HMMT February

Algebra Difficulty 4.9 AIME Find the answer

$a$ and $b$ are integers such that $a+\sqrt{b}=\sqrt{15+\sqrt{216}}$ . Compute $a / b$ .

Your answer

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

Solution

Squaring both sides gives $a^{2}+b+2 a \sqrt{b}=15+\sqrt{216}$ ; separating rational from irrational parts, we get $a^{2}+b=15,4 a^{2} b=216$ , so $a^{2}$ and $b$ equal 6 and $9 . a$ is an integer, so $a^{2}=9, b=6 \Rightarrow a / b=3 / 6=1 / 2$ . (We cannot have $a=-3$ , since $a+\sqrt{b}$ is positive.)

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