HMMT February

Algebra Difficulty 5.2 AIME, harder Find the answer

Find the number of sequences $a_{1}, a_{2}, \ldots, a_{10}$ of positive integers with the property that $a_{n+2}=a_{n+1}+a_{n}$ for $n=1,2, \ldots, 8$ , and $a_{10}=2002$ .

Your answer

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Solution

3 Let $a_{1}=a, a_{2}=b$ ; we successively compute $a_{3}=a+b ; \quad a_{4}=a+$ $2 b ; \quad \ldots ; \quad a_{10}=21 a+34 b$ . The equation $2002=21 a+34 b$ has three positive integer solutions, namely $(84,7),(50,28),(16,49)$ , and each of these gives a unique sequence.

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