Difference between revisions of "2010 AMC 12B Problems/Problem 25"
(→Solution) |
m (→See Also) |
||
(15 intermediate revisions by 13 users not shown) | |||
Line 1: | Line 1: | ||
− | == Problem | + | == 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 | ||
Line 10: | Line 10: | ||
== Solution == | == Solution == | ||
− | 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 | + | 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,...,70,72,74,...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 == | ||
− | After finding the prime factorization of <math>2010=2\cdot3\cdot5\ | + | 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>. |
+ | |||
+ | ==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== | ==See Also== | ||
{{AMC12 box|ab=B|year=2010|after=Last Problem|num-b=24}} | {{AMC12 box|ab=B|year=2010|after=Last Problem|num-b=24}} | ||
+ | {{MAA Notice}} |
Latest revision as of 21:26, 16 February 2021
Problem
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 + 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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.