Question

For every integer $n\ge2,$ let $\operatorname{pow}(n)$ be the largest power of the largest prime that divides $n.$ For example, $\operatorname{pow}(144)=\operatorname{pow}(2^4\cdot3^2)=3^2.$ What is the largest integer $m$ such that $2010^m$ divides $\prod_{n=2}^{5300}\operatorname{pow}(n)?$

$74$

$75$

$76$

$77$

$78$

Solution:

Since $2010=2\cdot3\cdot5\cdot67,$ write the product as $2^A3^B5^C67^D$ times a factor coprime to all four primes; then $m=\min(A,B,C,D).$

Prime $2:$ $\operatorname{pow}(n)$ is a power of $2$ only when $n=2^k.$ Since $2^{12}=4096\lt5300\lt2^{13},$ the values $k=1,\ldots,12$ contribute $A=1+2+\cdots+12=78.$

Prime $67:$ $\operatorname{pow}(n)=67$ when $67$ is the largest prime factor, i.e. $n=67j$ with $1\le j\le79$ and every prime factor of $j$ at most $67;$ excluding $j=67, 71, 73, 79$ leaves $75$ values. The one $n$ with $\operatorname{pow}(n)=67^2$ is $n=67^2\lt5300,$ adding $2.$ So $D=75+2=77.$

A similar count shows the exponents of $3$ and $5$ are each at least $77.$

Therefore $m=\min(78,B,C,77)=77.$

Thus, the correct answer is D.

2010 AMC 12B Problem 25

Solution:

Problem 25 in Other Years