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 number 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 $(321-3)+7+8+9=\boxed{342}$ .

-eric2020

A similar and simpler way to consider the initial manipulations is to observe that adding 1 to each term results in $(10+100+... 10^{320}+10^{321})$ . There are 321 terms, so it becomes $11...0$ , where there are 322 digits in $11...0$ . Then, subtract the 321 you initially added.

~ BJHHar

Solution 2

We can see that $9=9$ , $9+99=108$ , $9+99+999=1107$ , all the way to ten nines when we have $11111111100$ . Then, when we add more nines, we repeat the same process, and quickly get that the sum of digits is $\boxed{342}$ since we have to add $9\lfloor \log 321 \rfloor$ to the sum of digits, which is $9\lceil \frac{321}9 \rceil$ .

Solution 3

Observe how adding results in the last term but with a 1 concatenated in front and also a 1 subtracted (09, 108, 1107, 11106). Then for any index of terms, $n$ , the sum is $11...10-n$ , where the first term is of length $n+1$ . Here, that is $\boxed{342}$ .

~BJHHar

Video Solution

https://www.youtube.com/watch?v=JFHjpxoYLDk

2019 AIME I (Problems • Answer Key • Resources)
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