Find two positive numbers whos sum is 1000. one number must be a multiple of 19, and the other number must be a multiple of 47.

Step-by-step explanation:

The two numbers are 342 = 18·19, and 658 = 14·47.

Suppose the multipliers of 19 and 47 we want to find are "a" and "b", respectively. Then ...

  19a +47b = 1000

Solving for a, we get ...

  a = (1000 -47b)/19 = 52 -2b +(12 -9b)/19 = 52 -2b + a1 . . . . . defines a1

Solving for b in terms of a1, we have ...

  b = (12 -19a1)/9 = 1 -2a1 +(3 -a1)/9 = 1 -2a1 +b1 . . . . . defines b1

Solving for a1 in terms of b1, we have ...

  a1 = 3 -9b1 . . . . . . . always an integer for any integer b1

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Back-substituting this into the expression for b, we get ...

  b = (12 -19(3 -9b1))/9 = -5 +19b1 . . . . . for any integer b1

and substituting this into the expression for a gives ...

  a = (1000 -47(-5 +19b1))/19 = 65 -47b1

The smallest positive values of a and b are (a, b) = (18, 14), corresponding to the numbers 19·18 = 342 and 47·14 = 658.

Comment on the above

This is a version of the Extended Euclidean Algorithm for solving Diophantine equations. The basic idea is to keep reducing the fractional part of the solution until there is no fractional part.

2x^2+5x+17

your welcome. : )