Problem:
Set A consists of m consecutive integers whose sum is 2m, and set B consists of 2m consecutive integers whose sum is m. The absolute value of the difference between the greatest element of A and the greatest element of B is 99. Find m.
Solution:
Let the smallest elements of A and B be (n+1) and (k+1), respectively. Then
2mm=(n+1)+(n+2)+⋯+(n+m)=mn+21⋅m(m+1), and =(k+1)+(k+2)+⋯+(k+2m)=2km+21⋅2m(2m+1)
The second equation implies that k+m=0. Substitute this into ∣k+2m− (n+m)∣=99 to obtain n=±99. Now simplify the first equation to obtain 2=n+(m+1)/2, and substitute n=±99. This yields m=−195 or m=201. Because m>0,m=201.
OR
The mean of the elements in A is 2, and the mean of the elements in B is 1/2. Because the mean of each of these sets equals its median, and the median of A is an integer, m is odd. Thus A={2−2m−1,…,2,…,2+2m−1}, and B={−m+1,…,0,1,…,m}. Therefore ∣∣∣2+2m−1−m∣∣∣=99, which yields ∣∣∣23−m∣∣∣=99, so ∣3−m∣=198. Because m>0,m=201.
The problems on this page are the property of the MAA's American Mathematics Competitions