2014 AMC 12B Problems/Problem 25
Problem
Find the sum of all the positive solutions of
Solution
Rewrite as
. Now let
, and let
. We have
Notice that either
and
or
and
. For the first case,
only when
and
is an integer.
when
is an even multiple of
, and since
,
only when
is an odd divisor of
. This gives us these possible values for
:
For the case where
,
, so
, where m is odd.
must also be an odd multiple of
in order for
to equal
, so
must be odd. We can quickly see that dividing an even number by an odd number will never yield an odd number, so there are no possible values for
, and therefore no cases where
and
. Therefore, the sum of all our possible values for
is
See also
2014 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 24 |
Followed by Last Question |
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All AMC 12 Problems and Solutions |
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