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Difference between revisions of "2021 April MIMC 10 Problems/Problem 20"

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Given that <math>y=24\cdot 34\cdot 67\cdot 89</math>. Given that the product of the even divisors is <math>a</math>, and the product of the odd divisors is <math>b</math>. Find <math>a \colon b^4</math>.
 
Given that <math>y=24\cdot 34\cdot 67\cdot 89</math>. Given that the product of the even divisors is <math>a</math>, and the product of the odd divisors is <math>b</math>. Find <math>a \colon b^4</math>.
  
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<math>\textbf{(A)} ~512:1 \qquad\textbf{(B)} ~1024:1 \qquad\textbf{(C)} ~2^{64}:1 \qquad\textbf{(D)} ~2^{80}:1 \qquad\textbf{(E)} 2^{160}:1 \qquad</math>
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<math>\textbf{(A)} ~512:1 \qquad\textbf{(B)} ~1024:1 \qquad\textbf{(C)} ~2^{64}:1 \qquad\textbf{(D)} ~2^{80}:1 \qquad\textbf{(E)} ~2^{160}:1 \qquad</math>
 
==Solution==
 
==Solution==
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To be Released on April 26th.
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We can prime factorize the number first. <math>y=24\cdot34\cdot67\cdot89=2^3\cdot3\cdot2\cdot17\cdot67\cdot89=2^4\cdot3\cdot17\cdot67\cdot89</math>. All of the odd factors of <math>y</math> would be factors of <math>3\cdot17\cdot67\cdot89</math>. Therefore, there are <math>2\cdot2\cdot2\cdot2=16</math> odd factors of <math>y</math>. Let those factors form a set <math>A</math>, and all even factors would be <math>2A</math> (all elements in <math>A</math> multiplied by <math>2</math>), <math>4A</math>, <math>8A</math>, <math>16A</math>. Let the product of all odd factors in <math>A</math> be <math>b</math>, then the product of all even factors would be <math>a=2^{16}\cdot b\cdot4^{16}\cdot b\cdot8^{16}\cdot b\cdot16^{16}=2^{16}\cdot4^{16}\cdot8^{16}\cdot16^{16}\cdot b^4</math>. Therefore, the ratio of <math>a: b^4=2^{16}\cdot4^{16}\cdot8^{16}\cdot16^{16}:1=</math><math>\boxed{\textbf{(E)} 2^{160}:1}</math>.

Latest revision as of 12:59, 26 April 2021

Given that $y=24\cdot 34\cdot 67\cdot 89$. Given that the product of the even divisors is $a$, and the product of the odd divisors is $b$. Find $a \colon b^4$.

$\textbf{(A)} ~512:1 \qquad\textbf{(B)} ~1024:1 \qquad\textbf{(C)} ~2^{64}:1 \qquad\textbf{(D)} ~2^{80}:1 \qquad\textbf{(E)} ~2^{160}:1 \qquad$

Solution

We can prime factorize the number first. $y=24\cdot34\cdot67\cdot89=2^3\cdot3\cdot2\cdot17\cdot67\cdot89=2^4\cdot3\cdot17\cdot67\cdot89$. All of the odd factors of $y$ would be factors of $3\cdot17\cdot67\cdot89$. Therefore, there are $2\cdot2\cdot2\cdot2=16$ odd factors of $y$. Let those factors form a set $A$, and all even factors would be $2A$ (all elements in $A$ multiplied by $2$), $4A$, $8A$, $16A$. Let the product of all odd factors in $A$ be $b$, then the product of all even factors would be $a=2^{16}\cdot b\cdot4^{16}\cdot b\cdot8^{16}\cdot b\cdot16^{16}=2^{16}\cdot4^{16}\cdot8^{16}\cdot16^{16}\cdot b^4$. Therefore, the ratio of $a: b^4=2^{16}\cdot4^{16}\cdot8^{16}\cdot16^{16}:1=$$\boxed{\textbf{(E)} 2^{160}:1}$.

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