2002 AIME II 第 5 题

$\frac{6^a}{a^6} = \frac{2^a 3^a}{2^{6n} 3^{6m}},$ so $a^6$ fails to divide $6^a$ exactly when $6n \gt a$ or $6m \gt a$

If $m$ and $n$ are both at least $1,$ then $a \ge 6n$ and $a \ge 6m,$ so only pure powers of $2$ or of $3$ can work

With $a = 2^n 3^m,$ $$\frac{6^a}{a^6} = \frac{2^a 3^a}{2^{6n} 3^{6m}},$$ which fails to be an integer exactly when $6n \gt a$ or $6m \gt a.$

If $m, n \ge 1,$ then $a \ge 3 \cdot 2^n \ge 6n$ (since $2^n \ge 2n$) and similarly $a \ge 2 \cdot 3^m \ge 6m,$ so no such $a$ works. If $m = 0,$ the condition is $2^n \lt 6n,$ which holds for $n = 1,$ $2,$ $3,$ $4,$ giving $a = 2,$ $4,$ $8,$ $16.$ If $n = 0,$ the condition is $3^m \lt 6m,$ which holds for $m = 1,$ $2,$ giving $a = 3,$ $9.$ (For $a = 1$ the condition fails.)

The sum is $2 + 4 + 8 + 16 + 3 + 9 = 42.$