# 2010 AMC 12B Problems/Problem 25

## Problem 25

For every integer $n\ge2$, let $\text{pow}(n)$ be the largest power of the largest prime that divides $n$. For example $\text{pow}(144)=\text{pow}(2^4\cdot3^2)=3^2$. What is the largest integer $m$ such that $2010^m$ divides $\prod_{n=2}^{5300}\text{pow}(n)$? $\textbf{(A)}\ 74 \qquad \textbf{(B)}\ 75 \qquad \textbf{(C)}\ 76 \qquad \textbf{(D)}\ 77 \qquad \textbf{(E)}\ 78$

## Solution

Because 67 is the largest prime factor of 2010, it means that in the prime factorization of $\prod_{n=2}^{5300}\text{pow}(n)$, there'll be $p_1 ^{e_1} \cdot p_2 ^{e_2} \cdot .... 67^x ...$ where $x$ is the desired value we are looking for. Thus, to find this answer, we need to look for the number of times 67 would be incorporated into the giant product. Any number of the form $x \cdot 67$ would fit this form. However, this number tops at $71 = x$ because 71 is a higher prime than 67. $67^2$ itself must be counted twice because it's counted twice as a squared number. Any non-prime number that's less than 79 (and greater than 71) can be counted, and this totals 5. We have $70$ numbers (as 71 isn't counted - 1 through 70), an additional $1$ ( $67^2$), and $6$ values just greater than $71$ but less than $79$ (72, 74, 75, 76, 77, and 78). Thus, $70 + 1 + 6 = \boxed{77} \Rightarrow \boxed{D}$

## See Also

 2010 AMC 12B (Problems • Answer Key • Resources) Preceded byProblem 24 Followed byLast Problem 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AMC 12 Problems and Solutions
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