2019 AIME I Problems/Problem 1

Problem 1

Consider the integer $N = 9 + 99 + 999 + 9999 + \cdots + \underbrace{99\ldots 99}_\text{321 digits}.$ Find the sum of the digits of $N$ .

Solution

Let's express the terms in terms of $10^n$ . We can obtain $(10-1)+(10^2-1)+(10^3-1)+\cdots+(10^{321}-1)$ . By the commutative and associative property, we can group it into $(10+10^2+10^3+\cdots+10^{321})-321$ . We know the former will yield $1111....10$ , so we only have to figure out what the last few digits are. There are currently $321$ 1's. We know the last 4 digits are 1110, and that the others will not be affected if we subtract $321$ . If we do so, we get that $1110-321=789$ . This method will remove 3 1's, and add a 7, 8 and 9. Therefore, the sum of the digits is $(321-3)+7+8+9=\boxed{342}$ .

2019 AIME I (Problems • Answer Key • Resources)
Preceded by First Problem	Followed by Problem 2
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2019 AIME I Problems/Problem 1

Problem 1

Solution