2009 AMC 10A Problem 5

The number is too big to square by hand quickly, so start with smaller numbers made of ones and look for a pattern.

1. Small cases:

2. The pattern: with $n$ ones, the square's digits count up $1, 2, \ldots, n$ and then back down to $1$.
3. With nine ones this gives
$$111{,}111{,}111^2 = 12{,}345{,}678{,}987{,}654{,}321.$$
4. Add the digits. The numbers $1$ to $8$ appear twice and $9$ appears once:
$$2(1 + 2 + \cdots + 8) + 9 = 2 \times 36 + 9 = 81.$$

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

Why does the pattern keep working up to nine ones? In long multiplication of $111{,}111{,}111$ by itself, each column gets a stack of $1$s: one in the rightmost column, two in the next, and so on up to nine in the middle, then back down. Nine is still a single digit, so nothing carries and the column totals are exactly the digits. With ten or more ones there would be carrying and the pattern would break. Trying small cases is a good way to discover a pattern, but check why it holds before trusting it for a big case.

Since nothing carries, the digit sum equals the total number of $1$s written in all the columns of the long multiplication. Each of the $9$ ones in the top number times each of the $9$ ones in the bottom number puts one $1$ somewhere, so the total is $9 \times 9 = 81$.

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 $\boxed{\textbf{(E)}\ 81}$.