AMC 10 · 2020 · #6

Easy mode Grade 4
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Problem

A palindrome is a number that reads the same forward and backward, like 1595115951.

Megan is driving on a highway. She looks at her odometer and sees the number 1595115951 miles. Two hours later, she looks again. The odometer now shows the very next palindrome number that is bigger than 1595115951.

What was her average speed, in miles per hour, during those 22 hours?

Pick an answer.

(A)
50
(B)
55
(C)
60
(D)
65
(E)
70
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Toolkit + CCSS Solution

Understand

Restated: Megan's odometer reads $15951$, a palindrome. After driving for $2$ hours, it shows the next larger palindrome. Find her average speed in miles per hour.

Givens: Starting odometer reading is $15951$ (a $5$-digit palindrome); Time elapsed is $2$ hours; Ending odometer reading is the next palindrome after $15951$; Choices: (A) $50$, (B) $55$, (C) $60$, (D) $65$, (E) $70$

Unknowns: Average speed in miles per hour during the $2$-hour drive

Understand

Restated: Megan's odometer reads $15951$, a palindrome. After driving for $2$ hours, it shows the next larger palindrome. Find her average speed in miles per hour.

Givens: Starting odometer reading is $15951$ (a $5$-digit palindrome); Time elapsed is $2$ hours; Ending odometer reading is the next palindrome after $15951$; Choices: (A) $50$, (B) $55$, (C) $60$, (D) $65$, (E) $70$

Plan

Primary tool: #5 Look for a Pattern

Secondary: #7 Identify Subproblems, #3 Eliminate Possibilities

A $5$-digit palindrome has the shape $\overline{abcba}$ — the first three digits force the last two. Tool #5 says: read the palindrome pattern to find the smallest valid bump. Tool #7 then splits the problem into two clean parts: (1) find the next palindrome, (2) divide by $2$. Tool #3 cross-checks the speed against the five answer choices.

Execute — Answer: B

#5 Look for a Pattern 4.NBT.A.2 Step 1
  • Write the starting reading as $\overline{abcba}$ with $a = 1, b = 5, c = 9$.
  • To get the next palindrome, bump the middle digit $c$ by $1$ — that lifts the number by the smallest possible amount.
$$15951 = \overline{1\,5\,9\,5\,1}$$

💡 The middle digit is the cheapest knob to turn on a palindrome.

#5 Look for a Pattern 4.NBT.A.2 Step 2
  • But $c = 9$ already — bumping it triggers a carry into the thousands place.
  • The thousands digit $5$ becomes $6$, the middle digit resets to $0$, and the palindrome shape forces the hundreds digit to $0$ and the units digit to stay matching the ten-thousands digit.
$$\overline{1\,5\,9\,5\,1} \xrightarrow{\text{carry}} \overline{1\,6\,0\,6\,1} = 16061$$

💡 When a digit is already $9$, the bump cascades — like odometer rollover.

#7 Identify Subproblems 4.NBT.B.4 Step 3

Compute the distance driven by subtracting.

$$16061 - 15951 = 110 \text{ miles}$$

💡 Distance equals end reading minus start reading.

#7 Identify Subproblems 3.OA.A.3 Step 4

Divide distance by time to get average speed.

$$\dfrac{110 \text{ miles}}{2 \text{ hours}} = 55 \text{ mph}$$

💡 Speed is distance per unit time — divide.

#3 Eliminate Possibilities 4.NBT.A.2 Step 5
  • $55$ matches choice (B).
  • The other choices ($50, 60, 65, 70$) correspond to distances $100, 120, 130, 140$ — none of which lands on a palindrome reading from $15951$.
$$55 \Rightarrow \textbf{(B)}$$

💡 Match the computed speed against the listed choices.

[1] #5 4.NBT.A.2 Write the starting reading as $\overline{abcba}$ with $a = 1, b = 5, c = 9$. To
[2] #5 4.NBT.A.2 But $c = 9$ already — bumping it triggers a carry into the thousands place. The
[3] #7 4.NBT.B.4 Compute the distance driven by subtracting.
[4] #7 3.OA.A.3 Divide distance by time to get average speed.
[5] #3 4.NBT.A.2 $55$ matches choice (B). The other choices ($50, 60, 65, 70$) correspond to dist

Review

Reasonableness: Verify that $16061$ really is a palindrome: reading right-to-left gives $1, 6, 0, 6, 1$ — same as left-to-right. Also confirm no smaller palindrome sits between $15951$ and $16061$: any palindrome starting $15\_\_\_$ must end in $51$, and its middle digit would have to be $\geq 10$ to exceed $15951$, which is impossible. So $16061$ is genuinely the next one, distance $= 110$, speed $= 55$ mph. ✓

Alternative: Tool #3 (Eliminate) alone: each answer choice gives a candidate end-odometer ($16051, 16061, 16071, 16081, 16091$). Only $16061$ is a palindrome — instantly picks (B) without explicitly finding the next palindrome from scratch.

CCSS standards used (min grade 4)

  • 4.NBT.A.2 Read and write multi-digit whole numbers and compare using symbols (Reading $15951$ as $\overline{1\,5\,9\,5\,1}$ to inspect place values and matching $16061$ against the digit positions to confirm a palindrome and the answer choice.)
  • 4.NBT.B.4 Fluently add and subtract multi-digit whole numbers (Subtracting $16061 - 15951 = 110$ to find the distance driven.)
  • 3.OA.A.3 Solve multiplication and division word problems within 100 (Dividing distance $110$ miles by time $2$ hours to find average speed $55$ mph.)

⭐ This AMC 10 problem only needs Grade 4 place value you already know — to find the next palindrome after $15951$, bump the middle $9$ which forces a carry to $16061$. The drive is $110$ miles in $2$ hours, so the speed is $55$ mph.

⭐ This AMC 10 problem only needs Grade 4 place value you already know — to find the next palindrome after $15951$, bump the middle $9$ which forces a carry to $16061$. The drive is $110$ miles in $2$ hours, so the speed is $55$ mph.