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 "2015 AIME I Problems/Problem 14"

m
(Added solution.)
Line 6: Line 6:
  
 
By considering the graph of this function, it is shown that the graph is composed of trapezoids ranging from <math>a^2</math> to <math>(a+1)^2</math> with the top made of diagonal line <math>y=ax</math>. The width of each trapezoid is <math>3, 5, 7</math>, etc. Whenever <math>a</math> is odd, the value of <math>A(n)</math> increases by an integer value, plus <math>\frac{1}{2}</math>. Whenever <math>a</math> is even, the value of <math>A(n)</math> increases by an integer value. Since each trapezoid always has an odd width, every value of <math>n</math> is not an integer when <math>a \pmod{4} \equiv 2</math>, and is an integer when <math>a \pmod{4} \equiv 0</math>. Every other value is an integer when <math>a</math> is odd. Therefore, it is simply a matter of determining the number of values of <math>n</math> where <math>a \pmod{4} \equiv 0</math> (<math>(5^2-4^2)+(9^2-8^2)+...+(29^2-28^2)</math>), and adding the number of values of <math>n</math> where <math>a</math> is odd (<math>\frac{(2^2-1^2)+(4^2-3^2)+...+(30^2-29^2)+(1000-31^2)}{2}</math>). Adding the two values gives <math>231+252=483</math>.
 
By considering the graph of this function, it is shown that the graph is composed of trapezoids ranging from <math>a^2</math> to <math>(a+1)^2</math> with the top made of diagonal line <math>y=ax</math>. The width of each trapezoid is <math>3, 5, 7</math>, etc. Whenever <math>a</math> is odd, the value of <math>A(n)</math> increases by an integer value, plus <math>\frac{1}{2}</math>. Whenever <math>a</math> is even, the value of <math>A(n)</math> increases by an integer value. Since each trapezoid always has an odd width, every value of <math>n</math> is not an integer when <math>a \pmod{4} \equiv 2</math>, and is an integer when <math>a \pmod{4} \equiv 0</math>. Every other value is an integer when <math>a</math> is odd. Therefore, it is simply a matter of determining the number of values of <math>n</math> where <math>a \pmod{4} \equiv 0</math> (<math>(5^2-4^2)+(9^2-8^2)+...+(29^2-28^2)</math>), and adding the number of values of <math>n</math> where <math>a</math> is odd (<math>\frac{(2^2-1^2)+(4^2-3^2)+...+(30^2-29^2)+(1000-31^2)}{2}</math>). Adding the two values gives <math>231+252=483</math>.
 +
 +
=="Step" Solution==
 +
First, draw a graph of the function. Note that it is just a bunch of line segments. Since we only need to know whether or not <math>A(n)</math> is an integer, we can take the area of each piece from some <math>x</math> to <math>x+1</math> (mod 1), aka the piece from <math>2</math> to <math>3</math> has area <math>\frac{1}{2} (\mod 1)</math>. There are some patterns. Every time we increase <math>n</math> starting with <math>2</math>, we either add <math>0 (\mod 1)</math> or <math>\frac{1}{2} (\mod 1)</math>. We look at <math>\floor \sqrt x</math> for inspiration. Every time this floor (which is really the slope) is odd, there is always an addition of <math>\frac{1}{2} (\mod 1)</math>, and whenever that slope is even, that addition is zero.
 +
 +
Take a few cases. For slope <math>=1</math>, we see that only one value satisfies. Because the last value, <math>n=4</math>, fails, and the numbers <math>n</math> which have a slope of an even number don't change this modulus, all these do not satisfy the criterion. The pattern then comes back to the odds, and this time <math>\floor \frac{7}{2} + 1 = 4</math> values work. Since the work/fail pattern alternates, all the <math>n</math>s with even slope, <math>[17, 25]</math>, satisfy the criterion. This pattern is cyclic over period 4 of slopes.
 +
 +
Even summation of working cases: <math>9+17+25+...+57 = 231</math>.
 +
Odd summation: <math>1+4+5+8+9+12...+29</math> and plus the <math>20</math> cases from <math>n=[962, 1000]</math>: <math>252</math>. Answer is <math>\boxed{483}</math>.
  
 
==See Also==
 
==See Also==
 
{{AIME box|year=2015|n=I|num-b=13|num-a=15}}
 
{{AIME box|year=2015|n=I|num-b=13|num-a=15}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 19:55, 22 February 2018

Problem

For each integer $n \ge 2$, let $A(n)$ be the area of the region in the coordinate plane defined by the inequalities $1\le x \le n$ and $0\le y \le x \left\lfloor \sqrt x \right\rfloor$, where $\left\lfloor \sqrt x \right\rfloor$ is the greatest integer not exceeding $\sqrt x$. Find the number of values of $n$ with $2\le n \le 1000$ for which $A(n)$ is an integer.

Solution

By considering the graph of this function, it is shown that the graph is composed of trapezoids ranging from $a^2$ to $(a+1)^2$ with the top made of diagonal line $y=ax$. The width of each trapezoid is $3, 5, 7$, etc. Whenever $a$ is odd, the value of $A(n)$ increases by an integer value, plus $\frac{1}{2}$. Whenever $a$ is even, the value of $A(n)$ increases by an integer value. Since each trapezoid always has an odd width, every value of $n$ is not an integer when $a \pmod{4} \equiv 2$, and is an integer when $a \pmod{4} \equiv 0$. Every other value is an integer when $a$ is odd. Therefore, it is simply a matter of determining the number of values of $n$ where $a \pmod{4} \equiv 0$ ($(5^2-4^2)+(9^2-8^2)+...+(29^2-28^2)$), and adding the number of values of $n$ where $a$ is odd ($\frac{(2^2-1^2)+(4^2-3^2)+...+(30^2-29^2)+(1000-31^2)}{2}$). Adding the two values gives $231+252=483$.

"Step" Solution

First, draw a graph of the function. Note that it is just a bunch of line segments. Since we only need to know whether or not $A(n)$ is an integer, we can take the area of each piece from some $x$ to $x+1$ (mod 1), aka the piece from $2$ to $3$ has area $\frac{1}{2} (\mod 1)$. There are some patterns. Every time we increase $n$ starting with $2$, we either add $0 (\mod 1)$ or $\frac{1}{2} (\mod 1)$. We look at $\floor \sqrt x$ (Error compiling LaTeX. Unknown error_msg) for inspiration. Every time this floor (which is really the slope) is odd, there is always an addition of $\frac{1}{2} (\mod 1)$, and whenever that slope is even, that addition is zero.

Take a few cases. For slope $=1$, we see that only one value satisfies. Because the last value, $n=4$, fails, and the numbers $n$ which have a slope of an even number don't change this modulus, all these do not satisfy the criterion. The pattern then comes back to the odds, and this time $\floor \frac{7}{2} + 1 = 4$ (Error compiling LaTeX. Unknown error_msg) values work. Since the work/fail pattern alternates, all the $n$s with even slope, $[17, 25]$, satisfy the criterion. This pattern is cyclic over period 4 of slopes.

Even summation of working cases: $9+17+25+...+57 = 231$. Odd summation: $1+4+5+8+9+12...+29$ and plus the $20$ cases from $n=[962, 1000]$: $252$. Answer is $\boxed{483}$.

See Also

2015 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 13 		Followed by
Problem 15
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2015_AIME_I_Problems/Problem_14&oldid=92248"