2000 AIME II 第 4 题

Write $N = 2^a m$ with $m$ odd; the odd divisors of $N$ are exactly the divisors of $m$

Every even divisor is a power $2^k$ times an odd divisor, so there are $a \cdot d(m)$ of them; find the smallest odd $m$ with $d(m) = 6$

Write $N = 2^a m$ with $m$ odd. The odd divisors of $N$ are exactly the divisors of $m,$ so $d(m) = 6.$ Every even divisor is $2^k$ (for $1 \le k \le a$) times an odd divisor, so there are $a \cdot d(m) = 6a$ of them, and $6a = 12$ gives $a = 2.$

So $N = 4m$ where $m$ is the smallest odd number with exactly $6$ divisors. The shapes are $p^5$ (smallest $3^5 = 243$) and $p^2 q$ (smallest $3^2 \cdot 5 = 45$), so $m = 45$ and $N = 4 \cdot 45 = 180.$