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Difference between revisions of "2015 AIME I Problems/Problem 14"

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==Solution==
 
==Solution==
  
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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 to determine 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 add 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>.
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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 add 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>.
  
 
==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 23:10, 6 March 2016

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 add 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 $231+252=483$.

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

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