2010 AMC 10B Problems/Problem 21

Problem 21

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

$\textbf{(A)}\ \dfrac{1}{10} \qquad \textbf{(B)}\ \dfrac{1}{9} \qquad \textbf{(C)}\ \dfrac{1}{7} \qquad \textbf{(D)}\ \dfrac{1}{6} \qquad \textbf{(E)}\ \dfrac{1}{5}$

Solution #1

View the palindrome as some number with form (decimal representation): $a_3 \cdot 10^3 + a_2 \cdot 10^2 + a_1 \cdot 10 + a_0$ . But because the number is a palindrome, $a_3 = a_0, a_2 = a_1$ . Recombining this yields $1001a_3 + 110a_2$ . 1001 is divisible by 7, which means that as long as $a_2 = 0$ , the palindrome will be divisible by 7. This yields 9 palindromes out of 90 ( $9 \cdot 10$ ) possibilities for palindromes. However, if $a_2 = 7$ , then this gives another case in which the palindrome is divisible by 7. This adds another 9 palindromes to the list, bringing our total to $\frac{18}{90}=\boxed{\textbf{(E)}\ \frac15}$

Solution #1

It is known that the palindromes can be expressed as: $1000x+100y+10y+x$ (as it is a four digit palindrome it must be of the form $xyyx$ , where x and y are positive integers from [0,9]. Using the divisibility rules of 7, $100x+10y+y-2x$ = $98x+11y \equiv 0 \pmod 7$

The $98x$ is now irrelelvant

Thus we solve:

$11y \equiv 0 \pmod 7$

Which has two solutions: $0$ and $7$

There are thus, two options for $y$ out of the 10, so $2/10 = \boxed{\textbf{(E)}\ \frac15}$

2010 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 20	Followed by Problem 22
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2010 AMC 10B Problems/Problem 21

Contents

Problem 21

Solution #1

Solution #1

See also