HMMT 二月 2004 · GEN1 赛 · 第 8 题

HMMT February 2004 — GEN1 Round — Problem 8

You have a 10 × 10 grid of squares. You write a number in each square as follows: you write 1 , 2 , 3 , . . . , 10 from left to right across the top row, then 11 , 12 , . . . , 20 across the second row, and so on, ending with a 100 in the bottom right square. You then write a second number in each square, writing 1 , 2 , . . . , 10 in the first column (from top to bottom), then 11 , 12 , . . . , 20 in the second column, and so forth. When this process is finished, how many squares will have the property that their two numbers sum to 101?

解析

You have a 10 × 10 grid of squares. You write a number in each square as follows: you write 1 , 2 , 3 , . . . , 10 from left to right across the top row, then 11 , 12 , . . . , 20 across the second row, and so on, ending with a 100 in the bottom right square. You then write a second number in each square, writing 1 , 2 , . . . , 10 in the first column (from top to bottom), then 11 , 12 , . . . , 20 in the second column, and so forth. When this process is finished, how many squares will have the property that their two numbers sum to 101? Solution: 10 The number in the i th row, j th column will receive the numbers 10( i − 1) + j and 10( j − 1) + i , so the question is how many pairs ( i, j ) (1 ≤ i, j ≤ 10) will have 101 = [10( i − 1) + j ] + [10( j − 1) + i ] ⇔ 121 = 11 i + 11 j = 11( i + j ) . Now it is clear that this is achieved by the ten pairs (1 , 10) , (2 , 9) , (3 , 8) , . . . , (10 , 1) and no others.