AMC 10 · 2010 · #6

Grade 4 number-theory
place-valuedigit-sumsystematic-enumeration bound-inequality-then-enumerate ↑ Prerequisites: place-value
📏 Medium solution 💡 2 insights
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Problem
A three-digit palindrome becomes a four-digit palindrome when a small number is added. Find the sum of the original number's digits.

Pick an answer.

(A)
20
(B)
21
(C)
22
(D)
23
(E)
24
How to solve
Strategy Extreme Principle

The key pressure is the size jump: adding only 32 pushes a three-digit number up to four digits, so x must sit right at the top of the three-digit range. Pinning down that boundary shrinks the search to just a handful of large palindromes, and then each one can be tested directly by adding 32 and checking whether the result reads the same both ways.

1STEP 1

Force x near the top

Gaining a digit forces the number near the top.

x+32 ≥ 1000 → x ≥ 968
2STEP 2

List the palindromes there

Only four palindromes live there.

969, 979, 989, 999
3STEP 3

Add 32 and check each

Testing each finds exactly one that works.

969+32=1001, 979+32=1011, 989+32=1021, 999+32=1031
4STEP 4

Add the digits of x

Adding its digits gives 24, choice (E).

9+6+9=24
Answer
24
Check the found number against every rule: 969 is a three-digit palindrome, and 969+32=1001 is a four-digit palindrome, so both conditions hold. The digit sum 24 is the largest option, which fits because x was forced to be near 999 where the digits are big. Only choice (E) is reachable.
💡Key takeaway

When a small addition forces a number into more digits, the starting number must be sitting right at the edge, so look there first.

  • Force x near the top
  • List the palindromes there
  • Add 32 and check each
  • Add the digits of x