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

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==Solution==
 
==Solution==
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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<math> < 1</math>. Therefore, <math>m < 1800 mod n < 1800-m</math>.  
+
Let us take some cases. Since <math>m</math> and <math>n</math> are odds, and <math>200</math> is in the top row and <math>2000</math> in the bottom, <math>m</math> has to be <math>3</math>, <math>5</math>, <math>7</math>, or <math>9</math>. Also, taking a look at the diagram, the slope of the line connecting those centers has to have an absolute value of <math> < 1</math>. Therefore, <math>m < 1800 \mod n < 1800-m</math>.  
  
−
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 <math>167 - 2 = 165</math>.
+
If <math>m=3</math>, <math>n</math> can range from <math>667</math> to <math>999</math>. However, <math>900</math> divides <math>1800</math>, so looking at mods, we can easily eliminate <math>899</math> and <math>901</math>. Now, counting these odd integers, we get <math>167 - 2 = 165</math>.
  
−
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 <math>50 - 2 = 48</math>.
+
Similarly, let <math>m=5</math>. Then <math>n</math> can range from <math>401</math> to <math>499</math>. However, <math>450|1800</math>, so one can remove <math>449</math> and <math>451</math>. Counting odd integers, we get <math>50 - 2 = 48</math>.
  
−
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 <math>24 - 2 = 22</math>.
+
Take <math>m=7</math>. Then, <math>n</math> can range from <math>287</math> to <math>333</math>. However, <math>300|1800</math>, so one can verify and eliminate <math>299</math> and <math>301</math>. Counting odd integers, we get <math>24 - 2 = 22</math>.
  
−
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 <math>14 - 1 = 13</math>.
+
Let <math>m = 9</math>. Then <math>n can vary from </math>223<math> to </math>249<math>. However, </math>225|1800<math>. Checking that value and the values around it, we can eliminate </math>225<math>. Counting odd integers, we get </math>14 - 1 = 13$.
  
 
Add all of our cases to get <cmath> 165+48+22+13 = \boxed{248} </cmath>
 
Add all of our cases to get <cmath> 165+48+22+13 = \boxed{248} </cmath>
  
 
-Solution by thanosaops
 
-Solution by thanosaops
 +
 
==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 02:58, 8 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

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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