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Difference between revisions of "2020 AIME II Problems/Problem 12"

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This accounts for <math>165 + 48 + 22 + 13 = 248</math> rectangles.
 
This accounts for <math>165 + 48 + 22 + 13 = 248</math> rectangles.
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==Video Solution==
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https://www.youtube.com/watch?v=MrtKoO16XLQ ~ MathEx
  
 
==See Also==
 
==See Also==
 
{{AIME box|year=2020|n=II|num-b=11|num-a=13}}
 
{{AIME box|year=2020|n=II|num-b=11|num-a=13}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 22:46, 9 June 2020

Problem

Let $m$ and $n$ be odd integers greater than $1.$ An $m\times n$ rectangle is made up of unit squares where the squares in the top row are numbered left to right with the integers $1$ through $n$, those in the second row are numbered left to right with the integers $n + 1$ through $2n$, and so on. Square $200$ is in the top row, and square $2000$ is in the bottom row. Find the number of ordered pairs $(m,n)$ of odd integers greater than $1$ with the property that, in the $m\times n$ rectangle, the line through the centers of squares $200$ and $2000$ intersects the interior of square $1099$.

Solution

Let us take some cases. Since $m$ and $n$ are odds, and $200$ is in the top row and $2000$ in the bottom, $m$ has to be $3$, $5$, $7$, or $9$. Also, taking a look at the diagram, the slope of the line connecting those centers has to have an absolute value of $< 1$. Therefore, $m < 1800 \mod n < 1800-m$.

If $m=3$, $n$ can range from $667$ to $999$. However, $900$ divides $1800$, so looking at mods, we can easily eliminate $899$ and $901$. Now, counting these odd integers, we get $167 - 2 = 165$.

Similarly, let $m=5$. Then $n$ can range from $401$ to $499$. However, $450|1800$, so one can remove $449$ and $451$. Counting odd integers, we get $50 - 2 = 48$.

Take $m=7$. Then, $n$ can range from $287$ to $333$. However, $300|1800$, so one can verify and eliminate $299$ and $301$. Counting odd integers, we get $24 - 2 = 22$.

Let $m = 9$. Then $n$ can vary from $223$ to $249$. However, $225|1800$. Checking that value and the values around it, we can eliminate $225$. Counting odd integers, we get $14 - 1 = 13$.

Add all of our cases to get \[165+48+22+13 = \boxed{248}\]

-Solution by thanosaops

Solution 2 (Official MAA)

Because square $2000$ is in the bottom row, it follows that $\frac{2000}m \le n < \frac{2000}{m-1}$. Moreover, because square $200$ is in the top row, and square $2000$ is not in the top row, $1 < m \le 10$. In particular, because the number of rows in the rectangle must be odd, $m$ must be one of $3, 5, 7,$ or $9.$

For each possible choice of $m$ and $n$, let $\ell_{m,n}$ denote the line through the centers of squares $200$ and $2000.$ Note that for odd values of $m$, the line $\ell_{m,n}$ passes through the center of square $1100.$ Thus $\ell_{m,n}$ intersects the interior of cell $1099$ exactly when its slope is strictly between $-1$ and $1$. The line $\ell_{m,n}$ is vertical whenever square $2000$ is the $200$th square in the bottom row of the rectangle. This would happen for $m = 3, 5, 7, 9$ when $n = 900, 450, 300, 225$, respectively. When $n$ is 1 greater than or 1 less than these numbers, the slope of $\ell_{m,n}$ is $1$ or $-1$, respectively. In all other cases the slope is strictly between $-1$ and $1.$ The admissible values for $n$ for each possible value of $m$ are given in the following table. $\begin{tabular}{|c|c|c|c|c|} \hline m & minimum n & maximum n & avoided n & number of odd n\\ \hline 3 & 667 & 999 &$ (Error compiling LaTeX. Unknown error_msg)899, 900, 901$& 165\\ 5 & 400 & 499 &$ (Error compiling LaTeX. Unknown error_msg)449, 450, 451$& 48\\ 7 & 286 & 333 &$ (Error compiling LaTeX. Unknown error_msg)299, 300, 301$& 22\\ 9 & 223 & 249 &$ (Error compiling LaTeX. Unknown error_msg)224, 225, 226$& 13\\ \hline \end{tabular}$ (Error compiling LaTeX. Unknown error_msg) This accounts for $165 + 48 + 22 + 13 = 248$ rectangles.

Video Solution

https://www.youtube.com/watch?v=MrtKoO16XLQ ~ MathEx

See Also

2020 AIME II (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

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