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2010 AMC 12B Problems/Problem 25

Revision as of 03:58, 10 July 2016 by Rraj (talk | contribs) (→‎Solution)

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 is 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, $70 + 1 + 6 = \boxed{77} \Rightarrow \boxed{D}$

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 and 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. Details in the Discussions page of this Article: http://artofproblemsolving.com/wiki/index.php?title=Talk:2010_AMC_12B_Problems/Problem_25

See Also

2010 AMC 12B (Problems • Answer Key • Resources)
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All AMC 12 Problems and Solutions

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