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

m (→‎"Step" Solution)
m (→‎"Step" Solution)
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=="Step" Solution==
 
=="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.
+
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>\rfloor \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.
+
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>\rfloor \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>.
 
Even summation of working cases: <math>9+17+25+...+57 = 231</math>.

Revision as of 19:57, 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 $\rfloor \sqrt{x}$ 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 $\rfloor \frac{7}{2} + 1 = 4$ 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

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