2019 AIME I Problem 1

Question

Consider the integer

\begin{aligned} &N = 9 + 99 + 999 + 9999 \\ &\quad {}+ \cdots + \underbrace{99\ldots99}_{\text{321 digits}}. \end{aligned}

Find the sum of the digits of

N.

Answer

Each summand is

10^k - 1,

so

N

is a long string of ones (followed by a zero) minus

321

Subtracting

321

from

11\ldots110

disturbs only the last four digits; count the ones that survive

Each summand is

10^k - 1,

so

\begin{aligned} N &= \sum_{k=1}^{321} \left(10^k - 1\right) \\ &= \underbrace{11\ldots1}_{321}0 - 321. \end{aligned}

The subtraction changes only the last four digits:

1110 - 321 = 789,

so those four digits become

0789.

Thus

N

consists of

318

ones followed by

0789,

and the digit sum is

318 + 0 + 7 + 8 + 9 = 342.

Problem 1 in other years