Difference between revisions of "2010 AMC 10B Problems/Problem 21"
(→Solution) |
(Tag: Undo) |
||
Line 5: | Line 5: | ||
==Solution== | ==Solution== | ||
− | |||
The palindromes can be expressed as: <math>1000x+100y+10y+x </math> (since it is a four digit palindrome, it must be of the form <math>xyyx</math> , where x and y are integers from <math>[1, 9]</math> and <math>[0, 9]</math>, respectively.) | The palindromes can be expressed as: <math>1000x+100y+10y+x </math> (since it is a four digit palindrome, it must be of the form <math>xyyx</math> , where x and y are integers from <math>[1, 9]</math> and <math>[0, 9]</math>, respectively.) | ||
Revision as of 09:25, 5 August 2019
Problem 21
A palindrome between and
is chosen at random. What is the probability that it is divisible by
?
Solution
The palindromes can be expressed as: (since it is a four digit palindrome, it must be of the form
, where x and y are integers from
and
, respectively.)
We simplify this to:
.
Because the question asks for it to be divisible by 7,
We express it as .
Because ,
We can substitute for
We are left with
Since we can simplify the
in the expression to
.
In order for this to be true, must also be true.
Thus we solve:
Which has two solutions: and
There are thus two options for out of the 10, so the answer is
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.