HMMT 二月 2026 · COMB 赛 · 第 5 题
HMMT February 2026 — COMB Round — Problem 5
题目详情
- Let S be the set of positive integer divisors of 10 . Compute the number of subsets T of S such that 9 • for every element s of S , exactly one of s and 10 /s is in T , and • for every element t of T , all positive integer divisors of t are in T .
解析
- Let S be the set of positive integer divisors of 10 . Compute the number of subsets T of S such that 9 • for every element s of S , exactly one of s and 10 /s is in T , and • for every element t of T , all positive integer divisors of t are in T . Proposed by: Jackson Dryg Answer: 252 Solution: Represent the elements of S as cells in a 10 × 10 grid, where each cell corresponds to the product of the number left of its row and the number below its column. The shaded squares represent one possible value for T . ©2026 HMMT 9 5 8 5 7 5 6 5 5 5 4 5 3 5 2 5 1 5 0 5 0 1 2 3 4 5 6 7 8 9 2 2 2 2 2 2 2 2 2 2 Number the columns 0 , 1 , . . . , 9 from left to right. The second condition implies that for every shaded cell, all cells to the left and/or below that cell are also shaded. This means that the shaded cells in a column must be the lowest k cells in that column, for some k . Let the height of a column be the number of shaded cells in it. Then the column heights must be nonincreasing from left to right. The first condition implies that for all 0 ≤ i ≤ 9 , column i and column 9 − i have heights that add up to 10 . This means that if we choose the heights of columns 0 through 4 , the heights of the other columns are uniquely determined. By the second condition, the height of column 4 must be at least the height of column 5 . By the first condition, the heights of these two columns add to 10 , so the height of column 4 is at least 5 . Let the heights of the first five columns be a, b, c, d, e from left to right; then we know that 10 ≥ a ≥ b ≥ c ≥ d ≥ e ≥ 5 . Any ( a, b, c, d, e ) satisfying this inequality gives rise to exactly one valid subset T . Hence the answer is the number of such ( a, b, c, d, e ) . This is the same as the number of 5-tuples ( a + 4 , b + 3 , c + 2 , d + 1 , e ) ( ) 10 with 14 ≥ a + 4 > b + 3 > c + 2 > d + 1 > e ≥ 5 , which is = 252 . 5