2009 AMC 10A Problem 5

Attempt Problem 5 of the 2009 AMC 10A below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2009 AMC 10A solutions, or check the answer key.

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5.

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

1818

2727

4545

6363

8181

Answer: E
Concepts:digitspalindromepattern recognition
Difficulty rating: 1070
Solution:

The square of the nine-digit repunit is the palindrome 111,111,1112=12,345,678,987,654,321. \begin{aligned} &111{,}111{,}111^2 \\ &= 12{,}345{,}678{,}987{,}654{,}321. \end{aligned}

Its digits are 1,2,,9,8,,1,1, 2, \ldots, 9, 8, \ldots, 1, so the sum is 2(1+2++8)+9=236+9=81. \begin{aligned} &2(1 + 2 + \cdots + 8) + 9 \\ &= 2 \cdot 36 + 9 = 81. \end{aligned}

Thus, the correct answer is E.

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