2012 AIME I Problems/Problem 10
Problem 10
Let be the set of all perfect squares whose rightmost three digits in base
are
. Let
be the set of all numbers of the form
, where
is in
. In other words,
is the set of numbers that result when the last three digits of each number in
are truncated. Find the remainder when the tenth smallest element of
is divided by
.
Solution
It is apparent that for a perfect square to satisfy the constraints, we must have
or
Now in order for
to be a multiple of
at least one of
and
must be a multiple of
and since
and
are in different residue classes mod
one term must have all the factors of
and thus must be a multiple of
Furthermore, each of
and
must have at least two factors of
since otherwise
could not possibly be divisible by
So therefore the conditions are satisfied if either
or
is divisible by
or equivalently
Counting up from
to
we see that the tenth value of
is
and thus
See also
2012 AIME 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 | ||
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