HMMT 十一月 2017 · 团队赛 · 第 2 题
HMMT November 2017 — Team Round — Problem 2
题目详情
- [ 20 ] How many sequences of integers ( a , . . . , a ) are there for which − 1 ≤ a ≤ 1 for every i , and 1 7 i a a + a a + a a + a a + a a + a a = 4? 1 2 2 3 3 4 4 5 5 6 6 7
解析
- [ 20 ] How many sequences of integers ( a , . . . , a ) are there for which − 1 ≤ a ≤ 1 for every i , and 1 7 i a a + a a + a a + a a + a a + a a = 4? 1 2 2 3 3 4 4 5 5 6 6 7 Proposed by: mendel keller Answer: 38 For i = 1 , 2 , . . . , 6, let b = a a . From the problem condition each of b , b , . . . , b can only be − 1 , 0 , i i i +1 1 2 6 or 1. Since the sum of these six numbers is 4, either there are five 1s and a − 1 or there are four 1s and two 0s. In the first case, there are 6 ways to choose i such that b = − 1. Once that is fixed, determining the i value of a (one of 1 and − 1) will determine the value of all the remaining a ’s, so there are 6 · 2 = 12 1 i possible ways in this case. In the second case, since if one of b , b , b , b is zero, then one of the adjacent term to this zero term 2 3 4 5 must also be zero. Therefore the two zeroes must be next to each other or be b and b . 1 6 If b = b = 0, then a must be zero. a ’s value doesn’t matter, and a , a , . . . , a must have the same 1 2 2 1 3 4 7 sign. The same goes for b = b = 0, giving 3 · 2 · 2 = 12 possibilities in these two cases. 5 6 If b = b = 0 for i = 2 , 3 , 4, then a must be zero. Moreover, a , a , . . . , a must have the same i i +1 i +1 1 2 i sign, and so do a , . . . , a . this gives 2 · 2 · 3 = 12 possibilities in these three cases. i +2 7 If b = b = 0, then a = a = 0. Also, a , a , . . . , a must have the same sign so there are 2 possibilities. 1 6 1 7 2 3 6 Combining these cases gives 12 + 12 + 12 + 2 = 38 possible sequences in total.