2005 AIME I Problems/Problem 12
Problem
For positive integers let
denote the number of positive integer divisors of
including 1 and
For example,
and
Define
by
Let
denote the number of positive integers
with
odd, and let
denote the number of positive integers
with
even. Find
Solution
It is well-known that is odd if and only if
is a perfect square. (Otherwise, we can group divisors into pairs whose product is
.) Thus,
is odd if and only if there are an odd number of perfect squares less than
. So
and
are odd, while
are even, and
are odd, and so on.
So, for a given , if we choose the positive integer
such that
we see that
has the same parity as
.
It follows that the numbers between and
, between
and
, and so on, all the way up to the numbers between
and
have
odd. These are the only such numbers less than
(because
).
Solution 1
Notice that the difference between consecutive squares are consecutively increasing odd numbers. Thus, there are numbers between
(inclusive) and
(exclusive),
numbers between
and
, and so on. The number of numbers from
to
is
. Whenever the lowest square beneath a number is odd, the parity will be odd, and the same for even. Thus,
.
, the
accounting for the difference between
and
, inclusive. Notice that if we align the two and subtract, we get that each difference is equal to
. Thus, the solution is
.
Solution 2
Similarly, , where the
accounts for those numbers between
and
.
Thus .
Then, .
We can apply the formula
. From this formula, it follows that
and so that
. Thus,
.
Solution 3
Let denote the sum
. We can easily see from the fact "It is well-known that
is odd if and only if
is a perfect square.", that
.
.
. They ask for
, so our answer is
-Alexlikemath
See also
2005 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 11 |
Followed by Problem 13 | |
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