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Weekly Problem 41 - 2012

Stage: 4 Challenge Level: Challenge Level:1

Let the first two terms of the sequence be $a$ and $b$ respectively.

Then the next three terms are $a+b$, $a+2b$, $2a+3b$. So $2a+3b = 2004$.

For $a$ to be as large as possible, we need $b$ to be as small as possible, consistent with both being positive integers.

If $b=1$ then $2a=2001$, but $a$ is an integer, so $b\not=1$.

However, if $b=2$ then $2a=1998$, so the maximum possible value of $a$ is $999$.


This problem is taken from the UKMT Mathematical Challenges.

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