HMMT 二月 2024 · ALGNT 赛 · 第 2 题
HMMT February 2024 — ALGNT Round — Problem 2
题目详情
- Suppose a and b are positive integers. Isabella and Vidur both fill up an a × b table. Isabella fills it up with numbers 1 , 2 , . . . , ab , putting the numbers 1 , 2 , . . . , b in the first row, b + 1 , b + 2 , . . . , 2 b in the second row, and so on. Vidur fills it up like a multiplication table, putting ij in the cell in row i and column j . (Examples are shown for a 3 × 4 table below.) 1 2 3 4 1 2 3 4 5 6 7 8 2 4 6 8 9 10 11 12 3 6 9 12 Isabella’s Grid Vidur’s Grid Isabella sums up the numbers in her grid, and Vidur sums up the numbers in his grid; the difference between these two quantities is 1200 . Compute a + b .
解析
- Suppose a and b are positive integers. Isabella and Vidur both fill up an a × b table. Isabella fills it up with numbers 1 , 2 , . . . , ab , putting the numbers 1 , 2 , . . . , b in the first row, b + 1 , b + 2 , . . . , 2 b in the second row, and so on. Vidur fills it up like a multiplication table, putting ij in the cell in row i and column j . (Examples are shown for a 3 × 4 table below.) 1 2 3 4 1 2 3 4 5 6 7 8 2 4 6 8 9 10 11 12 3 6 9 12 Isabella’s Grid Vidur’s Grid Isabella sums up the numbers in her grid, and Vidur sums up the numbers in his grid; the difference between these two quantities is 1200. Compute a + b . Proposed by: Rishabh Das Answer: 21 n ( n +1) Solution: Using the formula 1 + 2 + · · · + n = , we get 2 ab ( ab + 1) a ( a + 1) b ( b + 1) ab (2( ab + 1) − ( a + 1)( b + 1)) − · = 2 2 2 4 ab ( ab − a − b + 1) = 4 ab ( a − 1)( b − 1) = 4 a ( a − 1) b ( b − 1) = · . 2 2 This means we can write the desired equation as a ( a − 1) · b ( b − 1) = 4800 . Assume b ≤ a , so we know b ( b − 1) ≤ a ( a − 1), so b ( b − 1) < 70. Thus, b ≤ 8. If b = 7 or b = 8, then b ( b − 1) has a factor of 7, which 4800 does not, so b ≤ 6. If b = 6 then b ( b − 1) = 30, so a ( a − 1) = 160, which can be seen to have no solutions. If b = 5 then b ( b − 1) = 20, so a ( a − 1) = 240, which has the solution a = 16, giving 5 + 16 = 21 . We need not continue since we are guaranteed only one solution, but we check the remaining cases 4800 for completeness. If b = 4 then a ( a − 1) = = 400, which has no solutions. If b = 3 then 12 4800 4800 a ( a − 1) = = 800 which has no solutions. Finally, if b = 2 then a ( a − 1) = = 2400, which has 6 2 no solutions. The factorization of the left side may come as a surprise; here’s a way to see it should factor without doing the algebra. If either a = 1 or b = 1, then the left side simplifies to 0. As a result, both a − 1 and b − 1 should be a factor of the left side.