2020 AMC 10B Problems/Problem 6

Revision as of 04:50, 21 January 2023 by Pi is 3.14 (talk | contribs) (Video Solution)

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

https://youtu.be/OHR_6U686Qg

~IceMatrix

https://youtu.be/0bfIDBkfcZs

~savannahsolver

Video Solution by OmegaLearn

https://youtu.be/ZhAZ1oPe5Ds?t=1361

~ pi_is_3.14

See Also

2020 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 5
Followed by
Problem 7
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All AMC 10 Problems and Solutions

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