AMC 10 · 2010 · #9
Easy mode Grade 4A number that reads the same forwards and backwards is called a palindrome. For example, 83438 is a palindrome. A three-digit number x is a palindrome. When you add 32 to it, the result x+32 is a four-digit palindrome. What is the sum of the digits of x?
Pick an answer.
AMC 10 2010 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: A palindrome reads the same forwards and backwards, like $83438$. A number $x$ is a three-digit palindrome, and $x+32$ is a four-digit palindrome. Find the sum of the digits of $x$.
Givens: $x$ is a three-digit palindrome.; $x+32$ is a four-digit palindrome.; A palindrome is unchanged when its digits are reversed.
Unknowns: The value of $x$, and then the sum of its digits.
Understand
Restated: A palindrome reads the same forwards and backwards, like $83438$. A number $x$ is a three-digit palindrome, and $x+32$ is a four-digit palindrome. Find the sum of the digits of $x$.
Givens: $x$ is a three-digit palindrome.; $x+32$ is a four-digit palindrome.; A palindrome is unchanged when its digits are reversed.
Plan
Primary tool: #14 Extreme Principle
Secondary: #3 Eliminate Possibilities, #6 Guess and Check
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.
Execute — Answer: E
4.NBT.A.2 Step 1 Force x near the top
- For $x+32$ to have four digits it must be at least $1000$.
- So $x+32\ge 1000$, which means $x\ge 968$.
- Since $x$ is three digits, $x$ is between $968$ and $999$.
💡 A small $+32$ can only cross into four digits if $x$ is already almost $1000$.
4.NBT.A.2 Step 2 List the palindromes there
- A three-digit palindrome has the form where the first and last digits match.
- Between $968$ and $999$, the ones that read the same reversed are $969$, $979$, $989$, and $999$.
- These four are the only candidates for $x$.
💡 In this range only numbers whose first and last digit agree are palindromes.
4.NBT.B.4 Step 3 Add 32 and check each
- Test each candidate by adding $32$ and seeing if the result is a palindrome.
- $969+32=1001$, which reads the same reversed.
- $979+32=1011$, $989+32=1021$, and $999+32=1031$ are not palindromes.
- Only $x=969$ works.
💡 Just try the few survivors; only one lands on a symmetric four-digit number.
4.NBT.B.4 Step 4 Add the digits of x
- With $x=969$, add its digits: $9+6+9=24$.
- So the sum of the digits of $x$ is $24$, which is choice (E).
💡 The question wants the digit sum, so just add the three digits of the winner.
4.NBT.A.2 For $x+32$ to have four digits it must be at least $1000$. So $x+32\ge 1000$, wh 4.NBT.A.2 A three-digit palindrome has the form where the first and last digits match. Bet 4.NBT.B.4 Test each candidate by adding $32$ and seeing if the result is a palindrome. $96 4.NBT.B.4 With $x=969$, add its digits: $9+6+9=24$. So the sum of the digits of $x$ is $24 Review
Reasonableness: 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.
Alternative: Reason from the four-digit side. The smallest four-digit palindrome is $1001$, and $x=1001-32=969$ is already a three-digit palindrome, so no larger four-digit palindrome is needed. This lands on $x=969$ directly, giving digit sum $24$.
CCSS standards used (min grade 4)
4.NBT.A.2Read, write, and compare multi-digit whole numbers using place value (Using place value to see $x\ge 968$ and to identify which numbers between $968$ and $999$ are palindromes.)4.NBT.B.4Fluently add and subtract multi-digit whole numbers (Adding $32$ to each candidate to test it, and adding the digits $9+6+9$ to get $24$.)
⭐ When a small addition forces a number into more digits, the starting number must be sitting right at the edge, so look there first.
⭐ When a small addition forces a number into more digits, the starting number must be sitting right at the edge, so look there first.
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