HMMT 十一月 2012 · GEN 赛 · 第 7 题
HMMT November 2012 — GEN Round — Problem 7
题目详情
英文原题
[6] Find the number of ordered 2012-tuples of integers ( x1,x2,... ,x 2012), with each integer between
0 and 2011 inclusive, such that the sum x1+ 2x2+ 3x3+· · ·+ 2012 x2012 is divisible by 2012.
解析
英文解析
- [ 6 ] Find the number of ordered 2012-tuples of integers ( x , x , . . . , x ), with each integer between
1 2 2012
0 and 2011 inclusive, such that the sum x + 2 x + 3 x + · · · + 2012 x is divisible by 2012.
1 2 3 2012
2011
Answer: 2012 We claim that for any choice of x , x , ..., x , there is exactly one possible
2 3 2012
value of x satisfying the condition. We have x + 2 x + ... + 2012 x ≡ 0 (mod 2012) or x ≡
1 1 2 2012 1 − (2 x + ... + 2012 x ) (mod 2012). Indeed, we see that the right hand side is always an integer
2 2012
between 0 and 2011, so x must equal this number.
20111
Now, there are 2012 choices for each of the 2011 variables x , ..., x , and each of the 2012 possible
2 2012
2011
combinations gives exactly one valid solution, so the total number of 2012-tuples is 2012 . π