Complete Residue System modulo m (Fall 2021 AMC 12A #25)
Fall 2021 AMC 12A #25
Problem: Let m\geq 5 be an odd integer, and let D(m) denote the number of quadruples (a_1, a_2, a_3, a_4) of distinct integers with 1\leq a_i \leq m for all i such that m divides a_1+a_2+a_3+a_4. There is a polynomial q(x) = c_3x^3+c_2x^2+c_1x+c_0 such that D(m) = q(m) for all odd integers m\geq 5. What is c_1?
Solution: First, we must find a way to account for the number of ordered quadruples (a_1, a_2, a_3, a_4) such that m divides a_1+a_2+a_3+a_4. We know that there are (m)(m-1)(m-2)(m-3) total ordered quadruples since all elements must be distinct.
First we can list out the sums of m quadruples of the form (a_1+n, a_2+n, a_3+n, a_4+n) where n takes on the values from 0 to m-1 and all elements of the quadruple are modulo m. This means that the m quadruples have the same difference between consecutive elements modulo m. If we let the sum S = a_1+a_2+a_3+a_4, we have
- (a_1, a_2, a_3, a_4) \Rightarrow S
- (a_1+1, a_2+1, a_3+1, a_4+1) \Rightarrow S + 4*1
- (a_1+2, a_2+2, a_3+2, a_4+2) \Rightarrow S + 4*2
- ...
- (a_1+(m-2), a_2+(m-2), a_3+(m-2), a_4+(m-2)) \Rightarrow S + 4*(m-2)
- (a_1+(m-1), a_2+(m-1), a_3+(m-1), a_4+(m-1)) \Rightarrow S + 4*(m-1),
Note: An alternate, but much longer way to solve this would be to find D(m) for m = 5,7,9,11 and solve for c_3, c_2, c_1, c_0 with a system of equations, but there is a high likelihood of making a mistake.
Comments
Post a Comment