Difference between revisions of "2023 AIME I Problems/Problem 4"
MRENTHUSIASM (talk | contribs) (→See also) |
m (→Solution 1) |
||
Line 6: | Line 6: | ||
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> | + | 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>. |
~chem1kall | ~chem1kall |
Revision as of 17:58, 8 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)
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.