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Let $ABC$ be a arbitrary triangle with integral sides such that the perimeter is $2006$. And one of the side $16$ times the other side. How many such triangles exist?

Attempt

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    $\begingroup$ Note that the length of the longest side of a triangle must be less than the sum of the lengths of the two shorter sides. This gives the additional requirements $15 x < y < 17 x$. $\endgroup$ Jan 15 '15 at 2:15
  • $\begingroup$ Note that $2006=(118)(17)$, So all integer solutions of $17x+y=2006$ have shape $y=17t$, $x=118-t$. Now we need to pick out the $t$ that give us a real triangle. $\endgroup$ Jan 15 '15 at 2:31
  • $\begingroup$ Be careful, a straight line is not a triangle :-) $\endgroup$
    – Joffan
    Jan 15 '15 at 2:44
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I'll offer some hints.

You've got the basic relationship:

$$17x + y = 2006,$$

where $x$ is the shortest side and $y$ is the side not constrained to be $16$ times the shortest side, $16x$.

Now consider this. What other relation constrains the sides? Hint: $(1, 16, 1989)$ aren't valid sides for a triangle, even though they add up to $2006$ and have one side sixteen times another. Why not?

This should make the number of triples manageable.

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