Printable

2010 AMC 10A Problem 9

2010 AMC 10A Problem 9AMC 10 Step by Step

Name: Date: September 13, 2026

Bases & Digits · difficulty 2 of 5 · about 2 min · position P1-10

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 $x$ ?

A)
$20$
B)
$21$
C)
$22$
D)
$23$
E)
$24$

Answer

Key insight

x + 32 has four digits, so x is at least 968; only four three-digit palindromes remain, and only 969 + 32 = 1001 is a palindrome.

Solution

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 $\boxed{\textbf{(E)}\ 24}$ .

Why this works

A change in the number of digits pins the value to a narrow window just below a power of $10$ . Once the window is small, listing the candidates is faster and safer than any algebraic setup. Look for size constraints before reaching for digit variables.

The trap

Trying to solve with digit variables instead of noticing x must lie in the tiny window 968 to 999.

Common mistakes

Trying to solve with digit variables instead of noticing x must lie in the tiny window 968 to 999.
Stopping at $x = 999$ (the largest palindrome) without checking that $1031$ is not a palindrome.

Techniques

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