2021 JMPSC Accuracy Problems/Problem 11

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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2021 JMPSC Accuracy Problems/Problem 11

Problem

Solution