Difference between revisions of "2009 AMC 10A Problems/Problem 5"

(Solution)
(Problem)
Line 1: Line 1:
 
==Problem==
 
==Problem==
What is the sum of the digits of the square of 111,111,111 ?
+
<!-- don't remove the following tag, for PoTW on the Wiki front page--><onlyinclude>What is the sum of the digits of the square of <math>111,111,111</math>?<!-- don't remove the following tag, for PoTW on the Wiki front page--></onlyinclude>
  
 
<math>\mathrm{(A)}\ 18\qquad\mathrm{(B)}\ 27\qquad\mathrm{(C)}\ 45\qquad\mathrm{(D)}\ 63\qquad\mathrm{(E)}\ 81</math>
 
<math>\mathrm{(A)}\ 18\qquad\mathrm{(B)}\ 27\qquad\mathrm{(C)}\ 45\qquad\mathrm{(D)}\ 63\qquad\mathrm{(E)}\ 81</math>

Revision as of 18:47, 27 March 2015

Problem

What is the sum of the digits of the square of $111,111,111$?

$\mathrm{(A)}\ 18\qquad\mathrm{(B)}\ 27\qquad\mathrm{(C)}\ 45\qquad\mathrm{(D)}\ 63\qquad\mathrm{(E)}\ 81$

Solution

Using the standard multiplication algorithm, $111111111^2=12345678987654321,$ whose digit sum is $81\longrightarrow \fbox{E}.$

See also

2009 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 4
Followed by
Problem 6
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 10 Problems and Solutions

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