2019 AIME I Problems/Problem 1: Difference between revisions

Latest revision as of 00:21, 18 June 2025

Problem

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 1

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 four 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 three $1$ 's, and add a $7$ , $8$ and $9$ . Therefore, the sum of the digits is $(321-3)+7+8+9=\boxed{342}$ .

-eric2020 -another Eric in 2020

A similar and simpler way to consider the initial manipulations is to observe that adding $1$ 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 (Pattern)

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

Solution 4 (Official MAA)

Write $\begin{align*}N &=(10-1)+(10^2-1)+\cdots+(10^{321}-1)\\ &=10+10^2+10^3+10^4+10^5+10^6+\cdots 10^{321}-321 \\ &=1110-321+10^4+10^5+10^6+\cdots+10^{321}\\ &=789+10^4+10^5+10^6+\cdots+10^{321}\\ \end{align*}$ The sum of the digits of $N$ is therefore equal to $7+8+9+(321-3)=342$ .

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

@@ Line 1: / Line 1: @@
-==Problem 1==
+==Problem==
 Consider the integer <cmath>N = 9 + 99 + 999 + 9999 + \cdots + \underbrace{99\ldots 99}_\text{321 digits}.</cmath>Find the sum of the digits of <math>N</math>.
@@ Line 7: / Line 7: @@
 -eric2020
--also Eric2020
+-another Eric in 2020
@@ Line 23: / Line 23: @@
 ~BJHHar
-==Video Solution==
+==Solution 4 (Official MAA)==
+Write <cmath>\begin{align*}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{align*}</cmath>
+The sum of the digits of <math>N</math> is therefore equal to <math>7+8+9+(321-3)=342</math>.
+==Video Solution #1(Using Smart Manipulation)==
+https://youtu.be/JQdad7APQG8?t=22
+==Video Solution 2==
 https://www.youtube.com/watch?v=JFHjpxoYLDk
-==Video Solution 2==
+==Video Solution 3==
 https://youtu.be/TSKcjht8Rfk
 ~IceMatrix
-==Video Solution 3==
+==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
 ==See Also==

2019 AIME I (Problems • Answer Key • Resources)
Preceded by First Problem	Followed by Problem 2
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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