And, by taking p = 1, 2, 3, 4, &c., we shall have y = 3, 6, 9, 12, 15, &c., and z = 6, 11, 16, 21, 26, &c. But from the first of the two given equations x=5+2y-z; whence, by substituting the above values for y and z, the results will give x = 5, 6, 7, 8, 9, &c. And therefore the first six values of x, y, and z, are as below : Where the law by which they can be continued is sufficiently obvious. EXAMPLES FOR PRACTICE. A 1. Given 3x = 8y - 16, to find the least values of x and y in whole numbers. Ans. x = 8, y = 5. 2. Given 14x = 5y +7, to find the least values of æ and y in whole numbers. Ans. x = 3, y = 7. 3. Given 27x = 1600 - 16y, to find the least values of x Ans. x = 48, y = 19. and y in whole numbers. 4. It is required to divide 100 into two such parts, that one of them may be divisible by 7, and the other by 11. Ans. The only parts are 56 and 44. 5. Given 9x + 13y = 2000, to find the greatest value of x, and the least value of y in whole numbers. Ans. x = 215, y = 5. 6. Given 11x + 5y = 254, to find all the possible values of x and y in whole numbers. Ans. x = 19, 14, 9, 4; у = 9, 20, 31, 42. 7. Given 17x + 19y + 21z = 400, to find all the answers in whole numbers which the question admits of. Ans. 10 different answers. 8. Given 5x + 7y+11z = 224, to find all the possible values x, y, and z, in whole positive numbers. Ans. The number of answers is 59. 9. It is required to find in how many different ways it is possible to pay 20l, in half-guineas and half-crowns, without using any other sort of coin. Ans. 7 different ways. 10. I owe my friend a shilling, and have nothing about me but guineas, and he has nothing but louis-d'ors; how must I contrive to acquit myself of the debt, the louis being valued at 17s. apiece, and the guineas at 21s. ? Ans. I must give him 13 guineas, and he must give me 16 louis. 11. How many gallons of British spirits, at 12s., 15s., and 18s. a gallon, must a rectifier of compounds take to make a mixture of 1000 gallons that shall be worth 17s. a gallon ? Ans. 111 at 12s.; 111 at 15s.; and 7777 at 18s. PROBLEM 2.-To find such a whole number as, being divided by other given numbers, shall leave given remainders. RULE 1. Call the number that is to be determined æ, the numbers by which it is to be divided a, b, c, &c., and the given remainders f, g, h, &c. 2. Subtract each of the remainders from z, and divide the differences by a, b, c, &c., and there will arise x-fx-gx-h &c., = whole numbers. x-f a = p, and substitute 3. Put the first of these fractions the value of x, as found in terms of p, from this equation, in the place of æ, in the second fraction. 4. Find the least value of p in this second fraction, by the last problem, which put = r, and substitute the value of æ, as found in terms of r, in the place of x in the third fraction. Find, in like manner, the least value of r, in this third fraction, which put = s. and substitute the value of x, as found in the terms of s, in the fourth fraction as before. Proceed in the same way with the next following fraction, and so on to the last; when the value of x, thus determined, will give the whole number required. EXAMPLES. 1. It is required to find the least whole number, which, being divided by 17, shall leave a remainder of 7, and when divided by 26, shall leave a remainder of 13. x-7 17 26 13 = whole numbers. And, putting = p, we shall have x = 17p + 7. Which value of æ, being substituted in the second fraction, Where, by rejecting p, there remains P+18 Therefore, p = 26r – 18 ; 26 = wh. wh. = r. Hence, if r be taken, = 1, we shall have p = 8. And consequently, x = 17p+7=17 x 8 + 7 = 143, the number sought. 2. It is required to find the least whole number, which, being divided by 11, 19, and 29, shall leave the remainders 3, 5, 10 respectively. Then Letr = the number required. x-3x-5 х 10 , 29 = whole numbers. p, we shall have x = 11p + 3. > Which value of x, being substituted in the second fraction, = P-4 3p 19 Or, by rejecting the 1, 19p 19 But 19p 19 18p X6 = 3p-4 = wh. 19 18p-5 19 is likewise = wh. 5 p+5 -24 18p-5 = wh. wh., which put = r. 19 19r-5, and x = 11 (195) + 3 = 209r And if this value be substituted for a in the third fraction, Or, by neglecting 7r - 2, we shall have the remaining part put = s. Then r = 29s+20; where, by taking s = 0, we shall haver 20. 4 And consequently, x=209r-52 = 209 × 20 52 = 4128, the number required. 3. 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. 4. It is required to find a number, which, being divided by 7, shall leave 5 for a remainder, and if divided by 9, the remainder shall be 2. Ans. 47, 110, &c. 5. It is required to 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. 6. It is required to find the least whole number, which, being divided by 7, 8, and 9, respectively, shall leave the Ans. 215. remainders 5, 7, and 8. 7. It is required to find the least whole number, which, being divided by each of the nine digits, 1, 2, 3, 4, 5, 6, 7, 8, 9, shall leave no remainders. Ans. 2520. 8. A person receiving a box of oranges observed, that when he told them out by 2, 3, 4, 5, and 6 at a time, he had none remaining; but when he told them out by 7 at a time, there remained 5; how many oranges were there in the box? Ans. 180. OF THE DIOPHANTINE ANALYSIS. THIS branch of Algebra, which is so called from its inventor, Diophantus, a Greek mathematician of Alexandria in Egypt, who flourished in or about the third century after Christ, relates chiefly to the finding of square and cube numbers, or to the rendering certain compound expressions free from surds: the method of doing which is by making such substitutions for the unknown quantity, as will reduce the resulting equation to a simple one, and then finding the value of that quantity in terms of the rest. It is to be observed, however, that questions of this kind do not always admit of answers in rationai numbers, and that when they are resolvable in this way, no rule can be given, that will apply in all the cases that may occur; but as far as respects a particular class of these problems relating to squares, they may generally be determined by means of some of the rules derived from the following formula : PROBLEM 1.- To find such values of x as will make √(ax2 + bx + c) rational, or ax2 + bx + c = a square.* RULE 1.-When the first term of the formula is wanting, or a = 0, put the side of the square sought == n; then bx + c == n2. n2 And consequently, by transposing c, and dividing by the coefficient b, we shall have x = ; where n may be any number taken at pleasure. b C 2. When the last term is wanting, or c = 0, put the side of the square sought = nx, or, for the sake of greater gene mx rality = ; then, in this case, we shall have ax2 + bx == n m2x2 n2 And consequently, by multiplying by n2, and dividing by 2 x, there will arise an2x + bn2 = m2x, and x = m2 bn2 where mand n, both in this and the following cases, may be any whole numbers whatever, that will give positive answers.† 3. When the coefficient a, of the first term, is a square number, put it = d2, and assume the side of the square sought And consequently, by cancelling dex3, and multiplying by * The coefficients a, b, of the unknown quantities, as well as the absolute term c, are here supposed to be all integers; for if they were fractions, they could be readily reduced to a common square denominator; which, being afterwards rejected, will not alter the nature of the question; since any square number, when multiplied or divided by a square number, is still a square. + The unknown quantity x, in this case, can always be found in integers when b is positive; and, in Case 4, next following, its integral value can always be determined, whether b be positive or negative. See Vol. II. of Bonnycastle's Treatise on Algebra, Art. (H.) |
