2021 April MIMC 10 Problems/Problem 20

Revision as of 13:58, 26 April 2021 by Cellsecret (talk | contribs) (Solution)

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 $a\colon b^4=2^{16}\cdot4^{16}\cdot8^{16}\cdot16^{16}\colon1=\fbox{\textbf{(E)}$ (Error compiling LaTeX. ! File ended while scanning use of \fbox.)2^{160}\colon1$}$ (Error compiling LaTeX. ! Extra }, or forgotten $.).

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