HMMT 二月 2009 · COMB 赛 · 第 6 题

HMMT February 2009 — COMB Round — Problem 6

[ 5 ] How many sequences of 5 positive integers ( a, b, c, d, e ) satisfy abcde ≤ a + b + c + d + e ≤ 10?

解析

[ 5 ] How many sequences of 5 positive integers ( a, b, c, d, e ) satisfy abcde ≤ a + b + c + d + e ≤ 10? Answer: 116 Solution: We count based on how many 1’s the sequence contains. If a = b = c = d = e = 1 then this gives us 1 possibility. If a = b = c = d = 1 and e 6 = 1, e can be 2 , 3 , 4 , 5 , 6. Each such sequence (1 , 1 , 1 , 1 , e ) can be arranged in 5 different ways, for a total of 5 · 5 = 25 ways in this case. If three of the numbers are 1, the last two can be (2 , 2), (3 , 3), (2 , 3), (2 , 4), or (2 , 5). Counting ordering, this gives a total of 2 · 10 + 3 · 20 = 80 possibilities. If two of the numbers are 1, the other three must be equal to 2 for the product to be under 10, and this yields 10 more possibilities. Thus there are 1 + 25 + 80 + 10 = 116 such sequences.