28. Forty-one persons, consisting of men, women and children, spent, among them, 40s, of which each man paid 4s, each woman 3s, and each child 4d. How many were there of each? Ans. 5 men, 3 women, and 33 children. 29. With guineas and moidores, the fewest, which way, Three huudred and fifty-one pounds can I pay? If paid every way 't will admit of, what sum Do the pieces amount to ?-my fortune's to come. Ans. 9 guineas and 233 moidores; and 37 ways, is = £12987. 30. How many ways can a refiner mix three kinds of silver, of 111, 134, and 17 pennyweights fine per oz. so as to form a mass of 75 oz. of 15 pennyweights fine per ounce? Ans. 8 ways. 31. A person bought 100 head of cattle for $200.; oxen at $20 apiece, cows at $10 apiece, calves at $4 apiece, and sheep at $1 apiece. How many of each sort did he purchase? Ans. The question admits of 10 different answers, one of which is 4 oxen, 1 cow, 5 calves, and 90 sheep. 32. A company of men and women spent 1000 cents at a tavern; the men paid each 19 cents, and each woman 13 cents. How many men and women were there? Let x the number of men, and y = the number of women. Then we shall have 19x+13y=1000. x =+ 13y-1000 19 And 17 is the least value of y; then diminishing x by the coefficient of y, and increasing y by the coefficient of x, we have 3-17, 36, 55, 74 Now, when x=41, y=17, so that the men spent 779 cents, and the women 221. When x=28, y=36, then the men spent 532 cents, and the women 468. When x=15, y=55, the men spent 285 cents, women 715. And when x=2, y=74, the men spent 38 cents, women 962. 33. To find a number which, being divided by 6, shall leave the remainder 2, and when divided by 13, shall leave the remainder 3. Ans. 68. 34. Find the least whole number which, being divided by 39, shall leave the remainder 16, and when divided by 56, the remainder shall be 27. Ans. 1147. 35. Find the least whole number which, being divided by 7, 8, and 9, respectively, shall leave the remainders 5, 7 and 8. Ans. 1727. 36. Find three numbers in the proportion of 5, 7 and 9, which, being divided by 11, 13 and 15, shall leave the remainders 1, 2, 3, respectively. Ans. 2685, 3759, 4833. Compound Indeterminate Equations. 163. Equations of this kind, not higher than the second degreewhich admit of answers in whole numbers, are chiefly such as consist of the products or squares of two unknown quantities, together with the quantities themselves; being usually, as far as regards the plan of this part of the present article, of one of the four general forms given below. RULE. If the proposed equation be of the form xy-ay+by+c, we shall have, for its solutions in whole numbers y=a+ ab +c x-b where x-b must be, a divisor of ab+c. 2. If the proposed equation be of the form +xy-ax+by+c, we shall have for its solution, in this case, y = a -b÷x+ c+b(a—b) x-b -- where x-b must be a divisor of cb(ab). 3. If the proposed equation be of the form xy+ay+b, we shall have, for its solution, in that case, y a a2 4b n- -a +, and 8n 2 x=2+y—n, where n—a must be an even number, and ʼn be so taken that 8n may be a divisor of a2-4b. 4. And if the proposed equation be of the form xay+by+c2 we shall have for its solution, in this case, y b-2cn 2 n-a and x = 9 +ny, where n must be some whole number between ✔a and 2c 1. Given xy-42-2x-3y, to find the several values of x and y in whole numbers. Here, by the first form, a 42, whence y-2+ 6-+-42 -2+ 48 2+3' -2, 6-3, and where it is plain *Every indeterminate equation of the second degree, containing two unknown quantities, can always be reduced to the form 2ayb, consequently, on the solution of this depends that of every possible case that can arise in indeterminate equations of this kind; for some examples of which, see the latter part of this present article. that x must be such a number, that, when added to 3, the sum. shall be a divisor of 48. But the divisors of 48, that will give quotients greater than 2, are 16, 12, 8, 6, 4, and 2. Therefore the integral values of the two unknown quantities, are x=16—3, or 13|=12-3, or 9|-8-3, or 5 | 6-3, or 3 | —4—3, or 1 y=18-2, or 1|=18-2, or 2|48-2, or 4 | 482, or 6=48—2, or 10, which are all the answers in whole positive numbers that the question admits of. 2. Given x2+xy=2x+3y+29, to find the values of x and y in whole positive numbers. Here, by the second form, a = 2, 29+3(23) b=3, and c=29. Whence y=—x—1+ 26 x-3' x-b + where it is plain, that x must be such a number, that, when diminished by 3, it shall be a divisor of 26. But the several divisors of 26 are 1, 2, 13, and 26, of which the only ones that will render the expression positive, are 1 and 2. Therefore 4 or 5, and y-4-1+2 = 21 | or—5—1+ 3263 7, which are all the answers in whole numbers that the question admits of. = 3. Given y+20y, to find the values of x and y in whole positive numbers. Here, by the third form, a = 20, and b -20 2 :0; -10, and x=10+y-n, where - 24. it is plain, that n must be some even number which is a divisor of 50. But the only number of this kind, that will give positive results, is 2. :: y = 50+1-10-16, and x= =10+16-2 5y-12y+64, to find the values of x and y in whole positive numbers. Here, by the 4th form, a= : 5, b——12, 4. Given 22 = -12-16n 16n+12 and x= 8+ny, where it is plain, that n must be some integer less than 5, which numbers are only 1 and 2. And x=8+1x7=15|=8+2×44-96. 5. It is required to find two numbers such, that their product, added to their sum, shall be 79. 1, 3, 4, 7, Ans. {39, 19, 15, 9. 6. Given xxy=4x+3y+27, to find the several values of x and y in whole numbers. Ans. 27, 11 and 5. x= 4, 5 and 6. 7. Given +100y+-1000, to find the least values of x and Ans. x 70 and x=30. y in whole numbers. 8. Given 50y+100y+100, to find the values of x and y in whole numbers. Ans. x 290 and y-40. by: In addition to the present article, it will here be proper to observe, that a more direct and general method of resolving indeter minate equations of the first degree, than that before given may be derived from the doctrine of continued fractions, as follows. 164. To determine, from the principles abovementioned, the values of x and y, in the equation ax ±c, which, when a and b are prime to each other, is always possible in whole Nos. RULE. Find the quotients arising from dividing the coefficient of y by that of x, (or b by a,) as in determining their common measure, and set their corresponding converging fractions under them, according to the method made use of in Art. 165. Then, if the last fraction but one, in this series, be denoted by p we shall have, by multiplying each of its terms by c, x=pc, or -pc, and y=qc, or qc, according as c is positive or negative; observing, when the values thus found, give a result with an opposite sign to that in the question, that they must be taken with their signs changed. And if any equi-multiples ma, mb of the coefficients a, b be now taken, we shall have, in general terms, x=bm±pc, and xam± qc; where, it is evident, by putting m-1,-2, -3, &c. 0, 1, 2, 3, &c., and taking pc and qc as before, that an indefinite number of positive integral answers to the question may be obtained. P q - the same Note. From the rule above given, it is obvious, that when C1, or ax- -by=±1, which is the form of the equation on which the solution of the general case depends, that taking fractions as before, we shall then have x-p or —p, and -q. And, for the general value of the same quantities, xbmp, and y = amq, where, taking p, q, and m, as above, it is plain that the number of answers, as in the former case, will be indefinite.* 1. Given 13x- 19y = y=q or 1, to find integers for x and y. The first general solution of this problem, from prinriples similar to those above followed, is commonly ascribed to Bachet de Mesiriac, the editor of Diophantus; who gave both the rule and its investigation, in the third edition of his Mathematical Recreations, entitled Problémes Plaisans et délectables qui se font par les Nombres, printed in 1624. But it is worthy of remark, that a rule of the same extent and import is given in the Hindu treatises on Algebra, by Brahmegupta, written in 696, and Bhascara, written in 1150. Here 13)19(1 13 Whence Quotients...1, 2, 6, Therefore xp=3 Converging fractions., t., 18; and y-q=2; which numbers give, by substitution, 13 x 3-19 X2-39-38-1. And if, according to the rule, there be taken the general values x19m+3, and 1)6(613m+2, we shall have, by putting m=0, 1, 2, 3, 4, &c. respectively. 6)13(2 12 6 0 (x=3, 22, 41, 60, 79, &c. which values are y=2, 15, 28, 41, 54, &c. } all deduced from the first two, by adding successively the coefficient of y to the last value of x, and the coefficient of x to that of y. 2. Given 15-17y = 1, to find the least integral values of x and y. Ans. x = = 8, and y - 7. 3. Given 14x-17y-1, to find the integral values of x and y. 14)17(1 14 3)14(4 12 2)3(1 2 1, 4, 1, 2, 5 6 Whence Converging fractions 3,,,,H Therefore xp=6, and y-q=5; which numbers give, by substitution, 14X6-17X5-8485-1, whence, as the result has here an opposite sign to that in the proposed equation, we must take x-p-6, and y-q-5, or 1)2(2 change the signs of the above values; in which case we shall have -14×6+17×5= -84+85-1. And if, agreeably to the rule, there be taken the general values x = 17m -6, and y-14m-5, we shall have, by putting m=1, 2, 3, 4 &c. as before, (x=11, 28, 45, 62, 79, &c. }, which values, both y= 9, 23, 37, 51, 65, &c. (' in this and the following examples, are derived from each other, as above. Ans. x = 5, and y - 11. 5. Given 13x-5y=-1, to find the integral values of x and y. Quotients. 1ly = of and y. 13)5(0 5)13(2 10 3)5(1 2)3(1 2 1)2(2 ...........0, 2, 1, 1, 2, Converging fractions...d, f, 1, 1, 3, fs. Therefore x=—p= -2, and y=-q=—5 ; which numbers give, by substitution, -13×2 +5×5-26+25-1. And consequently, taking the general values x=5m-2, and y=13m-5, we shall have, by putting m=1, 2, 3, 4, &c. as usual, x=3, 8, 13, 18, 23, &c. 19, to find the least integral values of 7. Given 18x-23y=4, to find the integral values of x and y. |