Difference between revisions of "2005 AIME I Problems/Problem 12"
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− | It follows that the numbers between <math>1^2</math> and <math>2^2</math>, between <math>3^2</math> and <math>4^2</math>, and so on, all the way up to the numbers between <math>43^2</math> and <math>44^2 = 1936</math> have <math>S(n)</math> odd, and that these are the only such numbers less than <math> | + | It follows that the numbers between <math>1^2</math> and <math>2^2</math>, between <math>3^2</math> and <math>4^2</math>, and so on, all the way up to the numbers between <math>43^2</math> and <math>44^2 = 1936</math> have <math>S(n)</math> odd, and that these are the only such numbers less than <math>2005</math> (because <math>45^2 = 2025 > 2005</math>). |
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>. | ||
− | Similarly, <math>b = (3^2 - 2^2) + (5^2 - 4^2) + \ldots + (45^2 - 44^2) - | + | Similarly, <math>b = (3^2 - 2^2) + (5^2 - 4^2) + \ldots + (45^2 - 44^2) - 19</math>, where the <math>-19</math> accounts for those numbers between <math>2005</math> and <math>2024</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 + | + | 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 must now 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 <math>1^2 + 3^2 + \ldots +(2n + 1)^2 = (1^2 + 2^2 + \ldots +(2n + 1)^2) - (2^2 + 4^2 + \ldots + (2n)^2) = \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, | We must now 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 <math>1^2 + 3^2 + \ldots +(2n + 1)^2 = (1^2 + 2^2 + \ldots +(2n + 1)^2) - (2^2 + 4^2 + \ldots + (2n)^2) = \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>|a - b| = \left| 2\cdot \frac{44\cdot23\cdot45}{3} - 2\cdot \frac{22 \cdot 43 \cdot 45}{3} - 45^2 + | + | <math>|a - b| = \left| 2\cdot \frac{44\cdot23\cdot45}{3} - 2\cdot \frac{22 \cdot 43 \cdot 45}{3} - 45^2 + 20\right| = |-25| = 025</math>. |
== See also == | == See also == |
Revision as of 13:02, 17 January 2007
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, and that these are the only such numbers less than
(because
).
Thus .
Similarly, , where the
accounts for those numbers between
and
.
Then .
We must now apply the formula . From this formula, it follows that
and so that
. Thus,
.