2009 AMC 10A Problem 5

Below is the professionally curated solution for Problem 5 of the 2009 AMC 10A, 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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Concepts:digitspalindromepattern recognition

Difficulty rating: 1070

5.

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

1818

2727

4545

6363

8181

Solution:

The square of the nine-digit repunit is the palindrome 111,111,1112=12,345,678,987,654,321.111{,}111{,}111^2 = 12{,}345{,}678{,}987{,}654{,}321.

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=81.2(1 + 2 + \cdots + 8) + 9 = 2 \cdot 36 + 9 = 81.

Thus, the correct answer is E.

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