Difference between revisions of "2023 AIME I Problems/Problem 4"
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We first rewrite <math>13!</math> as a prime factorization, which is <math>2^{10}\cdot3^5\cdot5^2\cdot7\cdot11\cdot13.</math> | We first rewrite <math>13!</math> as a prime factorization, which is <math>2^{10}\cdot3^5\cdot5^2\cdot7\cdot11\cdot13.</math> | ||
− | For the fraction to be a square, it needs each prime to be an even power. This means <math>m</math> must contain <math>7\cdot11\cdot13</math>. Also, <math>m</math> can contain any even power of <math>2</math> up to <math>10</math>, any odd power of <math>3</math> up to <math>5</math>, and any even power of <math>5</math> up to <math>2</math>. The sum of <math>m</math> is <cmath>(2^0+2^2+2^4+2^6+2^8+2^{10})(3^1+3^3+3^5)(5^0+5^2)\cdot7\cdot11\cdot13 = 1365\cdot273\cdot26\cdot7\cdot11\cdot13 = 2\cdot3^2\cdot5\cdot7^3\cdot11\cdot13^4.</cmath> Therefore, the answer is <math>1+2+1+3+1+4=\boxed{012}</math>. | + | For the fraction to be a square, it needs each prime to be an even power. This means <math>m</math> must contain <math>7\cdot11\cdot13</math>. Also, <math>m</math> can contain any even power of <math>2</math> up to <math>10</math>, any odd power of <math>3</math> up to <math>5</math>, and any even power of <math>5</math> up to <math>2</math>. The sum of <math>m</math> is <cmath>(2^0+2^2+2^4+2^6+2^8+2^{10})(3^1+3^3+3^5)(5^0+5^2)\cdot7\cdot11\cdot13 = </cmath> <cmath>1365\cdot273\cdot26\cdot7\cdot11\cdot13 = 2\cdot3^2\cdot5\cdot7^3\cdot11\cdot13^4.</cmath> Therefore, the answer is <math>1+2+1+3+1+4=\boxed{012}</math>. |
~chem1kall | ~chem1kall |
Revision as of 13:36, 9 February 2023
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 (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 conclude that, when we are given
, the desired sum is
. The problem gives us
, so the answer is
.
-InsetIowa9
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.