2009 AMC 10A Problem 5

P1-10~2 minPrint

Bases & Digits Compute small cases, spot the pattern, generalize

What is the sum of the digits of the square of $111,111,111$ ?

Show solution without answering

Key insight

Squaring 11, 111 and 1111 gives 121, 12321 and 1234321; the digits climb to the number of ones and come back down, with no carrying up to nine ones.

Solution

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

Small cases:

Number	Square
$11$	$121$
$111$	$12321$
$1111$	$1234321$

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 this works

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.

Alternative approach

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 $9 \times 9 = 81$ .

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

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

Review in the AoPS books

Introduction to Number Theory, Chapter 8: Base Numbers
Prealgebra, Chapter 15: Problem-Solving Strategies (§15.1)
Introduction to Number Theory, Chapter 9: Base Number Arithmetic
Introduction to Number Theory, Chapter 12: Introduction to Modular Arithmetic (§12.6)