Difference between revisions of "2023 AIME I Problems/Problem 4"
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+ | ==Solution 4== | ||
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+ | We have <math>\frac{13!}{m}=a^2</math> for some integer <math>a</math>. Writing <math>13!</math> in terms of its prime factorization, we have <cmath>\frac{13!}{m}=\frac{2^{10}\cdot3^5\cdot5^2\cdot7^1\cdot11^1\cdot13^1}{m}=a^2.</cmath> For a given prime <math>p</math>, let the exponent of <math>p</math> in the prime factorization of <math>m</math> be <math>k</math>. Then we have <cmath>\frac{10-2k}{2}+\frac{5-k}{2}+\frac{2-k}{2}+\frac{1-k}{2}+\frac{1-k}{2}+\frac{1-k}{2}\geq 2.</cmath> Simplifying the left-hand side, we get <cmath>k\geq 4.</cmath> Thus, the exponent of each prime factor in <math>m</math> is at least <math>4</math>. Also, since <math>13</math> is prime and appears in the prime factorization of <math>13!</math>, it follows that <math>13</math> must divide <math>m</math>. | ||
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+ | Thus, <math>m</math> is of the form <math>2^43^45^47^4\cdot11^413^{2f}</math> for some nonnegative integer <math>f</math>. There are <math>(4+1)(4+1)(4+1)(4+1)(1+1)(2f+1)=4050(2f+1)</math> such values of <math>m</math>. The sum of all possible values of <math>m</math> is <cmath>2^43^45^47^4\cdot11^4\sum_{f=0}^{6}13^{2f}(2f+1)=2^43^45^47^4\cdot11^4\left(\sum_{f=0}^{6}13^{2f}(2f)+\sum_{f=0}^{6}13^{2f}\right).</cmath> The first sum can be computed using the formula for the sum of the first <math>n</math> squares: <cmath>\sum_{f=0}^{6}13^{2f}(2f)=\sum_{f=0}^{6}(169)^f\cdot 2f=\frac{1}{4}\left[\left(169^7-1\right)+2\left(169^6-1\right)+3\left(169^5-1\right)+\cdots+12\left(169^1-1\right)\right].</cmath> Using the formula for the sum of a geometric series, we can simplify this as <cmath>\sum_{f=0}^{6}13^{2f}(2f)=\frac{169^7-1}{4}+\frac{169^5-1}{2}+\frac{169^3-1}{4}=2613527040.</cmath> The second sum can be computed using the formula for the sum of a geometric series: <cmath>\sum_{f=0}^{6}13^{2f}=110080026.</cmath> Thus, the sum of all possible values of <math>m</math> is <cmath>2^43^45^47^4\cdot11^4(2613527040+110080026)=2^33^45^37^411^413^2\cdot 29590070656,</cmath> so <math>a+b+c+d+e+f=3+4+3+4+4+2=\boxed{012}</math>. | ||
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==Video Solution by TheBeautyofMath== | ==Video Solution by TheBeautyofMath== |
Revision as of 04:46, 18 February 2023
Contents
[hide]Problem
The sum of all positive integers such that
is a perfect square can be written as
where
and
are positive integers. Find
Solution 1
We first rewrite as a prime factorization, which is
For the fraction to be a square, it needs each prime to be an even power. This means must contain
. Also,
can contain any even power of
up to
, any odd power of
up to
, and any even power of
up to
. The sum of
is
Therefore, the answer is
.
~chem1kall
Solution 2
The prime factorization of is
To get
a perfect square, we must have
, where
,
,
.
Hence, the sum of all feasible is
Therefore, the answer is
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Solution 3 (Educated Guess and Engineer's Induction)
Try smaller cases. There is clearly only one that makes
a square, and this is
. Here, the sum of the exponents in the prime factorization is just
. Furthermore, the only
that makes
a square is
, and the sum of the exponents is
here. Trying
and
, the sums of the exponents are
and
. Based on this, we (incorrectly!) conclude that, when we are given
, the desired sum is
. The problem gives us
, so the answer is
.
-InsetIowa9
However!
The induction fails starting at !
The actual answers for small
are:
In general, if p is prime,
are "lucky", and the pattern breaks down after
-"fake" warning by oinava
Solution 4
We have for some integer
. Writing
in terms of its prime factorization, we have
For a given prime
, let the exponent of
in the prime factorization of
be
. Then we have
Simplifying the left-hand side, we get
Thus, the exponent of each prime factor in
is at least
. Also, since
is prime and appears in the prime factorization of
, it follows that
must divide
.
Thus, is of the form
for some nonnegative integer
. There are
such values of
. The sum of all possible values of
is
The first sum can be computed using the formula for the sum of the first
squares:
Using the formula for the sum of a geometric series, we can simplify this as
The second sum can be computed using the formula for the sum of a geometric series:
Thus, the sum of all possible values of
is
so
.
Video Solution by TheBeautyofMath
I also somewhat try to explain how the formula for sum of the divisors works, not sure I succeeded. Was 3 AM lol. https://youtu.be/MUYC2fBF2U4
~IceMatrix
See also
2023 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 3 |
Followed by Problem 5 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.