2009 AMC 10A Problems/Problem 5

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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
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All AMC 10 Problems and Solutions
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