AMC 10 · 2010 · #9

Grade 4 number-theory

Archetype Small-Domain Constrained Search └ Sub-archetype Answer-Choice Plug-In

place-valuedigit-sumsystematic-enumeration bound-inequality-then-enumerate ↑ Prerequisites: place-value

📏 Medium solution 💡 2 insights

Problem

A $\text{palindrome}$ , 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 $x$ ?

Pick an answer.

(A)

(B)

(C)

(D)

(E)

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.

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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

#14 Extreme Principle 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 $999$.

$$x+32\ge 1000 \Rightarrow x\ge 968$$

💡 A small $+32$ can only cross into four digits if $x$ is already almost $1000$.

#3 Eliminate Possibilities 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 $x$.

$$969,\ 979,\ 989,\ 999$$

💡 In this range only numbers whose first and last digit agree are palindromes.

#6 Guess and Check 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.

$$969+32=1001,\quad 979+32=1011,\quad 989+32=1021,\quad 999+32=1031$$

💡 Just try the few survivors; only one lands on a symmetric four-digit number.

#6 Guess and Check 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).

$$9+6+9=24$$

💡 The question wants the digit sum, so just add the three digits of the winner.

[1] #14 4.NBT.A.2 For $x+32$ to have four digits it must be at least $1000$. So $x+32\ge 1000$, wh

[2] #3 4.NBT.A.2 A three-digit palindrome has the form where the first and last digits match. Bet

[3] #6 4.NBT.B.4 Test each candidate by adding $32$ and seeing if the result is a palindrome. $96

[4] #6 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.2 Read, 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.4 Fluently 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.

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: E

Force x near the top

List the palindromes there

Add 32 and check each

Add the digits of x

Review

CCSS standards used (min grade 4)

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