How many solutions in set N ∪ {0} have equation x1 + x2 + x3 = 100.

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- Jun 25th 2010, 12:16 PM #1

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- Jun 25th 2010, 12:25 PM #2

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lets fix a value of x1, say 5. How many solutions are left?

The new equation is x2+x3=100-5

Which has 96 solutions. These correspond to x2 taking every value between 0 and 95, and x3=95-x2.

Generalising, we could fix x1 at any value between 0 and 100.

So the number of solutions is:

$\displaystyle \sum_{k=0}^{k=100} 101- k$

Can you finish from there?

- Jun 25th 2010, 12:27 PM #3

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