2010 AIME II Problems/Problem 3

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Problem

Let $K$ be the product of all factors $(b-a)$ (not necessarily distinct) where $a$ and $b$ are integers satisfying $1\le a < b \le 20$. Find the greatest positive integer $n$ such that $2^n$ divides $K$.

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

See also

2010 AIME II (ProblemsAnswer KeyResources)
Preceded by
Problem 2
Followed by
Problem 4
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions

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