Problem:
Find the number of positive integers m for which there exist nonnegative integers x0,x1,…,x2011, such that
mx0=k=1∑2011mxk
Solution:
The value m=1 does not satisfy the given equation, and thus m≥2. Subtracting 2011 from both sides produces the equation
k=1∑2011(mxk−1)=mx0−2011=mx0−1−2010
Because m−1 divides the left side, it must divide the right side, and therefore it divides 2010. Thus if m satisfies the original equation, then m−1 must be a factor of 2010. Conversely, suppose that 2010=(m−1)⋅n for some positive integer n. Let x0=n,x1=x2=⋯=xm=0, and divide the remaining (m−1)(n−1) numbers into n−1 blocks of length m−1. Let all xi 's in the r th block equal r, where 1≤r≤n−1. Then
k=1∑2011mxk=m+(m−1)m+(m−1)m2+⋯+(m−1)mn−1=mn=mx0.
Thus the given equation has a solution exactly when m−1 divides 2010. Because 2010=2⋅3⋅5⋅67, there are 16 positive integer factors of 2010 and hence 16 values of m for which the equation has solutions.
The problems on this page are the property of the MAA's American Mathematics Competitions