Difference between revisions of "2010 AMC 12B Problems/Problem 25"
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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. | 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. | ||
− | 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,...78</math>, since <math>x=71,79</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>. | + | 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>. |
== Similar Solution == | == Similar Solution == |
Revision as of 14:53, 9 January 2017
Problem 25
For every integer , let
be the largest power of the largest prime that divides
. For example
. What is the largest integer
such that
divides
?
Solution
Because 67 is the largest prime factor of 2010, it means that in the prime factorization of , there'll be
where
is the desired value we are looking for. Thus, to find this answer, we need to look for the number of times
is incorporated into the giant product.
All numbers , given
such that for any integer
between
and
, prime
must be less than
, contributes a 67 to the product. Considering
, the possible values of x are
, since
are primes that are greater than 67. However,
contributes two
s to the product, so we must count it twice. Therefore, the answer is
.
Similar Solution
After finding the prime factorization of , divide
by
and add
divided by
in order to find the total number of multiples of
between
and
.
Since
,
, and
are prime numbers greater than
and less than or equal to
, subtract
from
to get the answer
.
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) | |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.