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

(Problem 1)
(Problem 1)
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==Problem 1==
 
==Problem 1==
What is the sum of the digits of <math>9+99+999+...+99..99</math> (321 9's)?
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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==

Revision as of 22:21, 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

2019 AIME I (ProblemsAnswer KeyResources)
Preceded by
First Problem
Followed by
Problem 2
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