Difference between revisions of "2021 JMPSC Accuracy Problems/Problem 11"
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==Solution 1== | ==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>, which works. |
~Bradygho | ~Bradygho | ||
− | |||
==Solution 2== | ==Solution 2== |
Revision as of 12:31, 11 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
, which works.
~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
Solution 3 (A Little Bashy)
Note , so the divisors are
. We see the set
is the largest 4-digit set we can form, so the answer is
~Geometry285