AIME 2010 II · 第 3 题

Solution

In general, there are $20-n$ pairs of integers $(a, b)$ that differ by $n$ because we can let $b$ be any integer from $n+1$ to $20$ and set $a$ equal to $b-n$. Thus, the product is $(1^{19})(2^{18})\cdots(19^1)$ (or alternatively, $19! \cdot 18! \cdots 1!$.)

When we count the number of factors of $2$, we have 4 groups, factors that are divisible by $2$ at least once, twice, three times and four times.

• Numbers that are divisible by $2$ at least once: $2, 4, \cdots, 18$

Exponent corresponding to each one of them $18, 16, \cdots 2$

Sum $=2+4+\cdots+18=\frac{(20)(9)}{2}=90$

• Numbers that are divisible by $2$ at least twice: $4, 8, \cdots, 16$

Exponent corresponding to each one of them $16, 12, \cdots 4$

Sum $=4+8+\cdots+16=\frac{(20)(4)}{2}=40$

• Numbers that are divisible by $2$ at least three times: $8,16$

Exponent corresponding to each one of them $12, 4$

Sum $=12+4=16$

• Number that are divisible by $2$ at least four times: $16$

Exponent corresponding to each one of them $4$

Sum $=4$

Summing these give an answer of $\boxed{150}$.

Video Solution

For those of you who want a video solution: https://www.youtube.com/watch?v=NL-9fJBE3HI&list=PLpoKXj-PWCba2d6OG3-ExCZKTLRVYoAkq&index=6