Assuming p=0, 1, 2, 3, 4, y=5, 16, 27, 38, 49. x=82, 65, 48, 31, 14. Which are all the possible values in whole positive mumbers. 3. Given 7x+9y=2342, to find the number of values of x and y in whole positive numbers. .. y=7p+2, and x=332--9p. From the first of these expressions we perceive, that the least value of p is 0, and from the second, that the greatest value of p is 36. Hence the required number is 37. 3. Given 5x+7y+9z=337, to find the number of values of x, y, and z, in whole positive numbers. 5y-5+10z4-4y-8z_y-1+2z_wh=p. 5 y=5p-2z+1, x=66-7ptz. Assuming now z successively equal to 1, 2, 3, &c. we shall have the corresponding values of x and y, the greatest and least values of p, and the number of answers as follows: Here we may observe, there are seven successive values of z, (beginning with the 5th,) which produce no change in the greatest value of p; and there are alternately, two and three equal least values of p; and this - order will evidently continue as long as successive values of z be assumed. The last three columns of the above table may, therefore, be continued without the former ones. Gr. val. of p, 9, 9, 9, 9, 10, 10, 10, 10, 10, 10, 10, 11, 11, 11. Least do. 1122 2 3 3 4 4 4 5 5 6 6 No. of Ans. 9988 9 8 8 7 7 7 6 7 6 6 Gr. val. of p, Gr. val. of p No. of Ans. 11 11 11 11 12 12 12 12 12 12 12 13 13 13 13 13 13 13 14 14 14 14 The number of answers, or sum of the numbers in the last column, is 169. 1 4. Given 14x=5y+19, to find the least possible values of x and y, in whole positive numbers. Ans. x=6, y=13. 5. Given 11x+5y=254, to find all the values of x and y in whole positive numbers. Ans. x=19, 14, 9, 4. y=9, 20, 31, 42. 6. Given 9x+13y=2000, required the number of va lues of x and y? Ans. 17. 7. Required to divide 100 into two parts, so that one of them may be divisible by 7, and the other by 11? The parts are 56 and 44. 8. Given 17x+19y+21z=400, required the number of values of x, y, and z. Ans. 10. 9. Required to pay 1000 dollars, in French crowns, and five franc pieces, so that the number of coins used shall be the least possible, what number of each kind will be necessary, and how many ways can that sum be paid in those coins, the French crown being 110 cents, and the five franc pieces 93 cents? Ans. 833 crowns, 90 five franc pieces, and 9 different [ways. 10. How many gallons of liquor at 12 cents, 15 cents, and 18 cents per gallon, may be mixed, to compose 300 gallons at 17 cents per gallon ?* Ans. { at 12 cents, 1, 2, 3, &c. to 49. 11. In how many ways can £1053 sterling be paid * When more equations than one are given, one, at least, of the unknown quantities may be eliminated, and the equations reduced to one. 1 without using any coins besides guineas and moidores; the guinea being 218. sterling, and the moidore 27? Ans. 112 ways. 12. A foreigner having a bill of 81 75, to pay at a hotel, offers napoleans in payment, the landlord agrees to receive them on condition that Turkish sequins shall be taken as change; how many pieces must be used, a napolean being worth $7 25, and a sequin rated at 82 10. Ans. 35 napoleans, and 120 sequins. 13. Given 7x+9y+23z=9999; of how many values will x, y, and z admit? Ans. 34365. CASE 2. 129. To find a whole number which being divided by any given numbers, shall leave given remainders. Denoting the required number by x, the given divisors by a, b, c, &c. and the remainders by f, g, h, &c. we numbers. x-f -, &c. severally equal to whole Put =p, find the value of x, and substitute it for a & in the second fraction; reduce this new fraction, as in the former case, so that the co-efficient of p may be a unit, and put the fraction thus obtained =q; find the value of p, and thence of x, in terms of q and given numbers. Substitute the value of x in the third equation, and so proceed. The last value of x will be the number sought. EXAMPLES. 1. Required a number which being divided by 7, 9, 10, and 11, shall leave the remainders 5, 4, 7, and 9, respectively. : Let x= the number sought. Then х-5 х 4 х-7 х-9 9 .7P+1x428p+4=3p+P+4=wh. p+4 9 9 9 9=q, and p=9q-4...x=63q-23. And consequently, 1*=wh=r. 10 q=10r, and x=630r-23. 630-32-577-3+3+1=wh. 11 11 Sr+1 =wh. Andr+x4=12+4 =r+1+"=wh. 11 a * If a fractional expression 7=wh, and we divide the terms into their primes, it is plain that all the prime numbers contained in the denominator, must be also contained in the numerator; thus, 3q 3.q 10 =25=wh, where the primes 2 and 5 not being contained in the 3, must be contained in q, and therefore, |