2010 AIME II Problems/Problem 3
Contents
Problem
Let be the product of all factors (not necessarily distinct) where and are integers satisfying . Find the greatest positive integer such that divides .
Solution
In general, there are pairs of integers that differ by because we can let be any integer from to and set equal to . Thus, the product is (or alternatively, .)
When we count the number of factors of , we have 4 groups, factors that are divisible by at least once, twice, three times and four times.
- Numbers that are divisible by at least once:
- Exponent corresponding to each one of them
- Sum
- Numbers that are divisible by at least twice:
- Exponent corresponding to each one of them
- Sum
- Numbers that are divisible by at least three times:
- Exponent corresponding to each one of them
- Sum
- Number that are divisible by at least four times:
- Exponent corresponding to each one of them
- Sum
Summing these give an answer of .
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 (Problems • Answer Key • Resources) | ||
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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