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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.

1 Answer

5 votes

Answer:

  • 342
  • 658

Explanation:

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

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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.

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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.

User CamQuest
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