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How many integer solutions are there to a+b+c=1000 and a/5+b/6+c/7=160. Don't use that much calculation

User Vjardel
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1 Answer

4 votes

Answer:

  • 59 Natural number solutions
  • 61 Whole number solutions
  • infinitely many integer solutions

Explanation:

You want the number of integer solutions to the system of equations ...

  • a + b + c = 1000
  • a/5 + b/6 + c/7 = 160

Substitution

To ensure that a/5 and b/6 are integers, we can let a = 5n and b = 6m, where m and n are integers. Then the system of equations becomes a relation between m and n.

Substituting for 'a' and 'b' in the two equations, we have ...

5n +6m +c = 1000 ⇒ c = 1000 -5n -6m

(5n)/5 +(6m)/6 + c/7 = 160

Substituting the value of c in this second equation gives ...

n + m + (1000 -5n -6m)/7 = 160

7n +7m +1000 -5n -6m = 1120 . . . . . . multiply by 7

2n +m = 120 . . . . . . . . . . . . . . . . subtract 1000

Then we have ...

m = 120 -2n = 2(60 -n)

Solutions

If we want positive integer solutions, then n may take on values 1 .. 59. If we allow 0 as a solution value, then n may take on values 0 .. 60. If solutions may be positive or negative, then there are infinitely many solutions.

  • 59 Natural number solutions
  • 61 Whole number solutions
  • infinitely many integer solutions

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Additional comment

The solutions are ...

(a, b, c) = (5n, 12(60 -n), 7(40+n)) . . . . . for n a suitable integer

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User Salome
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