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

Revision as of 22:17, 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 know the difference between two perfect squares are consecutive odd numbers. Since we want all $(n+1)^2-n^2$ where $n$ is odd in order to find $a$, we want all $2n+1$ for odd numbers from $1$ to $43$. We can easily evaluate this sum, which is $990$. Obviously $a+b=2005$, so $b=1015$ and $|a - b|=\boxed{025}$.

Alternatively 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=43069"