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

The above sequence is equivalent to $(10^1-1)+(10^2-1)+(10^3-1)+\cdots+(10^(321)-1)$ . We can take out the ones so that this becomes $(10^1)+(10^2)+(10^3)+\cdots+(10^321)-321$ . Now, notice that $(10^1)+(10^2)+(10^3)+\cdots+(10^(321))$ is equal to $111...11110$ , with 321 ones and one zero as digits. Subtract 321 to obtain $111...10790$ , with 318 ones, 2 zeros, a 7 and a 9 as digits. This sums to $\boxed{334}$ .

2019 AIME I (Problems • Answer Key • Resources)
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2019 AIME I Problems/Problem 1

Problem 1

Solution