Problem:
Consider the integer
N=9+99+999+9999+⋯+321 digits 99…99
Find the sum of the digits of N.
Solution:
Write
N=(10−1)+(102−1)+⋯+(10321−1)=10+102+103+104+105+106+⋯+10321−321=1110−321+104+105+106+⋯+10321=789+104+105+106+⋯+10321
The sum of the digits of N is therefore equal to 7+8+9+(321−3)=342.
The problems on this page are the property of the MAA's American Mathematics Competitions