Difference between revisions of "2023 AIME I Problems/Problem 4"
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~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ||
− | ==Solution 3 Guess | + | ==Solution 3 (Educated Guess and Engineer's Induction)== |
Try smaller cases. There is clearly only one <math>m</math> that makes <math>\frac{2!}{m}</math> a square, and this is <math>m=2</math>. Here, the sum of the exponents in the prime factorization is just <math>1</math>. Furthermore, the only <math>m</math> that makes <math>\frac{3!}{m}</math> a square is <math>m = 6 = 2^13^1</math>, and the sum of the exponents is <math>2</math> here. Trying <math>\frac{4!}{m}</math> and <math>\frac{5!}{m}</math>, the sums of the exponents are <math>3</math> and <math>4</math>. Based on this, we (incorrectly!) conclude that, when we are given <math>\frac{n!}{m}</math>, the desired sum is <math>n-1</math>. The problem gives us <math>\frac{13!}{m}</math>, so the answer is <math>13-1 = \boxed{012}</math>. | Try smaller cases. There is clearly only one <math>m</math> that makes <math>\frac{2!}{m}</math> a square, and this is <math>m=2</math>. Here, the sum of the exponents in the prime factorization is just <math>1</math>. Furthermore, the only <math>m</math> that makes <math>\frac{3!}{m}</math> a square is <math>m = 6 = 2^13^1</math>, and the sum of the exponents is <math>2</math> here. Trying <math>\frac{4!}{m}</math> and <math>\frac{5!}{m}</math>, the sums of the exponents are <math>3</math> and <math>4</math>. Based on this, we (incorrectly!) conclude that, when we are given <math>\frac{n!}{m}</math>, the desired sum is <math>n-1</math>. The problem gives us <math>\frac{13!}{m}</math>, so the answer is <math>13-1 = \boxed{012}</math>. |
Revision as of 14:11, 13 February 2023
Contents
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
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.