HMMT 二月 2016 · 代数 · 第 7 题

英文解析

7. Determine the smallest positive integer n ≥ 3 for which
   10 n 170
   A ≡ 2 (mod 2 )
   10 20 10 nwhere A denotes the result when the numbers 2 , 2 , . . . , 2 are written in decimal notation andconcatenated (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. Thusthe conclusion holds for n ≤ 16.)
   For any positive integer n ≤ 16,
   ∑ ∑nn
   10 i (3 j +1)
   j = i +1
   A = 2 × 10 .
   i =1
   ∑Letn
   10 i (3 j +1)
   j = i +1
   A = 2 × 10
   for 1 ≤ i ≤ n , then we know thatin − 1
   ∑
   10 n
   A − 2 = Aii =1
   ∑ ∑andn n
   10 i + (3 j +1) (3 j +1) u vi ij = i +1 j = i +1
   A = 2 × 5 = 2 × 5
   ∑ ∑in nwhere u = 10 i + (3 j + 1), v = (3 j + 1). We have thati ij = i +1 j = i +1
   u − u = 10 − (3 i + 1) = 3(3 − i ) .
   i i − 1
   3 n +5 n +122
   Thus, for 1 ≤ i ≤ n − 1, u is minimized when i = 1 or i = n − 1, with u = andi 1
   u = 13 n − 9. When n = 5,2
   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 .
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