HMMT 二月 2006 · GEN1 赛 · 第 4 题
HMMT February 2006 — GEN1 Round — Problem 4
题目详情
英文原题
- Find
√
√
√
31 + 31 + 31 + . . .
√
.
√
√
1 + 1 + 1 + . . .
解析
英文解析
- Find
√
√
√
31 + 31 + 31 + . . .
√
.
√
√
1 + 1 + 1 + . . .
√
Answer: 6 − 5
Solution: Let the numerator be x and the denominator y . Then x = 31 + x , so, as 2
x > 0, we have
√
√
1 + 1 + 4 · 31 1 + 5 5
x = = .
2 2
Similarly we compute that
√
√
1 + 1 + 4 · 1 1 + 5
y = = ,
2 2
so that
√ √ √ √
√
x 1 + 5 5 1 + 5 5 1 − 5 − 24 + 4 5
√ √ √
= = · = = 6 − 5 .
y − 4
1 + 5 1 + 5 1 − 5 1