HMMT 二月 2016 · 代数 · 第 7 题
HMMT February 2016 — Algebra — Problem 7
题目详情
- Determine the smallest positive integer n ≥ 3 for which 10 n 170 A ≡ 2 (mod 2 ) 10 20 10 n where A denotes the result when the numbers 2 , 2 , . . . , 2 are written in decimal notation and concatenated (for example, if n = 2 we have A = 10241048576). !
解析
- Determine the smallest positive integer n ≥ 3 for which 10 n 170 A ≡ 2 (mod 2 ) 10 20 10 n where A denotes the result when the numbers 2 , 2 , . . . , 2 are written in decimal notation and concatenated (for example, if n = 2 we have A = 10241048576). Proposed by: Evan Chen Answer: 14 Note that 10 n n n 3 n 2 = 1024 = 1 . 024 × 10 . 10 n So 2 has roughly 3 n + 1 digits for relatively small n ’s. (Actually we have that for 0 < x < 1, 2 2 (1 + x ) = 1 + 2 x + x < 1 + 3 x. 2 2 2 2 2 16 Therefore, 1 . 024 < 1 . 03 < 1 . 09, 1 . 09 < 1 . 27, 1 . 27 < 1 . 81 < 2, and 2 = 4, so 1 . 024 < 4. Thus the conclusion holds for n ≤ 16.) For any positive integer n ≤ 16, n ∑ ∑ n 10 i (3 j +1) j = i +1 A = 2 × 10 . i =1 Let ∑ n 10 i (3 j +1) j = i +1 A = 2 × 10 i for 1 ≤ i ≤ n , then we know that n − 1 ∑ 10 n A − 2 = A i i =1 and ∑ ∑ n n 10 i + (3 j +1) (3 j +1) u v i i j = i +1 j = i +1 A = 2 × 5 = 2 × 5 i ∑ ∑ n n where u = 10 i + (3 j + 1), v = (3 j + 1). We have that i i j = i +1 j = i +1 u − u = 10 − (3 i + 1) = 3(3 − i ) . i i − 1 2 3 n +5 n +12 Thus, for 1 ≤ i ≤ n − 1, u is minimized when i = 1 or i = n − 1, with u = and i 1 2 u = 13 n − 9. When n = 5, n − 1 10 n 10 46 20 39 30 29 40 16 A − 2 = A + A + A + A = 2 × 10 + 2 × 10 + 2 × 10 + 2 × 10 1 2 3 4 57 170 is at most divisible by 2 instead of 2 . For all other n ’s, we have that u 6 = u , so we should 1 n − 1 have that both 170 ≤ u and 170 ≤ u . Therefore, since 170 ≤ u , we have that 14 ≤ n . We can 1 n − 1 n − 1 see that u > 170 and 14 < 16 in this case. Therefore, the minimum of n is 14 . 1 !