Difference between revisions of "2023 AIME I Problems/Problem 4"
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==Problem 4== | ==Problem 4== | ||
− | + | The sum of all positive integers <math>m</math> such that <math>\frac{13!}{m}</math> is a perfect square can be written as <math>2^a3^b5^c7^d11^e13^f,</math> where <math>a,b,c,d,e,</math> and <math>f</math> are positive integers. Find <math>a+b+c+d+e+f.</math> | |
− | ==Solution== | + | |
− | We first rewrite 13! as a prime factorization, which is <math>2^{10}\cdot3^5\cdot5^2\cdot7\cdot11\cdot13.</math> | + | ==Solution 1== |
− | 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 2 up to 10, any odd power of 3 up to 5, and any even power of 5 up to 2. The sum of <math>m</math> is < | + | |
+ | 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>(7\cdot11\cdot13)(2^0+2^2+2^4+2^6+2^8+2^{10})(3^1+3^3+3^5)(5^0+5^2) = 1365\cdot26\cdot270\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 | ||
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==Solution 2== | ==Solution 2== | ||
− | The prime factorization of <math>13!</math> is < | + | The prime factorization of <math>13!</math> is <cmath>2^{10} \cdot 3^5 \cdot 5^2 \cdot 7 \cdot 11 \cdot 13.</cmath> |
To get <math>\frac{13!}{m}</math> a perfect square, we must have <math>m = 2^{2x} \cdot 3^{1 + 2y} \cdot 5^{2z} \cdot 7 \cdot 11 \cdot 13</math>, where <math>x \in \left\{ 0, 1, \cdots , 5 \right\}</math>, <math>y \in \left\{ 0, 1, 2 \right\}</math>, <math>z \in \left\{ 0, 1 \right\}</math>. | To get <math>\frac{13!}{m}</math> a perfect square, we must have <math>m = 2^{2x} \cdot 3^{1 + 2y} \cdot 5^{2z} \cdot 7 \cdot 11 \cdot 13</math>, where <math>x \in \left\{ 0, 1, \cdots , 5 \right\}</math>, <math>y \in \left\{ 0, 1, 2 \right\}</math>, <math>z \in \left\{ 0, 1 \right\}</math>. | ||
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\begin{align*} | \begin{align*} | ||
1 + 2 + 1 + 3 + 1 + 4 | 1 + 2 + 1 + 3 + 1 + 4 | ||
− | & = \boxed{ | + | & = \boxed{012} . |
\end{align*} | \end{align*} | ||
</cmath> | </cmath> | ||
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) |
Revision as of 15:46, 8 February 2023
Problem 4
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)