# 2020 AMC 10B Problems/Problem 6

## Problem

Driving along a highway, Megan noticed that her odometer showed $15951$ (miles). This number is a palindrome-it reads the same forward and backward. Then $2$ hours later, the odometer displayed the next higher palindrome. What was her average speed, in miles per hour, during this $2$-hour period? $\textbf{(A)}\ 50 \qquad\textbf{(B)}\ 55 \qquad\textbf{(C)}\ 60 \qquad\textbf{(D)}\ 65 \qquad\textbf{(E)}\ 70$

## Solution

In order to get the smallest palindrome greater than $15951$, we need to raise the middle digit. If we were to raise any of the digits after the middle, we would be forced to also raise a digit before the middle to keep it a palindrome, making it unnecessarily larger.

So we raise $9$ to the next largest value, $10$, but obviously, that's not how place value works, so we're in the $16000$s now. To keep this a palindrome, our number is now $16061$.

So Megan drove $16061-15951=110$ miles. Since this happened over $2$ hours, she drove at $\frac{110}{2}=\boxed{\textbf{(B) }55}$ mph. ~quacker88

## Video Solution (HOW TO CRITICALLY THINK!!!)

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~savannahsolver

## Video Solution

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 