2009 AMC 10A Problem 5

Try small cases: $11^2 = 121$, $111^2 = 12321$, $1111^2 = 1234321$. A string of $n$ ones squared (for $n \le 9$) gives the palindrome $12\ldots n\ldots21$: the digits climb from $1$ to $n$ and then descend back to $1$.

With $n = 9$,
$$111{,}111{,}111^2 = 12{,}345{,}678{,}987{,}654{,}321.$$

Its digit sum is
$$2(1 + 2 + \cdots + 8) + 9 = 2 \cdot 36 + 9 = 81.$$

The answer is $\boxed{\textbf{(E)}\ 81}$.

Multiplying $111{,}111{,}111$ by itself stacks nine shifted copies of the number; column $k$ from the right receives one $1$ for each way to write $k$ as a sum of two positions, which is $1, 2, \ldots, 9, \ldots, 2, 1$. Because no column reaches $10$, there are no carries, and the pattern from small cases extends exactly up to nine ones.

Since there are no carries, the digit sum equals the total number of $1 \times 1$ products in the long multiplication, which is $9 \cdot 9 = 81$. Checking that all choices are multiples of $9$ (as they must be, since the number is divisible by $9$) does not narrow it down, so the pattern is needed.

• Trusting a half-remembered pattern like 123456789 (sum 45) and forgetting the descending half of the palindrome.
• Extending the palindrome pattern to ten or more ones, where carries break it; here nine ones is exactly the last safe case.