Suppose $N=1+11+101+1001+10001+\ldots+1 \overbrace{000 \ldots 0000}^{50 \text { zeroes }}$.

When $N$ is calculated and written as a single integer, the sum of its digits is
(A) 58
(B) 99
(C) 55
(D) 50
(E) 103
There are 52 terms in the sum: the number 1 , the number 11 , and the 50 numbers starting with a 1 , ending with a 1 and with 1 to 50 zeroes in between. The longest of these terms thus has 52 digits ( 50 zeroes and 2 ones).

When the units digits of all 52 terms are added up, their sum is 52 , so the units digit of $N$ is 2 , and a 5 carried to the tens digit.

In the tens digit, there is only 1 non-zero digit: the 1 in the number 11 . Therefore, using the carry, the tens digit of $N$ is $1+5=6$.

In each of positions 3 to 52 from the right-hand end, there is only one non-zero digit, which is a 1 .

Therefore, the digit in each of these positions in $N$ is also a 1. (There is no carrying to worry about.)

Therefore, $N=11 \cdots 1162$, where $N$ has $52-2=50$ digits equal to 1 .

This tells us that the sum of the digits of $N$ is $50(1)+6+2=58$.

ANSwer: (A)