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AIME 2020 II · 第 12 题

AIME 2020 II — Problem 12

专题
Contest Math
难度
L4
来源
AIME

题目详情

Problem

Let mm and nn be odd integers greater than 1.1. An m×nm\times n rectangle is made up of unit squares where the squares in the top row are numbered left to right with the integers 11 through nn, those in the second row are numbered left to right with the integers n+1n + 1 through 2n2n, and so on. Square 200200 is in the top row, and square 20002000 is in the bottom row. Find the number of ordered pairs (m,n)(m,n) of odd integers greater than 11 with the property that, in the m×nm\times n rectangle, the line through the centers of squares 200200 and 20002000 intersects the interior of square 10991099.

解析

Solution 1

Let us take some cases. Since mm and nn are odds, and 200200 is in the top row and 20002000 in the bottom, mm has to be 33, 55, 77, or 99. Also, taking a look at the diagram, the slope of the line connecting those centers has to have an absolute value of <1< 1. Therefore, m<1800modn<1800mm < 1800 \mod n < 1800-m.

If m=3m=3, nn can range from 667667 to 999999. However, 900900 divides 18001800, so looking at mods, we can easily eliminate 899899 and 901901. Now, counting these odd integers, we get 1672=165167 - 2 = 165.

Similarly, let m=5m=5. Then nn can range from 401401 to 499499. However, 4501800450|1800, so one can remove 449449 and 451451. Counting odd integers, we get 502=4850 - 2 = 48.

Take m=7m=7. Then, nn can range from 287287 to 333333. However, 3001800300|1800, so one can verify and eliminate 299299 and 301301. Counting odd integers, we get 242=2224 - 2 = 22.

Let m=9m = 9. Then nn can vary from 223223 to 249249. However, 2251800225|1800. Checking that value and the values around it, we can eliminate 225225. Counting odd integers, we get 141=1314 - 1 = 13.

Add all of our cases to get

165+48+22+13=248165+48+22+13 = \boxed{248} -Solution by thanosaops

Solution 2 (Official MAA)

Because square 20002000 is in the bottom row, it follows that 2000mn<2000m1\frac{2000}m \le n < \frac{2000}{m-1}. Moreover, because square 200200 is in the top row, and square 20002000 is in the bottom top row, 1<m101 < m \le 10. In particular, because the number of rows in the rectangle must be odd, mm must be one of 3,5,7,3, 5, 7, or 9.9.

For each possible choice of mm and nn, let m,n\ell_{m,n} denote the line through the centers of squares 200200 and 2000.2000. Note that for odd values of mm, the line m,n\ell_{m,n} passes through the center of square 1100.1100. Thus m,n\ell_{m,n} intersects the interior of cell 10991099 exactly when its slope is strictly between 1-1 and 11. The line m,n\ell_{m,n} is vertical whenever square 20002000 is the 200200th square in the bottom row of the rectangle. This would happen for m=3,5,7,9m = 3, 5, 7, 9 when n=900,450,300,225n = 900, 450, 300, 225, respectively. When nn is 1 greater than or 1 less than these numbers, the slope of m,n\ell_{m,n} is 11 or 1-1, respectively. In all other cases the slope is strictly between 1-1 and 1.1. The admissible values for nn for each possible value of mm are given in the following table.

mminimumnmaximumnavoidednnumberofoddn3667999899,900,9011655400499449,450,451487286333299,300,301229223249224,225,22613\begin{array}{|c|c|c|c|c|}\hline m & minimum n & maximum n & avoided n & number of odd n\\\hline 3&667&999&899, 900, 901&165\\\hline 5&400&499&449, 450, 451&48\\\hline 7&286&333&299, 300, 301&22\\\hline 9&223&249&224, 225, 226&13\\\hline \end{array} This accounts for 165+48+22+13=248165 + 48 + 22 + 13 = 248 rectangles.

Video Solution 1

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

Video Solution 2

https://youtu.be/v58SLOoAKTw

Video Solution 3

https://youtu.be/pu_79SSh3mM

~MathProblemSolvingSkills.com