AMC 10 · 2010 · #9
Grade 4 number-theoryPick an answer.
AMC 10 2010 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
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.
Force x near the top
Four digits means x+32 is at least 1000, so x is at least 968 — and at most 999.
A small +32 can only cross into four digits if x is already almost 1000.
4.NBT.A.2Extreme PrincipleList the palindromes there
First and last digit must match, so between 968 and 999 the palindromes are 969, 979, 989, 999.
In this range only numbers whose first and last digit agree are palindromes.
In this range only the numbers whose first and last digits agree are palindromes.
▸ Why?
A number is its digits sitting in fixed places, so reversing it swaps the outer two places.
▸ Why?
Each digit must land on the one it is paired with, so the outer pair has to match exactly.
Add 32 and check each
Adding 32 gives 1001, 1011, 1021, 1031 — only 969 lands on a palindrome, 1001.
Just try the few survivors; only one lands on a symmetric four-digit number.
4.NBT.B.4Guess And CheckAdd the digits of x
With x=969, the digits give 9+6+9=24, which is choice (E).
The question wants the digit sum, so just add the three digits of the winner.
4.NBT.B.4Guess And CheckWhen 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