A triangle has integral sides. Its perimeter is 2006. One of the sides is sixteen times another side. The number of such triangles is [ ]

Solution

Let the sides be $$a$$, $$16a$$ and $$c$$. Then $$17a + c = 2006$$, so $$c = 2006 - 17a$$. The triangle inequality $$c < a + 16a$$ gives $$2006 < 34a$$, so $$a \ge 60$$, and $$16a < a + c$$ gives $$32a < 2006$$, so $$a \le 62$$. Hence $$a = 60, 61, 62$$, giving the triangles $$(60, 960, 986)$$, $$(61, 976, 969)$$ and $$(62, 992, 952)$$, so there are 3 such triangles.

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