Difference between revisions of "2010 AMC 10B Problems/Problem 21"
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Using the divisibility rules of 7, <math>100x+10y+y-2x</math> = <math>98x+11y \equiv 0 \pmod 7</math> | Using the divisibility rules of 7, <math>100x+10y+y-2x</math> = <math>98x+11y \equiv 0 \pmod 7</math> | ||
| − | The <math>98x</math> is now irrelelvant | + | Since <math>98 \equiv 0 \pmod 7</math>, The <math>98x</math> is now irrelelvant |
Thus we solve: | Thus we solve: | ||
Revision as of 22:24, 12 June 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, 
= 
Since 
, 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. 
