PUMaC 2023 · 数论（B 组） · 第 3 题

PUMaC 2023 — Number Theory (Division B) — Problem 3

解析

Find the integer x for which 135 + 138 = x − 1. Proposed by Sunay Joshi Answer: 172 We claim that x = 172. First, we show that x lies in the interval [138 , 184]. That the lower bound holds is clear. To see the upper bound, note that √ 3 3 3 3 3 x = 135 + 138 + 1 ≤ 2 · 138 → x ≤ 2 · 138 √ 3 3 3 Let 2 = 1 + h ; then (1 + h ) = 2. By Bernoulli’s Inequality, 1 + 3 h ≤ (1 + h ) , implying that √ 3 1 4 4 h ≤ and hence 2 ≤ . Our upper bound is therefore x ≤ · 138 = 184, as claimed. 3 3 3 3 3 3 Next, we consider the equation x = 135 + 138 + 1 modulo p for p = 5 , 11. We choose these 3 values of p because for p ̸ = 1 (mod 3), the cubing map x 7 → x is a bijection from Z /p Z to 3 3 3 itself. For p = 5, we compute x ≡ 0 + 3 + 1 ≡ 3 (mod 5), hence x ≡ 2 (mod 5). For p = 11, 3 3 3 we compute x ≡ 3 +6 +1 ≡ 2 (mod 11), hence x ≡ 7 (mod 11). By the Chinese Remainder Theorem, it follows that x ≡ 7 (mod 55), and it is easy to see that the unique number of the form 55 k + 7 in the interval [138 , 184] is 172. Remark: This identity was discovered by Ramanujan.