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Difference between revisions of "2010 AMC 12B Problems/Problem 25"

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== Problem 25 ==
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== Problem ==
 
For every integer <math>n\ge2</math>, let <math>\text{pow}(n)</math> be the largest power of the largest prime that divides <math>n</math>. For example <math>\text{pow}(144)=\text{pow}(2^4\cdot3^2)=3^2</math>. What is the largest integer <math>m</math> such that <math>2010^m</math> divides
 
For every integer <math>n\ge2</math>, let <math>\text{pow}(n)</math> be the largest power of the largest prime that divides <math>n</math>. For example <math>\text{pow}(144)=\text{pow}(2^4\cdot3^2)=3^2</math>. What is the largest integer <math>m</math> such that <math>2010^m</math> divides
  
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<math>\textbf{(A)}\ 74 \qquad \textbf{(B)}\ 75 \qquad \textbf{(C)}\ 76 \qquad \textbf{(D)}\ 77 \qquad \textbf{(E)}\ 78</math>
 
<math>\textbf{(A)}\ 74 \qquad \textbf{(B)}\ 75 \qquad \textbf{(C)}\ 76 \qquad \textbf{(D)}\ 77 \qquad \textbf{(E)}\ 78</math>
  
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<center>
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== Solution ==  
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Because 67 is the largest prime factor of 2010, it means that in the prime factorization of <math>\prod_{n=2}^{5300}\text{pow}(n)</math>, there'll be <math>p_1 ^{e_1} \cdot p_2 ^{e_2} \cdot .... 67^x ...</math> where <math>x</math> is the desired value we are looking for. Thus, to find this answer, we need to look for the number of times <math>67</math> is incorporated into the giant product.
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== Solution ==
 
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WARNING: THIS IS A SOMEWHAT TIME-CONSUMING METHOD. IF SOMEONE HAS A MORE ELEGANT SOLUTION, PLEASE POST IT.
 
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Because 67 is the largest prime factor of 2010, it means that in the prime factorization of <math>\prod_{n=2}^{5300}\text{pow}(n)</math>, there'll be <math>p_1 ^{e_1} \cdot p_2 ^{e_2} \cdot .... 67^x ...</math> where <math>x</math> 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. We can consider several cases.
 
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'''Case 1:'''
 
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<math>p_1 \cdot 67</math>
 
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<math>p</math> can be any prime from 2 to 67. Listing all the primes and enumerating them (from 2 to 67) yields 19 primes. This results in <math>19 + 1 = 20</math> times where 67 is counted in in the form <math>p_1 \cdot 67</math>.
 
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'''Case 2:'''
 
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<math>p_1 ^{e_1} \cdot 67</math>
 
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<math>p</math> again can be any prime from 2 to 67. Looking at <math>\lfloor{\frac{5300}{67}}\rfloor = 79</math> helps show whether or not the number multiplied by 67 is larger than 5300. <math>2^{x}</math> must be less than 76. This gives from <math>2^2</math> to <math>2^6</math>, <math>3^2</math> and <math>3^3</math>, <math>5^2</math>,  and <math>7^2</math>, <math>11^2</math>, ... <math>67^2</math>, totaling 20.
 
  
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'''Case 3:'''
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All numbers <math>n=67 \cdot x</math>, given <math>x = p_1 ^ {e_1} \cdot p_2 ^{e_2} \cdot ... \cdot p_m ^ {e_m}</math> such that for any integer <math>x</math> between <math>1</math> and <math>m</math>, prime <math>p_x</math> must be less than <math>67</math>, contributes a 67 to the product. Considering <math>67 \cdot 79 < 5300 < 67 \cdot 80</math>, the possible values of x are <math>1,2,...,70,72,74,...78</math>, since <math>x=71,73,79</math> are primes that are greater than 67. However, <math>\text{pow}\left(67^2\right)</math> contributes two <math>67</math>s to the product, so we must count it twice. Therefore, the answer is <math>70 + 1 + 6 = \boxed{77} \Rightarrow \boxed{D}</math>.
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<math>p_1 \cdot p_2 ... \cdot 67</math>
 
  
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<math>p</math> could theoretically be any prime. 2 can pair with any prime from 3 to 37, totaling 11 primes. 3 can pair with any prime from 5 to 23, totaling 7 primes. 5 can pair with any prime from 7 to 13, totaling 3. These, altogether, sum to 21.  
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== Similar Solution ==
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After finding the prime factorization of <math>2010=2\cdot3\cdot5\cdot67</math>, divide <math>5300</math> by <math>67</math> and add <math>5300</math> divided by <math>67^2</math> in order to find the total number of multiples of <math>67</math> between <math>2</math> and <math>5300</math>. <math>\lfloor\frac{5300}{67}\rfloor+\lfloor\frac{5300}{67^2}\rfloor=80</math> Since <math>71</math>,<math>73</math>, and <math>79</math> are prime numbers greater than <math>67</math> and less than or equal to <math>80</math>, subtract <math>3</math> from <math>80</math> to get the answer <math>80-3=\boxed{77}\Rightarrow\boxed{D}</math>.
  
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'''Case 4:'''
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==Need Discussion + Clarification?==
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<math>p_1 ^ {e_1} \cdot p_2 ^ {e_2} \cdot ... 67</math>
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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
  
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<math>p</math>'s combination with any other number cannot exceed 76, that is the general rule.
 
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For example, for 2, <math>2^2 \cdot 3, 2^3 \cdot 3, 2^4 \cdot 3, 2^2 \cdot 5, 2^3 \cdot 5, 2^2 \cdot 7, 2^3 \cdot 7, 2^2 \cdot 11, 2^2 \cdot 13, 2^2 \cdot 17,</math> and <math>2^2 \cdot 19.</math> (Alternatively, a speedier method would have been to isolate <math>2^2</math>, find the lowest and highest primes, and count much quicker). Repeating the same process for <math>3^{e_2}</math> and <math>5^{e_3}</math> quickly arrives at 3 and 2, respectively. Adding these all together yields 16.
 
  
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Response to concern:
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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.
  
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Adding all these numbers, <math>20 + 20 + 21 + 16 = \boxed{77} \Rightarrow D</math>
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==See Also==
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{{AMC12 box|ab=B|year=2010|after=Last Problem|num-b=24}}
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{{MAA Notice}}

Latest revision as of 20:26, 16 February 2021

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.

See Also

2010 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 24 		Followed by
Last 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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