Problem:
Consider the set of numbers {1,10,102,103,…,1010}. The ratio of the largest element of the set to the sum of the other ten elements of the set is closest to which integer?
Answer Choices:
A. 1
B. 9
C. 10
D. 11
E. 101
Solution:
The sum of the smallest ten elements is
1+10+100+⋯+1,000,000,000=1,111,111,111.
Hence the desired ratio is
1,111,111,11110,000,000,000=1,111,111,1119,999,999,999+1=9+1,111,111,1111≈(B)9
OR
The sum of a finite geometric series of the form a(1+r+r2+⋯+rn) is 1−ra(1−rn+1). The desired denominator 1+10+102+⋯+109 is a finite geometric series with a=1,r=10, and n=9. Therefore the ratio is
1+10+102+⋯+1091010=1−101(1−1010)1010=1010−11010⋅9≈10101010⋅9=(B)9
The problems on this page are the property of the MAA's American Mathematics Competitions