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AIME 2019 I · 第 1 题

AIME 2019 I — Problem 1

专题
Contest Math
难度
L4
来源
AIME

题目详情

Problem

Consider the integer

N=9+99+999+9999++9999321 digits.N = 9 + 99 + 999 + 9999 + \cdots + \underbrace{99\ldots 99}_\text{321 digits}. Find the sum of the digits of NN.

解析

Solution 1

Let's express the number in terms of 10n10^n. We can obtain (101)+(1021)+(1031)++(103211)(10-1)+(10^2-1)+(10^3-1)+\cdots+(10^{321}-1). By the commutative and associative property, we can group it into (10+102+103++10321)321(10+10^2+10^3+\cdots+10^{321})-321. We know the former will yield 1111....101111....10, so we only have to figure out what the last few digits are. There are currently 321321 1's. We know the last four digits are 11101110, and that the others will not be affected if we subtract 321321. If we do so, we get that 1110321=7891110-321=789. This method will remove three 11's, and add a 77, 88 and 99. Therefore, the sum of the digits is (3213)+7+8+9=342(321-3)+7+8+9=\boxed{342}.

-eric2020 -another Eric in 2020

A similar and simpler way to consider the initial manipulations is to observe that adding 11 to each term results in (10+100+...10320+10321)(10+100+... 10^{320}+10^{321}). There are 321321 terms, so it becomes 11...011...0, where there are 322322 digits in 11...011...0. Then, subtract the 321321 you initially added.

~ BJHHar

Solution 2

We can see that 9=99=9, 9+99=1089+99=108, 9+99+999=11079+99+999=1107, all the way to ten nines when we have 1111111110011111111100. Then, when we add more nines, we repeat the same process, and quickly get that the sum of digits is 342\boxed{342} since we have to add 9log3219\lfloor \log 321 \rfloor to the sum of digits, which is 932199\lceil \frac{321}9 \rceil.

Solution 3 (Pattern)

Observe how adding results in the last term but with a 11 concatenated in front and also a 11 subtracted (0909, 108108, 11071107, 1110611106). Then for any index of terms, nn, the sum is 11...10n11...10-n, where the first term is of length n+1n+1. Here, that is 342\boxed{342}.

~BJHHar

Solution 4 (Official MAA)

Write

N=(101)+(1021)++(103211)=10+102+103+104+105+106+10321321=1110321+104+105+106++10321=789+104+105+106++10321\begin{aligned}N &=(10-1)+(10^2-1)+\cdots+(10^{321}-1)\\ &=10+10^2+10^3+10^4+10^5+10^6+\cdots 10^{321}-321 \\ &=1110-321+10^4+10^5+10^6+\cdots+10^{321}\\ &=789+10^4+10^5+10^6+\cdots+10^{321}\\ \end{aligned} The sum of the digits of NN is therefore equal to 7+8+9+(3213)=3427+8+9+(321-3)=342.

Video Solution #1(Using Smart Manipulation)

https://youtu.be/JQdad7APQG8?t=22

Video Solution 2

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

Video Solution 3

https://youtu.be/TSKcjht8Rfk

~IceMatrix

Video Solution 4

https://youtu.be/9X18wCiYw9M

~Shreyas S

Video Solution 5

https://youtu.be/hzbzEAo9ezA

~savannahsolver

Video Solution 6 by Steakmath (fast and simple)

https://youtu.be/XNGRjxpPw6I