3. To divide a given number into two numbers such that one is a given ratio of the other plus a given difference1. Given number 80, ratio 3: I, difference 4. Lesser number x. Therefore the larger is 3x + 4, and 4x+4=80, so that x = 19. The numbers are 61, 19. 4. To find two numbers in a given ratio and such that their difference is also given. Given ratio 5: I, given difference 20. Numbers 5x, x. Therefore 4x= 20, x = 5, and the numbers are 25, 5. 5. To divide a given number into two numbers such that given fractions (not the same) of each number when added together produce a given number. Necessary condition. The latter given number must be such that it lies between the numbers arising when the given fractions respectively are taken of the first given number. First given number 100, given fractions and †, given Second part 5r. Therefore first part = 3 (30-x). The required parts are 75, 25. 6. To divide a given number into two numbers such that a given fraction of the first exceeds a given fraction of the other by a given number. Necessary condition. The latter number must be less than that which arises when that fraction of the first number is taken which exceeds the other fraction. Given number 100, given fractions and respectively, given excess 20. Second part 6r. Therefore the first part is 4 (x+20). the parts are 88, 12. 'Literally "to divide an assigned number into two in a given ratio and difference (év Xóyw kai vжepoxy rŷn doleion).” The phrase means the same, though it is not so clear, as Euclid's expression (Data, Def. 11 and passim) dobévti μelšwv îî èv Xbyw. According to Euclid's definition a magnitude is greater than a magnitude "by a given amount (more) than in a (certain) ratio" when the remainder of the first magnitude, after subtracting the given amount, has the said ratio to the second magnitude. This means that, if x, y are the magnitudes, d the given amourt, and ✯ the ratio, x-d=ky or x=ky+d. From the same (required) number to subtract two given. numbers so as to make the remainders have to one another a given ratio. Given numbers 100, 20, given ratio 3: I. Required number x. Therefore x − 20 = 3 (x - 100), and x = 140. 8. To two given numbers to add the same (required) number so as to make the resulting numbers have to one another a given ratio. Necessary condition. The given ratio must be less than the ratio which the greater of the given numbers has to the lesser. Given numbers 100, 20, given ratio 3: 1. Required number x. Therefore 3x+60 = x + 100, and x = 20. 9. From two given numbers to subtract the same (required) number so as to make the remainders have to one another a given ratio. Necessary condition. The given ratio must be greater than the ratio which the greater of the given numbers has to the lesser. Given numbers 20, 100, given ratio 6 : 1. Required number x. x = 4. Therefore 120 – 6x = 100-x, and Given two numbers, to add to the lesser and to subtract from the greater the same (required) number so as to make the sum in the first case have to the difference in the second case a given ratio. Given numbers 20, 100, given ratio 4: I. Required number x. Therefore (20+x) = 4 (100 – x), and x = 76. 11. Given two numbers, to add the first to, and subtract the second from, the same (required) number, so as to make the resulting numbers have to one another a given ratio. Given numbers 20, 100, given ratio 3 : 1. Required number. Therefore 3x - 300= x + 20, and x = 160. 12. To divide a given number twice into two numbers such that the first of the first pair may have to the first of the second pair a given ratio, and also the second of the second pair to the second of the first pair another given ratio. Given number 100, ratio of greater of first parts to lesser of second 2: I, and ratio of greater of second parts to lesser of first parts 3: 1. 13. To divide a given number thrice into two numbers such that one of the first pair has to one of the second pair a given ratio, the second of the second pair to one of the third pair another given ratio, and the second of the third pair to the second of the first pair another given ratio. Given number 100, ratio of greater of first parts to lesser of second 3: I, of greater of second to lesser of third 2:1, and of greater of third to lesser of first 4:1. x lesser of third parts. Therefore greater of second parts = 2x, lesser of second = 100-2x, greater of first = 300 — 6x. Hence lesser of first = 6x-200, so that greater of third = 24t – 8oo. Therefore 25x800 100, x = 36, and the respective divisions are (84, 16), (72, 28), (64, 36). 14. To find two numbers such that their product has to their sum a given ratio. [One is arbitrarily assumed.] Necessary condition. The assumed value of one of the two must be greater than the number representing the ratio1. Ratio 3 I, one of the numbers, 12 the other (> 3). the numbers are 4, 12. 15. To find two numbers such that each after receiving from the other a given number may bear to the remainder a given ratio. Let the first receive 30 from the second, the ratio being then 21, and the second 50 from the first, the ratio being then 3:1; take x + 30 for the second. 1 Literally "the number homonymous with the given ratio.” 16. To find three numbers such that the sums of pairs are given numbers. Necessary condition. Half the sum of the three given numbers must be greater than any one of them singly. Let (1)+(2) = 20, (2) + (3) = 30, (3) + (1) = 40. r the sum of the three. Therefore the numbers are The sum x = 3x – 90, and x = 45. The numbers are 15, 5, 25. 17. To find four numbers such that the sums of all sets of three are given numbers. Necessary condition. One-third of the sum of the four must be greater than any one singly. Sums of threes 22, 24, 27, 20 respectively. r the sum of all four. Therefore the numbers are x-22, x-24, x-27, x-20. Therefore 4-93 = x, x = 31, and the numbers are 9, 7, 4, II. 18. To find three numbers such that the sum of any pair exceeds the third by a given number. Given excesses 20, 30, 40. 2x the sum of all three. We have (1) + (2) = (3) + 20. Adding (3) to each side, we have: twice (3) +20=2x, and (3) = x − 10. Similarly the numbers (1) and (2) are x-15, x- 20 respectively. Therefore 3- 45 = 2x, x = 45, and the numbers are 30, 25, 35. [Otherwise thus1. As before, if the third number (3) is x, (1)+(2) = x + 20. Next, if we add the equations (1) + (2) − (3) = 20 'Tannery attributes the alternative solution of 1. 18 (as of 1. 19) to an old scholiast. 19. To find four numbers such that the sum of any three exceeds the fourth by a given number. Necessary condition. Half the sum of the four given differences must be greater than any one of them. Given differences 20, 30, 40, 50. 2x the sum of the required numbers. x-15, x-20, x-25, x IO. Therefore 4x-70= 2x, and x = 35. The numbers are 20, 15, 10, 25. [Otherwise thus1. If the fourth number (4) is x, (1) + (2) + (3) = x + 20. Therefore the Put (2)+(3) equal to half the sum of the two excesses 20 and 30, i.e. 25 [this is equivalent to adding the two equations (1)+(2)+(3) − (4) = 20, (2)+(3)+(4) − (1) = 30]. It follows by subtraction that (1)=x-5. Next we add the equations beginning with (2) and (3) respectively, and we obtain The numbers are accordingly 20, 15, 10, 25.] 20. To divide a given number into three numbers such that the sum of each extreme and the mean has to the other extreme a given ratio. Given number 100; and let (1)+(2) = 3. (3) and (2)+(3) = 4. (1). 1 Tannery attributes the alternative solution of 1. 19 (as of 1. 18) to an old scholiast. |