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

Latest revision as of 18:58, 5 August 2023

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 1

Let $n\ge 2$ and define $a(n) = \left\lfloor \sqrt n \right\rfloor$ . For $2\le n \le 1000$ , we have $1\le a(n)\le 31$ .

For $a^2 \le x < (a+1)^2$ we have $y=ax$ . Thus $A(n+1)-A(n)=a(n+\tfrac 12) = \Delta_n$ (say), and $\Delta_n$ is an integer if $a$ is even; otherwise $\Delta_n$ is an integer plus $\tfrac 12$ .

If $a=1$ , $n\in \{1,2,3\}$ and $\Delta_n$ is of the form $k(n)+\tfrac 12$ so $A(n)$ is an integer when $n$ is even.

If $a=2$ , $n\in\{4,\ldots , 8\}$ and $\Delta_n$ is an integer for all $n$ . Since $A(3)$ is not an integer, so $A(n)$ is not an integer for any $n$ .

If $a=3$ , $n\in\{9,\ldots , 15\}$ and $\Delta_n$ is of the form $k(n)+\tfrac 12$ . Since $A(8)$ is of the form $k+\tfrac 12$ so $A(n)$ is an integer only when $n$ is odd.

If $a=4$ , $n\in\{16,\ldots , 24\}$ and $\Delta_n$ is an integer for all $n$ . Since $A(15)$ is an integer so $A(n)$ is an integer for all $n$ .

Now we are back to where we started; i.e., the case $a=5$ will be the same as $a=1$ and so on. Thus, $\begin{align} a(n)\equiv 1\pmod 4 \qquad &\Longrightarrow \qquad A(n) \in \mathbb{Z} \textrm{ for even } n, \\ a(n)\equiv 2\pmod 4 \qquad &\Longrightarrow \qquad A(n) \not\in \mathbb{Z} \textrm{ for any } n, \\ a(n)\equiv 3\pmod 4 \qquad &\Longrightarrow \qquad A(n) \in \mathbb{Z} \textrm{ for odd } n, \\ a(n)\equiv 0\pmod 4 \qquad &\Longrightarrow \qquad A(n) \in \mathbb{Z} \textrm{ for all } n. \end{align}$

For each $a$ there are $2a+1$ corresponding values of $n$ : i.e., $n\in \{a^2, \ldots , (a+1)^2-1\}$ .

Thus, the number of values of $n$ corresponding to $(4)$ (i.e., $a(n)\equiv 0\pmod 4$ ) is given by $\sum_{\substack{a=4k \\ a\le 31}}(2a+1) = \sum_{k=1}^7 (8k+1)=231.$

The cases $(1)$ and $(3)$ combine to account for half the values of $n$ corresponding to odd values of $a(n)$ ; i.e., $\frac 12 \cdot \sum_{\substack{a=2k+1 \\ a\le 31}} (2a+1) = \sum_{k=0}^{15} (2k+\tfrac 32) = 264$ However, this also includes the odd integers in $\{1001, \ldots , 1023\}$ . Subtracting $12$ to account for these, we get the number of values of $n$ corresponding to cases $(1)$ and $(3)$ to be $264-12=252$ .

Adding the contributions from all cases we get our answer to be $231+252= \boxed{483}$ .

Solution 2

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 $231+252=\boxed{483}$ .

"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 $\lfloor \sqrt{x} \rfloor$ 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 $\lfloor \frac{7}{2} \rfloor + 1 = 4$ 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 $\boxed{483}$ .

Video Solution

https://youtu.be/tup11-90Bqw

~MathProblemSolvingSkills.com

@@ Line 3: / Line 3: @@
 For each integer <math>n \ge 2</math>, let <math>A(n)</math> be the area of the region in the coordinate plane defined by the inequalities <math>1\le x \le n</math> and <math>0\le y \le x \left\lfloor \sqrt x \right\rfloor</math>, where <math>\left\lfloor \sqrt x \right\rfloor</math> is the greatest integer not exceeding <math>\sqrt x</math>. Find the number of values of <math>n</math> with <math>2\le n \le 1000</math> for which <math>A(n)</math> is an integer.
-==Solution==
+==Solution 1==
+Let <math>n\ge 2</math> and define <math>a(n) = \left\lfloor \sqrt n \right\rfloor</math>. For <math>2\le n \le 1000</math>, we have <math>1\le a(n)\le 31</math>.
+For <math>a^2 \le x < (a+1)^2</math> we have <math>y=ax</math>. Thus <math>A(n+1)-A(n)=a(n+\tfrac 12) = \Delta_n</math> (say), and <math>\Delta_n</math> is an integer if <math>a</math> is even; otherwise <math>\Delta_n</math> is an integer plus <math>\tfrac 12</math>.
+If <math>a=1</math>, <math>n\in \{1,2,3\}</math> and <math>\Delta_n</math> is of the form <math>k(n)+\tfrac 12</math> so <math>A(n)</math> is an integer when <math>n</math> is even.
+If <math>a=2</math>, <math>n\in\{4,\ldots , 8\}</math> and <math>\Delta_n</math> is an integer for all <math>n</math>. Since <math>A(3)</math> is not an integer, so <math>A(n)</math> is not an integer for any <math>n</math>.
+If <math>a=3</math>, <math>n\in\{9,\ldots , 15\}</math> and <math>\Delta_n</math> is of the form <math>k(n)+\tfrac 12</math>. Since <math>A(8)</math> is of the form <math>k+\tfrac 12</math> so <math>A(n)</math> is an integer only when <math>n</math> is odd.
+If <math>a=4</math>, <math>n\in\{16,\ldots , 24\}</math> and <math>\Delta_n</math> is an integer for all <math>n</math>. Since <math>A(15)</math> is an integer so <math>A(n)</math> is an integer for all <math>n</math>.
+Now we are back to where we started; i.e., the case <math>a=5</math> will be the same as <math>a=1</math> and so on. Thus,
+<cmath>
+\begin{align}
+a(n)\equiv 1\pmod 4 \qquad &\Longrightarrow \qquad A(n) \in \mathbb{Z} \textrm{ for even } n, \\
+a(n)\equiv 2\pmod 4 \qquad &\Longrightarrow \qquad A(n) \not\in \mathbb{Z} \textrm{  for any } n, \\
+a(n)\equiv 3\pmod 4 \qquad &\Longrightarrow \qquad A(n) \in \mathbb{Z} \textrm{ for odd  } n, \\
+a(n)\equiv 0\pmod 4 \qquad &\Longrightarrow \qquad A(n) \in \mathbb{Z} \textrm{ for all } n.
+\end{align}
+</cmath>
+For each <math>a</math> there are <math>2a+1</math> corresponding values of <math>n</math>: i.e., <math>n\in \{a^2, \ldots , (a+1)^2-1\}</math>.
+Thus, the number of values of <math>n</math> corresponding to <math>(4)</math> (i.e., <math>a(n)\equiv 0\pmod 4</math>) is given by <cmath>\sum_{\substack{a=4k \\ a\le 31}}(2a+1) = \sum_{k=1}^7 (8k+1)=231.</cmath>
+The cases <math>(1)</math> and <math>(3)</math> combine to account for half the values of <math>n</math> corresponding to odd values of <math>a(n)</math>; i.e.,
+<cmath>\frac 12 \cdot \sum_{\substack{a=2k+1 \\ a\le 31}} (2a+1) = \sum_{k=0}^{15} (2k+\tfrac 32) = 264</cmath>However, this also includes the odd integers in <math>\{1001, \ldots , 1023\}</math>. Subtracting <math>12</math> to account for these, we get the  number of values of <math>n</math> corresponding to cases <math>(1)</math> and <math>(3)</math> to be <math>264-12=252</math>.
+Adding the contributions from all cases we get our answer to be <math>231+252= \boxed{483}</math>.
+==Solution 2==
 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=\boxed{483}</math>.
@@ Line 14: / Line 47: @@
 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>.
+==Video Solution==
+https://youtu.be/tup11-90Bqw
+~MathProblemSolvingSkills.com
 ==See Also==
 {{AIME box|year=2015|n=I|num-b=13|num-a=15}}
+[[Category:Intermediate Geometry Problems]]
 {{MAA Notice}}

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