Difference between revisions of "2021 JMPSC Accuracy Problems/Problem 11"
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+ | ==Solution 4 (Very algebraic)== | ||
+ | If <math>a</math> divides <math>252,</math> then <math>3a</math> and <math>4a</math> must also divide <math>252.</math> This implies that <math>\frac{252}{3a},\frac{252}{4a}</math> are both integers, and that <math>a</math> divides and multiplying, we have that <math>a</math> divides <math>84</math> and <math>63.</math> The greatest common divisor of <math>84</math> and <math>63</math> is <math>21,</math> and we can check that indeed <math>a=\boxed{21}.</math> ~samrocksnature | ||
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+ | ==See also== | ||
+ | #[[2021 JMPSC Accuracy Problems|Other 2021 JMPSC Accuracy Problems]] | ||
+ | #[[2021 JMPSC Accuracy Answer Key|2021 JMPSC Accuracy Answer Key]] | ||
+ | #[[JMPSC Problems and Solutions|All JMPSC Problems and Solutions]] | ||
+ | {{JMPSC Notice}} |
Latest revision as of 01:17, 12 July 2021
Contents
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
Solution 4 (Very algebraic)
If divides
then
and
must also divide
This implies that
are both integers, and that
divides and multiplying, we have that
divides
and
The greatest common divisor of
and
is
and we can check that indeed
~samrocksnature
See also
The problems on this page are copyrighted by the Junior Mathematicians' Problem Solving Competition.