2009 AMC 10A Problem 5

Question

What is the sum of the digits of the square of

111{,}111{,}111?

18

27

45

63

81

Try squaring the shorter repunits

11,

111,

1111

to spot a pattern

The square of the nine-digit repunit is

12345678987654321

The square of the nine-digit repunit is the palindrome

\begin{aligned} &111{,}111{,}111^2 \\ &= 12{,}345{,}678{,}987{,}654{,}321. \end{aligned}

Its digits are

1, 2, \ldots, 9, 8, \ldots, 1,

so the sum is

\begin{aligned} &2(1 + 2 + \cdots + 8) + 9 \\ &= 2 \cdot 36 + 9 = 81. \end{aligned}

Thus, the correct answer is E.

Problem 5 in other years