2010 AMC 12B Problem 11

Attempt Problem 11 of the 2010 AMC 12B below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2010 AMC 12B solutions, or check the answer key.

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11.

A palindrome between $1000$ and $10{,}000$ is chosen at random. What is the probability that it is divisible by $7?$

$\dfrac{1}{10}$

$\dfrac{1}{9}$

$\dfrac{1}{7}$

$\dfrac{1}{6}$

$\dfrac{1}{5}$

Answer: E

Concepts:palindrome divisibility basic probability

Difficulty rating: 1500

Small Hint:

A four-digit palindrome $\overline{abba}$ equals $1001a+110b$

Big Hint:

$1001=7\cdot11\cdot13$ is divisible by $7,$ but $110$ is not

Solution:

A four-digit palindrome has the form $\overline{abba}=1001a+110b$ with $1\le a\le9$ and $0\le b\le9.$

Since $1001=7\cdot11\cdot13$ is divisible by $7$ and $110$ is not, the number is divisible by $7$ exactly when $b$ is divisible by $7,$ that is $b=0$ or $b=7.$

For each $a,$ that is $2$ of the $10$ choices of $b,$ a probability of $\dfrac{2}{10}=\dfrac15.$

Thus, the correct answer is E.

Problem 10#10

Full Exam

Problem 12#12

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