Problem:
For each integer n≥2, let Sn be the sum of all products jk, where j and k are integers and 1≤j<k≤n. What is the sum of the 10 least values of n such that Sn is divisible by 3?
Answer Choices:
A. 196
B. 197
C. 198
D. 199
E. 200
Solution:
To get from Sn to Sn+1, we add 1(n+1)+2(n+1)+⋯+n(n+1)=(1+2+⋯+n)(n+1)=2n(n+1)2.
Now, we can look at the different values of nmod3. For n≡0(mod3) and n≡2(mod3), then we have 2n(n+1)2≡0(mod3). However, for n≡1(mod3), we have
21⋅22≡2(mod3)
Clearly, S2≡2(mod3). Using the above result, we have S5≡1(mod3), and S8,S9, and S10 are all divisible by 3. After 3⋅3=9, we have S17,S18, and S19 all divisible by 3, as well as S26,S27,S28, and S35. Thus, our answer is 8+9+10+17+18+19+26+27+28+35=27+54+81+35=162+35=B. 197.
The problems on this page are the property of the MAA's American Mathematics Competitions