. SMS EEGLE. Le produit de Tun ou de Tautre de deux nombres est 1'dgal de leur valeur diminue' par F6gal de la moitie de leur difference. LIGHTNING CALCULATOR. 79 POUR TROUVER LA VALEUR DE LA MON- NAIE COURANTE QUAND LE PRIX DE L'OR EST FAIT. Quand Tor est a 111-g, qu' est la valeur de $1.00 enmonnaiecourante? Nous prenons 109, le nornbre de cent dans la piastre, cornme le numerateur, et la valeur de For comnie le clenominateur. Noun simplifions la fraction en multipliant le numerateur et denominateur par 9, et nous aurons le -f^ d'une piastre, ou 90 cents, la valeur de la monnaie cou- rante. Quand la monnaie -courante vaut 75 cents, quel est la valeur de Tor? \ff=$, f X 1 ? cents egale REGLE. Nous prenons 100, le nombre de cent dans une piastre, pour le numdrateur, et la valeur de For ou de Targent courant, quelque soit la cause, pour le denominateur. Nous simplifions ia fraction en ajoutant un zero au numerateur et en divisant par le denominateur, NOUVELLE METHODS DECIMAL DE CAL- CULER L'INTfiRET ET DE L'EXPLI- QUER. REGLE. Revers^ le prix, ajoutez un zero, et mettez devant le point. Immediatement au-dessus de ces caracteres, placez le montant sur lequel I'interSt es^ demande, le centieme etant toujours dans la co- lonne du prix. RAISONS. Posez le prix pour trouver la regie. \ a R A t . t OF THE TT "NT T ^r T 1 -R QTT 1 ^ 80 LIGHTNING CALCULATOR. Quand la piastre fait uncent. Ajoutez un zero, vous trouverez la regie, quand une piastre fait dix cents. Mettez devant le point, vous trouverez la regie, Quand une piastre fait un mille. Quand une piastre fait un cent, vous rechangez le point deux places a la gauche: parce^ue un centieme du principale egale Finterefc. Quand une piastre fait dix cents, rechangez le point une place a la gauche, parceque le dixieme du principal est Tinteret. Le prix reiiverse invariablement represente le temps quand une piastre gagne un cent, ou cent piastres une piastre. Ainsi, pour trouver quel que soit le nom- bre donne de centiemes, placez le immediatement dessous le prix renverse et vous aurez la reponse en piastres et declines. J.. A. HETOEBSCOT, A. M. 7 Phrenologist and Phreno-Magnetic Healer. CERTIFICATES OF APPROVAL. I have examined the new methods of calcula- tion by Prof. J. A. Henderson, they are inval- uable to business men, and will prove a light in science to all coming generations. A. J. WAENEB, Pres. Elmira Commercial College. Henderson's methods are the finest known for lightning multiplication. .Prof. D. K. FOKD, Female College, Elmira. I have examined Prof. J. A. Henderson's new methods of calculation; they are remarkable for originality and of great practical value. His methods of calculating interest are peculiarly clear and comprehensive in their adaptation to all possible cases. Rev. Dr. O. P. FITZGEBALD, Ex. State Superintendent, Col. CERTIFICATES OF APPROVAL.- Mr. J. A. Henderson has taught mathematics in Delhi Academy for a year. We consider him an excellent mathematical teacher. J. L. SAWYER, Principal of Delhi^.Academy . Delhi, Oct. 1862. P. S. J. A. H., taught analytical Trignome- try, University Algebra, Intellectual Arithmetic and English Grammar in Delhi Academy, New York. John Alexander Henderson, A. M., attended Union College and graduated with me in class "64." He is an excellent scholar among the first and his character is above reproach. ELISHA CURTIS, A. M., Principal of Sodus Academy. I have known Prof. J. A. Henderson from earliest boyhood; his character has always been beyond reproach. As a mathematician he has scarcely an equal ; as a teacher he has been emi- nently successful; as a phrenologist, he is con- sidered by many not a whit behind Fowler & Wells. New York. Rev A. G. KING, of U. P. Church, N. Y., 1869. EXTBACT FBOM J. A. HENDERSON'S PHBENO- MEDICAL CHABT. Diet, exercise, rest, light, good water and pure air develop the mental, motive and vital forces. Young maiden and young man, make these your physicians, for they insure health, success in business, give you the key to philosophy and are the handmaids of Christianity. Add in- telligence and contentment, the two great pillars of felicity, and you do much to sustain the moral government of the domestic circle, the moral government of the human family, and the moral government of the Creator. Strong faith makes a stout heart; active hope, a healthy liver; rounded up veneration, excellent functions of digestion; large firmness, strong vertebrae; fine conscientious gives not only a pure mind, but healthy kidneys eliminating all surplus secretion from the brain and body; large combativeness develops a fine osseous force; destructiveness an excellent muscular force. The mind is the root, the body the trunk, and the sciences the branches. The seat of the mind is the brain and nerves. The organs, forty in number, are the instruments used in framing constitutions and building up science. They are also the instruments used in CERTIFICATES OF APPKOVAL. building up the constitution of the body. There- fore beware that you build in no error, for dis- ease will surely follow, which is the result of insulted law. It is this law insulted that binds the body with disease; break the bands and your blood will flow like wine, and disease dis- appear like mist before the morning sun. Mind is light. Hail ! holy light which lighteth everyone, in tJiee is the life of the blood, flesh and spirit, from thee the face gets its form and beauty, the eye its light, the tongue the word, the muscle its action, and every function its health and de- velopment. Hence keep all the organs of the mind and functions of the body well ventilated by being correct in diet, exercise, rest, light, good water and pure air, and the result is sound flesh and pure blood. The child is the zero power of its parents, that is a unit of the parents, plus or minus the surrounding influences. A well balanced child has its eyes in the center of its head, that is the same distance from the point of the chin to the optics, as from the optics to the upper part of the organ of benevolence. A wise man's eyes are in the center of the head. The face is an index of the strength of the will of the flesh; the brain an index of the strength of the will of the mind. Hence when Nature establishes an equilibrium between the two forces, the master and the servant, the will of the mind and the will of the flesh, w r e have harmony of character, prudence, wisdom and proficiency. The history of aM nations illus- trates the truth of the above propositions very clearly. J. A. Henderson is preparing a lightning . of discerning character. METODO IKSTANTAMO DE CALCULAB Intelectnal y Practical PROFESOR J. A. HENDERSON, SRADUADO EN EL COLEGIO UNION, Y AUTOR DEL CALCULADOR, LIBRO DE "CUADRO ILOS- TRANDO RAICES, Y CARTA PRENA-MEDICAL. Examinacions tocante at salted negocios y otras cosas, 5&2 Calle de Market, Cuatro ^K. 16. SAN FRANCISCO: IMPRENTA C08MOPOLITANA, CALLE DE CLAY, NO. -505. 1872. K EL CALCUIADOK INSTANTAflEO. El Alfabeto numerico, segun se~escribe y se lee: r I' I- I I I O p. B PI O W O ' El segundo es dos veces el primero, __ l_l_JLl_ l_.L_.Ll__Lel tercero tres ^7890. veces el pri- l' l' l' l' l' l' l' l' i' i mero, etc., hasta el ultimo. KEGLA DE ESTTEEES. REGLA. El reciproco de la tasa 6 la tasa in- vertida, indica el tiempo cuando en todos casos. Se puede mudar el tiempo decimal dos lugares hacia la izquierda; diez veces aquel tiempo uii lugar hacia la izquierda y un desimo del mismo tiempo tres lugares hacia la izquierda. Aumen- tase 6 disminuyase el resultado para que coii- cuerde con el tiempo dado. 90 EL CALCULADOB INSTANTANEO Tasa por ano. ...... 9 " invertido....... . . 1 Con diez desimos*,,. 9 $1 9 7 dias. 40 dias 400 dias | por mes... ......... 3 % 8 3 EJEMPLO-^ . $15 4 9 8 dias. 80 dias 800 dias "o i Unidades es la suma de^='l J=^ EJEMPLO. De f restase J= DIVISION DE QUEBEADOS. KEGLA Reduzcanse los niimeros fraccionarios a* la forma de quebrados impropios; multipli- quese el dividiendo por el divisor invertido 6 multiplfquese tanto el numerador y denomina- dor por el mas comma multiplicador de los de- nominadores de los partes fraccionales. Dividase 5J por 2J-, multiplfquese tanto el numerador y denominador por 6, el mas comun de 2 y 3. PABA HALLAB EL VALOB DE MONEDA O PAPEL MONEDA CUANDO SE CONOCE EL PBEGIO DEL OBO. EEGLA.---Tomamos 100, el numero de centa- vos que contiene el peso fuerte para numerador y el valor del oro, 6 papel moneda, segun sea el caso, 'para denominador. Simplifiquese el quebrado anadiendo ceros, al numerador y dividiendo por el denominador. Cuando el oro esta* al valor de 109J. <: Cuanto es el valor de $1 . 00 moneda papel ? EJEMPLO.-H =^-l^-$91 ** ~ 982 491~ 401. 94 EL CALCULADOK INSTANTANEQ Cuando el papel moneda esta al valor de 75 centavos, cuanto es el valor del oro ? EJEMPLO ^=t | de 100 centavos igual a $1. 33J. METODO DE EXTRAER LA KAlZ CUADRADA. EEGLA. Dividase cualquier niimero por el cuadrado de tres, estraigase la rafz cuadrada del cociente y tenemos un tercio de la raiz del niimero. Dividase cualquier niimero por el cuadrado de cuatro, estraigase la raiz cuadrada del cociente y tenemos un cuarto de la raiz cuadrada del niimero, etc, 32 1 100 15625(100+20+5 Primer divisor probante 200 10000 Segundo verdadero 220 5625 probante 240 4400 verdadero 235 1225 1225; RAIZ CUBICA. Anadase a cada verdadero divisor, segun se presenten dos veces la supe^cie de un lado del pequeno tubo, y uno a cada uno de los tres rectangolos para el divisor probante, porque eso hara los tres lados del cubico entero. BE HENDERSON. 95 REGLA. Dividase cnalquier mimero por el cu- bico de 2, extraigase la raiz ciibica del cociente, y tenemos la raiz cubica del mimero. Dividase cnalquier mimero por el ciibico de 3, 6 27, ex- traigase la raiz cubica del cociente y tenemos un tercio de la raiz del mimero. Dividase cual- quier mimero por el ciibico de 4, 6 64, extrai- gase la raiz cubica del cociente y tenemos un cuarto de la raiz del mimero. Dividase cual- quier mimero por el cubo de 5, 6 125 extraigase la raiz cubica del cociente y tenemos un quinto de la raiz del mimero. 1728) 110592 (64 ' 10368 6912 6912 PARA ENCONTRAR EL NUMERO DE PIES UBICOS~QUE CONTIENE EL BUSHEL. EEGLA. Mudase el punto decimal un lugar hacia la izquierda y multipliquese el cociente por 8. EJEMPLO. Dentro de un vacillo de 800.9 pies ciibicos, raudase el punto un lugar hacia la iz- quierda, y tenemos 80.09; multipliquese por 8 y tenemos 640 . 72 el numero de bushels. N. B. Anadase uno por cada tres cientos . 96 EL CALCULATOR INSTANTANEO EJEBCICIO MENTAL. El mimero medio de dos numeros es la mitad de sus sumas 6 el numero igualmente distante de los dos rnimeros. El producto de dos nii- meros, es el cuadrado del numero medio dismi- nuido por el cuadrado de la mitad de su dife- rencia. PEOBLEMA. 19 por. 21, 18 por 22, etc., hasta 15. Asi: el numero medio es 20, el cuadrado de 20 es 400, 400 el cuadrado de 1 es 399; el producto, 18 por 22 es el cuadrado de 200, 400 el cuadrado de 2, 4, 396. 17 por 23 es 391, 16 por 24, 384; 15 por 25, 375. El complimento de un numero es la diferencia de aquel numero y algun otro numero especial mayor. El suplemento de un numeroes la dife- rencia de aquel mimero y algun otro numero especial menor. Asi el complimento de 99 es la diferencia de 99 y 100, el cual es uno. El suplemento de 101 es la diferencia de 101 y 100, el cual es 1. ticue X 31. i^enbcrson, un& urn fcicfel&e leidjt Bcgmflufj ju tuadjen. Die $?aregel betfelben finb fotgenber 2Irt : 1. Das numerirte Sllpljatet femten ju lernen/ wie folgt : 1234567890 2* X)a^ jiipulirte Sntereffe na^ ximjubre^en, mt folgt : 3 pr. fit. per SHonat ober 5 ; umgebre^t ~ 5Wonat f 10 age. 3. giir ba^ jel)nfad)e eine 9luK . ober Zero ju anenren, mie folgt : i 9Jionatxbet 10 ober f S)?onat ^ 100 Sage. 4, giit beit jefynten Sfytil eine S^utt (Zero) ju prafixiren, wic folgt : ^bet 10 ober 9Honat - 1 Sa, Die riinbe ftnb folgenbe : Umbrefjen ber Sntereffen bcmonjitht immct bte B^^, tDenn em Staler ($ 1.00) einen Sent (1 ct.) ma$t. 98 anenren einer 9Zutt (0) aerjetynfadjt bic Beit, gteicfy bieS 1 Sent ju ] gents. >a prajmren eines ^unites ober 9lutt (0) aerliirjt bte gtit je^nfad^ f ba^er an^ bet Sent ate etne $tttte ober ^ Sent gere^ttcttt>etbenniu. 3 pr. eld)er Sei^tigleit btefe 9J?etl;obe atte 99 JRecfottimgcn lofH inbem fte bas Scfyanerige burd) Decimals angreift at man baljer ausgefunben menu $ 1. 00 ei* uen Sent macfyt, fo nnrb Sebermann flat einfe* 1)011, ba er nur bie ftoftl bet Better aU Sent^ ju betrad;ten fyat, um bie Sofung berfelien ju ftnben, ottte man nun iDiinfi^en au^juftnben bie Sntereffen i^on enter utnmc ju 3 per Sent per Sftonat fur 1 33 Sage, fo gtbt bte $enntnt btefer 9)Jet|)obe efcenfalt^ etne letcfyte Sofung. [3nm 93eifpiel] Die Sntereffen on $ 1,00 ju 3 per Sent per Sftonat : 3 per St. per SRonat ift ^ J SKonat * 10 Sage (unb ntacfyt 1 Sent in 10 Sagen) fotgtt^ fur 1 33 Sage fiir 100 Sage t>a3 setynfadje i\ 10 Sagen ot). iSt^. = 10 30 ba^3x 10 L - 3 3 3 mal b. loten Jfeeil , 10 1. n \ * 0,3 Kent* 13,3 HENDERSON'S CHART, For Computing Time and Interest, Squaring and Mul- tiplying Numbers, Dividing Fractions, etc. Copyrighted January 22, 1872. Price, $1.00. A BOOK OF BLOCKS, ILLUSTRATING ROOTS. For Schools, Academies and Colleges. Copyrighted May 11, 1872. Price, 5.00. HENDERSON'S INTELLECTUAL AND PRACTICAL LIGHTNING CALCULATOR. Copyrighted October 24, 1872. Price, $1.00. NEW DECIMAL METHOD OF COM- PUTING AND IMPARTING INTEREST, ILLUSTRATED BY AN ENGRAVING. Copyrighted November 20, 1872. Price, 25 cents. AUTHOR'S NOTICE. Instruction gratis to those who purchase the above of the Author at his office, No. 542 Market street, San Francisco. Office Hours From 10 to 11 A.M., and from 4 to 5 P.M. THIS BOOK IS DUE ON THE LAST DATE STAMPED BELOW AN INITIAL FINE OF 25 CENTS WILL BE ASSESSED FOR FAILURE TO RETURN THIS BOOK ON THE DATE DUE. THE PENALTY WILL INCREASE TO SO CENTS ON THE FOURTH DAY AND TO $1.OO ON THE SEVENTH DAY OVERDUE. MAR 12 - LD 21-100ra-7,'39(4<"