Problem: Given the progression 10111​,10112​,10113​,10114​,⋯,1011n​. The least positive integer n such that the product of the first n terms of the progression exceeds 100,000 is
Answer Choices:
A. 7
B. 8
C. 9
D. 10
E. 11
Solution:
Since 100,000=105, the sum of the first n exponents must exceed 5. i.e. n(n+ 1) /22>5∴n(n+1)>110∴n=11.