2020 CIME I Problems/Problem 10
Problem 10
Let be the divisors of a positive integer
. Let
be the sum of all positive integers
satisfying
Find the remainder when
is divided by
.
Solution
Note that is even, as if
is odd then the RHS is even and the LHS is odd. This implies that the first two divisors of
are
and
. Now, there are three cases (note that
represents a prime):
When
. Then,
is always odd, so this is a contradiction.
When
. This implies that
or
. Then,
.
When
. If
then
. If
then
. This means
as
, however, this fails.
So, the sum of all possible values of are
, so the remainder is
. ~rocketsri
See also
2020 CIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 9 |
Followed by Problem 11 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All CIME Problems and Solutions |
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