Problem: If a,ba, ba,b and ccc are positive integers less than 101010, then (10a+b)(10a+c)(10a+b)(10a+c)(10a+b)(10a+c) equals 100a(a+1)+bc100a(a+1)+bc100a(a+1)+bc if
Answer Choices:
A. b+c=10b+c=10b+c=10
B. b=cb=cb=c
C. a+b=10a+b=10a+b=10
D. a=ba=ba=b
E. a+b+c=10a+b+c=10a+b+c=10 Solution:
100a2+10a(b+c)+bc=100a2+100a+bc100 a^{2}+10 a(b+c)+b c=100 a^{2}+100 a+b c100a2+10a(b+c)+bc=100a2+100a+bc. For equality we must have b+c=10b+c=10b+c=10.