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"

(→‎Solution)
(→‎Solution)
Line 4: Line 4:
  
 
==Solution==
 
==Solution==
−
The above sequence is equivalent to <math>(10^1-1)+(10^2-1)+(10^3-1)+\cdots+(10^(321)-1)</math>. We can take out the ones so that this becomes <math>(10^1)+(10^2)+(10^3)+\cdots+(10^321)-321</math>. Now, notice that <math>(10^1)+(10^2)+(10^3)+\cdots+(10^(321))</math> is equal to <math>111...11110</math>, with 321 ones and one zero as digits. Subtract 321 to obtain <math>111...10790</math>, with 318 ones, 2 zeros, a 7 and a 9 as digits. This sums to <math>\boxed{334}</math>.
+
Let's express the terms 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>.
  
 
{{AIME box|year=2019|n=I|before=First Problem|num-a=2}}
 
{{AIME box|year=2019|n=I|before=First Problem|num-a=2}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 21:29, 14 March 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

Let's express the terms 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 $(321-3)+7+8+9=\boxed{342}$.

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=104417"