2009 AMC 10A Problems/Problem 5: Difference between revisions

Revision as of 09:29, 21 March 2009

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

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

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, $(9)^2 = 81$

2009 AMC 10A (Problems • Answer Key • Resources)
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

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 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==
 {{AMC10 box|year=2009|ab=A|num-b=4|num-a=6}}