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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 (ProblemsResources)
Preceded by
Problem 4
Followed by
Problem 6
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