Difference between revisions of "2005 AIME I Problems/Problem 12"
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Thus <math>a = (2^2 - 1^2) + (4^2 - 3^2) + \ldots + (44^2 - 43^2)</math>. | Thus <math>a = (2^2 - 1^2) + (4^2 - 3^2) + \ldots + (44^2 - 43^2)</math>. | ||
− | + | Then, <math>|a - b| = |2(2^2 + 4^2 + \ldots + 44^2) - 2(1^2 + 3^2 + 5^2 + \ldots 43^2) + 1^2 - 45^2 + 19|</math>. | |
− | + | We can apply the formula <math>1^2 + 2^2 + \ldots + n^2 = \frac{n(n + 1)(2n + 1)}{6}</math>. From this formula, it follows that <math>2^2 + 4^2 + \ldots + (2n)^2 = \frac{2n(n + 1)(2n + 1)}{3}</math> and so that | |
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:<math>1^2 + 3^2 + \ldots +(2n + 1)^2 = (1^2 + 2^2 + \ldots +(2n + 1)^2) - (2^2 + 4^2 + \ldots + (2n)^2)</math> | :<math>1^2 + 3^2 + \ldots +(2n + 1)^2 = (1^2 + 2^2 + \ldots +(2n + 1)^2) - (2^2 + 4^2 + \ldots + (2n)^2)</math> | ||
:<math>= \frac{(2n + 1)(2n + 2)(4n + 3)}{6} - \frac{2n(n + 1)(2n + 1)}{3} = \frac{(n + 1)(2n + 1)(2n + 3)}{3}</math>. Thus, | :<math>= \frac{(2n + 1)(2n + 2)(4n + 3)}{6} - \frac{2n(n + 1)(2n + 1)}{3} = \frac{(n + 1)(2n + 1)(2n + 3)}{3}</math>. Thus, |
Revision as of 22:21, 13 November 2011
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
Contents
[hide]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,
.
See also
2005 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 11 |
Followed by Problem 13 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |