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

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-Solution by thanosaops
 
-Solution by thanosaops
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==See Also==
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{{AIME box|year=2020|n=II|num-b=11|num-a=13}}
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{{MAA Notice}}

Revision as of 23:02, 7 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 is 3, n can be 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 be 5. Then n can range from 401 to 499. However, 450 divides 1800, so one can remove 449 and 451. Counting odd integers, we get $50 - 2 = 48$.

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

Let m be 9. Then n can vary from 223 to 249. However, 225 divides 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

See Also

2020 AIME II (ProblemsAnswer KeyResources)
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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