Difference between revisions of "1956 AHSME Problems/Problem 48"
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Revision as of 16:49, 29 January 2020
Problem 48
If is a positive integer, then
can be a positive integer, if and only if
is:
Solution
Lets begin by noticing that:
Therefore, in order for to be a positive integer,
must be a non-negative integer. Since the bottom the the fraction is an odd number, we can multiply the top of
by 2 without changing whether it is an integer or not. Therefore, in order for
to be an integer,
must also be an integer. As a result,
must be a factor of 65, or
. Therefore
must be at least 3, and less than or equal to 35. So the answer which best fits these constraints is
.