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Difference between revisions of "2021 JMPSC Accuracy Problems/Problem 11"

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~Bradygho
 
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Notice that <math>252=2^2\cdot 3^2\cdot 7</math>. Because <math>b=2a</math> and <math>d=4a,</math> it is invalid for <math>a</math> to be a multiple of <math>2</math>. With similar reasoning, <math>a</math> must have at most one factor of <math>3</math>. Thus, <math>a=\boxed{21}</math>.
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(With <math>a=21</math>, we have <math>b=42, c=63, d=84,</math> which is valid)
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~Apple321

Revision as of 22:57, 10 July 2021

Problem

If $a : b : c : d=1 : 2 : 3 : 4$ and $a$, $b$, $c$, and $d$ are divisors of $252$, what is the maximum value of $a$?

Solution

$a$ must be a number such that $2a \mid 252$, $3a \mid 252$, $4a \mid 252$. Thus, we must have $12a \mid 252$. This implies the maximum value of $a$ is $252/12 = \boxed{21}$

~Bradygho

Notice that $252=2^2\cdot 3^2\cdot 7$. Because $b=2a$ and $d=4a,$ it is invalid for $a$ to be a multiple of $2$. With similar reasoning, $a$ must have at most one factor of $3$. Thus, $a=\boxed{21}$.

(With $a=21$, we have $b=42, c=63, d=84,$ which is valid)

~Apple321

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