By proceeding, therefore, as in the former rule, we shall And consequently, putting p=0, we shall have the least value of y z — 1; where z may be any number, from 1 up to 13, that will answer the conditions of the question. When, therefore, z == 2, we have y=1, Hence, by taking z = 2, 3, 4, 5, &c., the corresponding values of x and y, together with those of z, will be found to be as below: х Which are all the integral values of x, y, and z, that can be obtained from the given equation. Note 2.-If there be three unknown quantities, and only two equations for determining them, as ax + by + cz= d, and ex +fy + gz h, exterminate one of these quantities in the usual way, and find the values of the other two from the resulting equation, as before. Then, if the values, thus found, be separately substituted in either of the given equations, the corresponding values of the remaining quantities will likewise be determined: thus: 6. Let there be given x ·2y + z = 5, and 2x+y-z=7, to find the values of x, y, and z. Here, by multiplying the first of these equations by 2, and subtracting from the second the product, we shall have And, by taking p 1, 2, 3, 4, &c., we shall have y 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 ; 3, 6, 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. Ans. x 3, y 2. Given 14x= 5y +7, to find the least values of x and y in whole numbers. 3. Given 27x = 1600 7. 16y, to find the least values of x Ans. x 48, y 48, y = 19. 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. and y in whole numbers. Ans. The only parts are 56 and 44. 2000, to find the greatest value of x, Ans. x 5. Given 9x + 13y and the least value of y in whole numbers. 215, y 215, y = 5. 254, to find all the possible values of 6. Given 11x+5y * and y Ans. x = 19, 14, 9, 4; y 9, 20, 31, 42. 7. Given 17+ 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 and 2, in whole positive numbers. values x, y, Ans. The number of answers is 59. 9. It is required to find in how many different ways it is possible to pay 201, 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 x, 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 x, and divide the differences by a, b, c, &c., and there will arise x - f x g x - h α 3. Put the first of these fractions p, and substitute the value of x, as found in terms of p, from this equation, in the place of x, in the second fraction. 4. Find the least value of p in this second fraction, by the last problem, which putr, and substitute the value of x, 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. Let x= the number required. х х Then and 1'7 X And, putting 17 26 13 whole numbers. p, we shall have x 17p + 7. Which value of x, being substituted in the second fraction, p+18 wh. = r. 1, we shall have p p= 8. Where, by rejecting p, there remains Therefore, p=26r 18; Hence, if r be taken, 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. the number required. Then Let x х 3 And, putting 11 29 whole numbers. p, we shall have x11p+3. Which value of x, being substituted in the second fraction, gives 11p 2 wh. And, by rejecting p, there will remain wh. 19 18p - 5 Also, by multiplication, X6 = 19 p And if this value be substituted for x in the third fraction, there will arise 209r 52. Or, by neglecting 7r2, we shall have the remaining part Then r=29s+20; where, by taking s=0, we shall have r= put = s. 20. x = 209r 52 209 X 20 52 = 4128, the number required. And consequently, 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 remainders 5, 7, and 8. Ans. 215. 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 |