HMMT 十一月 2021 · 冲刺赛 · 第 19 题

HMMT November 2021 — Guts Round — Problem 19

[11] Integers 0 ≤ a, b, c, d ≤ 9 satisfy 6 a + 9 b + 3 c + d = 88 a − b + c − d = − 6 a − 9 b + 3 c − d = − 46 Find 1000 a + 100 b + 10 c + d .

解析

[11] Integers 0 ≤ a, b, c, d ≤ 9 satisfy 6 a + 9 b + 3 c + d = 88 a − b + c − d = − 6 a − 9 b + 3 c − d = − 46 Find 1000 a + 100 b + 10 c + d . Proposed by: Akash Das Answer: 6507 Solution: Let N = abcd be an at most-four digit number. Note that the first equation gives us N ≡ 4 (mod 7). The second equation gives us N ≡ 6 (mod 11). The third equation gives us N ≡ 7 1 (mod 13). Using CRT, we get N ≡ ≡ 501 (mod 1001). Thus, we have N = 501 + 1001 k for some 2 integer 0 ≤ k ≤ 9. The only value of k that satisfies the first equation is k = 6, which yields N = 6507.