# 2010 AMC 12B Problems/Problem 25

(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

## Problem

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 $67$ is incorporated into the giant product.

All numbers $n=67 \cdot x$, given $x = p_1 ^ {e_1} \cdot p_2 ^{e_2} \cdot ... \cdot p_m ^ {e_m}$ such that for any integer $x$ between $1$ and $m$, prime $p_x$ must be less than $67$, contributes a 67 to the product. Considering $67 \cdot 79 < 5300 < 67 \cdot 80$, the possible values of x are $1,2,...,70,72,74,...78$, since $x=71,73,79$ are primes that are greater than 67. However, $\text{pow}\left(67^2\right)$ contributes two $67$s to the product, so we must count it twice. Therefore, the answer is $70 + 1 + 6 = \boxed{77} \Rightarrow \boxed{D}$.

## Similar Solution

After finding the prime factorization of $2010=2\cdot3\cdot5\cdot67$, divide $5300$ by $67$ and add $5300$ divided by $67^2$ in order to find the total number of multiples of $67$ between $2$ and $5300$. $\lfloor\frac{5300}{67}\rfloor+\lfloor\frac{5300}{67^2}\rfloor=80$ Since $71$, $73$, and $79$ are prime numbers greater than $67$ and less than or equal to $80$, subtract $3$ from $80$ to get the answer $80-3=\boxed{77}\Rightarrow\boxed{D}$.

## Need Discussion + Clarification?

How do we know that we only have to check 67? There is no solid relationship between 67 being the largest prime factor in 2010 and 67 giving the smallest result of 77. Ok, 67 is waaaay larger than any of the other prime factors of 2010. 2001's prime factors are close so there might be some ambiguity but if you can't see why given intervals of 67, 2s 3s and 5s don't need to be accounted for, you probably can't solve this problem .-.. Details in the Discussions page of this Article: http://artofproblemsolving.com/wiki/index.php?title=Talk:2010_AMC_12B_Problems/Problem_25

Response to concern: When I solved it, I actually ended up trying 2, 3, and 5 just to be safe. It actually ends up not being that bad (since for each one, we only need to consider numbers that only have that prime and lower primes in its prime factorization, so you can just divide 5300 by your prime, find how many numbers less than that new number are only divisible by lower primes, divide by your prime again, and repeat). All said and done, it only ended up taking about 7-8 minutes of mental math (ran out of paper oops), and I think that reasonably speaking, many who are good enough to have a decent chance of solving this 25 can check 2, 3, and 5 on paper in 5 minutes.

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 