2007 AIME I Problems/Problem 11
Contents
Problem
For each positive integer , let
denote the unique positive integer
such that
. For example,
and
. If
find the remainder when
is divided by 1000.
Solution
and
. Therefore
if and only if
is in this range, or
. There are
numbers in this range, so the sum of
over this range is
.
, so all numbers
to
have their full range. Summing this up with the formula for the sum of the first
squares (
), we get
. We need only consider the
because we are working with modulo
.
Now consider the range of numbers such that . These numbers are
to
. There are
(1 to be inclusive) of them.
, and
, the answer.
Solution 2
Let be in the range of
. Then, we need to find the point where the value of
flips from
to
. This will happen when
exceeds
or
. Thus, if
then
. For
, then
. There are
terms in the first set of
, and
terms in the second set. Thus, the sum of
from
is
or
. For the time being, consider that
. Then, the sum of the values of
is
. We can collapse this to
. Now, we have to consider
from
. Considering
from just
, we see that all of these values have
. Because there are
values of
in that range, the sum of
in that range is
. Adding this to
we get
or
mod
. Now, take the range
. There are
values of
in this range, and each has
. Thus, that contributes
or
to the sum. Finally, adding
and
we get
.
See also
2007 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 10 |
Followed by Problem 12 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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