Difference between revisions of "2021 JMPSC Accuracy Problems/Problem 11"
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If <math>a : b : c : d=1 : 2 : 3 : 4</math> and <math>a</math>, <math>b</math>, <math>c</math>, and <math>d</math> are divisors of <math>252</math>, what is the maximum value of <math>a</math>? | If <math>a : b : c : d=1 : 2 : 3 : 4</math> and <math>a</math>, <math>b</math>, <math>c</math>, and <math>d</math> are divisors of <math>252</math>, what is the maximum value of <math>a</math>? | ||
− | ==Solution== | + | ==Solution #1== |
<math>a</math> must be a number such that <math>2a \mid 252</math>, <math>3a \mid 252</math>, <math>4a \mid 252</math>. Thus, we must have <math>12a \mid 252</math>. This implies the maximum value of <math>a</math> is <math>252/12 = \boxed{21}</math> | <math>a</math> must be a number such that <math>2a \mid 252</math>, <math>3a \mid 252</math>, <math>4a \mid 252</math>. Thus, we must have <math>12a \mid 252</math>. This implies the maximum value of <math>a</math> is <math>252/12 = \boxed{21}</math> | ||
~Bradygho | ~Bradygho | ||
+ | |||
+ | ==Solution #2== | ||
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>. | 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) | (With <math>a=21</math>, we have <math>b=42, c=63, d=84,</math> which is valid) | ||
~Apple321 | ~Apple321 |
Revision as of 23:58, 10 July 2021
Problem
If and
,
,
, and
are divisors of
, what is the maximum value of
?
Solution #1
must be a number such that
,
,
. Thus, we must have
. This implies the maximum value of
is
~Bradygho
Solution #2
Notice that . Because
and
it is invalid for
to be a multiple of
. With similar reasoning,
must have at most one factor of
. Thus,
.
(With , we have
which is valid)
~Apple321