Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2019 AIME I Problems/Problem 1"

m (→‎Solution 3)
m (→‎Solution: LaTeX)
Line 3: Line 3:
 
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>.
 
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>.
  
−
==Solution==
+
==Solution 1==
−
Let's express the number in terms of <math>10^n</math>. We can obtain <math>(10-1)+(10^2-1)+(10^3-1)+\cdots+(10^{321}-1)</math>. By the commutative and associative property, we can group it into <math>(10+10^2+10^3+\cdots+10^{321})-321</math>. We know the former will yield <math>1111....10</math>, so we only have to figure out what the last few digits are. There are currently <math>321</math> 1's. We know the last 4 digits are 1110, and that the others will not be affected if we subtract <math>321</math>. If we do so, we get that <math>1110-321=789</math>. This method will remove 3 1's, and add a 7, 8 and 9. Therefore, the sum of the digits is <math>(321-3)+7+8+9=\boxed{342}</math>.
+
Let's express the number in terms of <math>10^n</math>. We can obtain <math>(10-1)+(10^2-1)+(10^3-1)+\cdots+(10^{321}-1)</math>. By the commutative and associative property, we can group it into <math>(10+10^2+10^3+\cdots+10^{321})-321</math>. We know the former will yield <math>1111....10</math>, so we only have to figure out what the last few digits are. There are currently <math>321</math> 1's. We know the last four digits are <math>1110</math>, and that the others will not be affected if we subtract <math>321</math>. If we do so, we get that <math>1110-321=789</math>. This method will remove three <math>1</math>'s, and add a <math>7</math>, <math>8</math> and <math>9</math>. Therefore, the sum of the digits is <math>(321-3)+7+8+9=\boxed{342}</math>.
  
 
-eric2020
 
-eric2020
Line 10: Line 10:
  
  
−
A similar and simpler way to consider the initial manipulations is to observe that adding 1 to each term results in <math>(10+100+... 10^{320}+10^{321})</math>. There are 321 terms, so it becomes <math>11...0</math>, where there are 322 digits in <math>11...0</math>. Then, subtract the 321 you initially added.  
+
A similar and simpler way to consider the initial manipulations is to observe that adding <math>1</math> to each term results in <math>(10+100+... 10^{320}+10^{321})</math>. There are <math>321</math> terms, so it becomes <math>11...0</math>, where there are <math>322</math> digits in <math>11...0</math>. Then, subtract the <math>321</math> you initially added.  
  
 
~ BJHHar
 
~ BJHHar

Revision as of 11:45, 26 December 2019

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 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


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

Video Solution

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

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2019_AIME_I_Problems/Problem_1&oldid=113444"