Difference between revisions of "2010 AMC 10B Problems/Problem 21"
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<math>\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}</math> | <math>\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}</math> | ||
| − | ==Solution | + | ==Solution== |
It is known that the palindromes can be expressed as: <math>1000x+100y+10y+x </math> (as it is a four digit palindrome it must be of the form <math>xyyx</math> , where x and y are positive integers from [0,9]. | It is known that the palindromes can be expressed as: <math>1000x+100y+10y+x </math> (as it is a four digit palindrome it must be of the form <math>xyyx</math> , where x and y are positive integers from [0,9]. | ||
Revision as of 23:44, 21 January 2014
Problem 21
A palindrome between 
and 
is chosen at random. What is the probability that it is divisible by 
?
Solution
It is known that the palindromes can be expressed as: 
(as it is a four digit palindrome it must be of the form 
, where x and y are positive integers from [0,9].
Using the divisibility rules of 7, 
= 
The 
is now irrelelvant
Thus we solve:
Which has two solutions: 
and
There are thus, two options for 
out of the 10, so 
See also
| 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. 
