Problem: Given the sequence 10111​,10112​,10113​,…,1011n​, the smallest value of n such that the product of the first n members of this sequence exceeds 100,000 is:
Answer Choices:
A. 7
B. 8
C. 9
D. 10
E. 11
Solution:
Let P=101/11⋅102/11⋯10n/11=10s where s=111+2+…+n​=111​⋅21​n(n+1)
For P>100000,10s>105, that is, s>5
∴22n2+n​>5,n2+n−110>0,(n+11)(n−10)>0,n>10