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

Line 7: Line 7:
 
Using the standard multiplication algorithm, <math>111111111^2=12345678987654321</math> whose digit sum is <math>81\longrightarrow \fbox{E}</math>
 
Using the standard multiplication algorithm, <math>111111111^2=12345678987654321</math> whose digit sum is <math>81\longrightarrow \fbox{E}</math>
  
 +
Or
 +
 +
Add up all the ones(thus deriving the sum of the number) of <math>111,111,111</math> gives us <math>1+1+1+\dots+1 = 9</math> Thus, <math>(9)^2 = 81</math>
 
==See also==
 
==See also==
 
{{AMC10 box|year=2009|ab=A|num-b=4|num-a=6}}
 
{{AMC10 box|year=2009|ab=A|num-b=4|num-a=6}}

Revision as of 09:29, 21 March 2009

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 $81\longrightarrow \fbox{E}$

Or

Add up all the ones(thus deriving the sum of the number) of $111,111,111$ gives us $1+1+1+\dots+1 = 9$ Thus, $(9)^2 = 81$

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
Invalid username
Login to AoPS