Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2005 AIME I Problems/Problem 12"

m (→‎Solution 2)
(→‎Solution 2)
Line 18: Line 18:
 
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>.
  
−
We know the difference between two consecutive perfect squares are consecutive odd numbers.
+
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>.
−
Since we want all <math>(n+1)^2-n^2</math> where <math>n</math> is odd in order to find <math>a</math>, we want all <math>2n+1</math> for odd numbers from <math>1</math> to <math>43</math>. We can easily evaluate this sum, which is <math>990</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  
−
Obviously <math>a+b=2005</math>, so <math>b=1015</math> and <math>|a - b|=\boxed{025}</math>.
 
−
 
 
−
 
 
−
Alternatively <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>.
 
−
Then 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  
 
 
:<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 $n,$ let $\tau (n)$ denote the number of positive integer divisors of $n,$ including 1 and $n.$ For example, $\tau (1)=1$ and $\tau(6) =4.$ Define $S(n)$ by $S(n)=\tau(1)+ \tau(2) + \cdots + \tau(n).$ Let $a$ denote the number of positive integers $n \leq 2005$ with $S(n)$ odd, and let $b$ denote the number of positive integers $n \leq 2005$ with $S(n)$ even. Find $|a-b|.$

Solution

It is well-known that $\tau(n)$ is odd if and only if $n$ is a perfect square. (Otherwise, we can group divisors into pairs whose product is $n$.) Thus, $S(n)$ is odd if and only if there are an odd number of perfect squares less than $n$. So $S(1), S(2)$ and $S(3)$ are odd, while $S(4), S(5), \ldots, S(8)$ are even, and $S(9), \ldots, S(15)$ are odd, and so on.

So, for a given $n$, if we choose the positive integer $m$ such that $m^2 \leq n < (m + 1)^2$ we see that $S(n)$ has the same parity as $m$.

It follows that the numbers between $1^2$ and $2^2$, between $3^2$ and $4^2$, and so on, all the way up to the numbers between $43^2$ and $44^2 = 1936$ have $S(n)$ odd. These are the only such numbers less than $2005$ (because $45^2 = 2025 > 2005$).

Solution 1

Notice that the difference between consecutive squares are consecutively increasing odd numbers. Thus, there are $3$ numbers between $1$ (inclusive) and $4$ (exclusive), $5$ numbers between $4$ and $9$, and so on. The number of numbers from $n^2$ to $(n + 1)^2$ is $(n + 1 - n)(n + 1 + n) = 2n + 1$. Whenever the lowest square beneath a number is odd, the parity will be odd, and the same for even. Thus, $a = [2(1) + 1] + [2(3) + 1] \ldots [2(43) + 1] = 3 + 7 + 11 \ldots 87$. $b = [2(2) + 1] + [2(4) + 1] \ldots [2(42) + 1] + 70 = 5 + 9 \ldots 85 + 70$, the $70$ accounting for the difference between $2005$ and $44^2 = 1936$, inclusive. Notice that if we align the two and subtract, we get that each difference is equal to $2$. Thus, the solution is $|a - b| = |b - a| = |2 \cdot 21 + 70 - 87| = \boxed{025}$.

Solution 2

Similarly, $b = (3^2 - 2^2) + (5^2 - 4^2) + \ldots + (45^2 - 44^2) - 19$, where the $-19$ accounts for those numbers between $2005$ and $2024$.

Thus $a = (2^2 - 1^2) + (4^2 - 3^2) + \ldots + (44^2 - 43^2)$.

Then, $|a - b| = |2(2^2 + 4^2 + \ldots + 44^2) - 2(1^2 + 3^2 + 5^2 + \ldots 43^2) + 1^2 - 45^2 + 19|$. We can apply the formula $1^2 + 2^2 + \ldots + n^2 = \frac{n(n + 1)(2n + 1)}{6}$. From this formula, it follows that $2^2 + 4^2 + \ldots + (2n)^2 = \frac{2n(n + 1)(2n + 1)}{3}$ and so that

$1^2 + 3^2 + \ldots +(2n + 1)^2 = (1^2 + 2^2 + \ldots +(2n + 1)^2) - (2^2 + 4^2 + \ldots + (2n)^2)$
$= \frac{(2n + 1)(2n + 2)(4n + 3)}{6} - \frac{2n(n + 1)(2n + 1)}{3} = \frac{(n + 1)(2n + 1)(2n + 3)}{3}$. Thus,

$|a - b| = \left| 2\cdot \frac{44\cdot23\cdot45}{3} - 2\cdot \frac{22 \cdot 43 \cdot 45}{3} - 45^2 + 20\right| = |-25| =\boxed{025}$.

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
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2005_AIME_I_Problems/Problem_12&oldid=43072"