Competition problem — Algebra — Maths Olympiad Prep

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Algebra Difficulty 4.9 AIME Find the answer

4. Given that $a$ and $b$ are integers. Then the number of ordered pairs $(a, b)$ that satisfy $a + b + ab = 2008$ is $\qquad$ groups.

Your answer

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

Solution

4. 12 .
$\begin{array}{r} \text { Since }(a+1)(b+1) \\ =2009=41 \times 7^{2} \text {, and given } \end{array}$

that $a, b$ take integer values, so, $a+1, b+1$ have 12 factor combinations, i.e., $1 \times 2009, 7 \times 287, 41 \times 49$ and their negatives as well as swaps.

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.