2010 AMC 10A Problem 9

P1-10~2 minPrint

Bases & Digits Bound the quantity above/below or estimate to pin it down Organized listing / direct enumeration

A palindrome, such as $83438$ , is a number that remains the same when its digits are reversed. The numbers $x$ and $x+32$ are three-digit and four-digit palindromes, respectively. What is the sum of the digits of $x$ ?

Show solution without answering

Key insight

A three-digit number plus 32 has four digits only if the number is at least 968. That leaves 969, 979, 989, 999, and only 969 + 32 = 1001 works.

Solution

The key fact is that adding just $32$ turns a three-digit number into a four-digit number. That can only happen near $1000$ .

1. Four-digit numbers start at $1000$ , so $x + 32 \ge 1000$ , which means $x \ge 968$ . Also $x \le 999$ because it has three digits.
2. A three-digit palindrome between $968$ and $999$ must start and end with $9$ , so it looks like $9\_9$ . The middle digit must make it at least $968$ , so the options are
$969,\quad 979,\quad 989,\quad 999.$
3. Add $32$ to each and check whether the result reads the same backwards:

$x$	$969$	$979$	$989$	$999$
$x + 32$	$1001$	$1011$	$1021$	$1031$
palindrome?	yes	no	no	no

So $x = 969$ , and its digit sum is $9 + 6 + 9 = 24$ .

The answer is $\boxed{\textbf{(E)}\ 24}$ .

Why this works

When a small addition changes the number of digits, the number must sit just below a power of $10$ like $1000$ . That squeezes the possibilities down to a short list you can simply check. Look for a size limit like this before setting up digit algebra.

Concise solution

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

Since $x$ has three digits and $x + 32$ has four, we need $x + 32 \ge 1000$ , i.e. $968 \le x \le 999$ .

The three-digit palindromes in that range have the form $9a9$ with $a \ge 6$ : they are $969, 979, 989, 999$ . Adding $32$ gives $1001, 1011, 1021, 1031$ . Only $1001$ reads the same backwards.

So $x = 969$ , and its digit sum is $9 + 6 + 9 = 24$ .

The answer is $\boxed{\textbf{(E)}\ 24}$ .

Review in the AoPS books

Introduction to Number Theory, Chapter 8: Base Numbers
Prealgebra, Chapter 15: Problem-Solving Strategies (§15.2)
Introduction to Number Theory, Chapter 9: Base Number Arithmetic
Introduction to Number Theory, Chapter 6: Special Numbers (§6.5)