Difference between revisions of "2020 AIME II Problems/Problem 12"
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==Solution== | ==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. | + | 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>. |
| − | 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. | + | 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>. |
| − | 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. | + | 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>. |
| − | 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. | + | 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>. |
| − | 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. | + | 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>. |
| − | Add all | + | Add all of our cases to get <cmath> 165+48+22+13 = \boxed{248} </cmath> |
| + | |||
| + | -Solution by thanosaops | ||
Revision as of 16:12, 7 June 2020
Let 
and 
be odd integers greater than 
An 
rectangle is made up of unit squares where the squares in the top row are numbered left to right with the integers 
through , those in the second row are numbered left to right with the integers 
through 
, and so on. Square 
is in the top row, and square 
is in the bottom row. Find the number of ordered pairs 
of odd integers greater than with the property that, in the rectangle, the line through the centers of squares and intersects the interior of square 
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
. Therefore, 
.
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 
.
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 
.
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 
.
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 
.
Add all of our cases to get ![\[165+48+22+13 = \boxed{248}\]](http://latex.artofproblemsolving.com/1/5/4/1549b814936cbc042d3fc0d4c16540c29a986541.png)
-Solution by thanosaops
