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

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, $100x+10y+y-2x$ = $98x+11y \equiv 0 \pmod 7$

Since $98 \equiv 0 \pmod 7$ , 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
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

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

Page

Toolbox

Search

2010 AMC 10B Problems/Problem 21

Problem 21

Solution

See also