Difference between revisions of "2013 AMC 12A Problems/Problem 22"
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<math> \textbf{(A)} \ \frac{8}{25} \qquad \textbf{(B)} \ \frac{33}{100} \qquad \textbf{(C)} \ \frac{7}{20} \qquad \textbf{(D)} \ \frac{9}{25} \qquad \textbf{(E)} \ \frac{11}{30}</math> | <math> \textbf{(A)} \ \frac{8}{25} \qquad \textbf{(B)} \ \frac{33}{100} \qquad \textbf{(C)} \ \frac{7}{20} \qquad \textbf{(D)} \ \frac{9}{25} \qquad \textbf{(E)} \ \frac{11}{30}</math> | ||
− | ==Solution== | + | ==Solution 1== |
Working backwards, we can multiply 5-digit palindromes <math>ABCBA</math> by <math>11</math>, giving a 6-digit palindrome: | Working backwards, we can multiply 5-digit palindromes <math>ABCBA</math> by <math>11</math>, giving a 6-digit palindrome: | ||
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So, the probability is <math>\frac{330}{900} = \frac{11}{30}</math> | So, the probability is <math>\frac{330}{900} = \frac{11}{30}</math> | ||
− | + | == Solution 2 == | |
+ | Let the palindrome be the form in the previous solution which is <math>XYZZYX</math>. It doesn't matter what <math>Z</math> is because it only affects the middle digit. There are <math>90</math> ways to pick <math>X</math> and <math>Y</math>, and the only answer choice with denominator a factor of <math>90</math> is <math>\boxed{\textbf{(E)} \ \frac{11}{30}}</math>. | ||
== See also == | == See also == | ||
{{AMC12 box|year=2013|ab=A|num-b=21|num-a=23}} | {{AMC12 box|year=2013|ab=A|num-b=21|num-a=23}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 18:12, 5 February 2016
Contents
[hide]Problem
A palindrome is a nonnegative integer number that reads the same forwards and backwards when written in base 10 with no leading zeros. A 6-digit palindrome is chosen uniformly at random. What is the probability that
is also a palindrome?
Solution 1
Working backwards, we can multiply 5-digit palindromes by
, giving a 6-digit palindrome:
Note that if or
, then the symmetry will be broken by carried 1s
Simply count the combinations of for which
and
implies
possible
(0 through 8), for each of which there are
possible C, respectively. There are
valid palindromes when
implies
possible
(0 through 7), for each of which there are
possible C, respectively. There are
valid palindromes when
Following this pattern, the total is
6-digit palindromes are of the form , and the first digit cannot be a zero, so there are
combinations of
So, the probability is
Solution 2
Let the palindrome be the form in the previous solution which is . It doesn't matter what
is because it only affects the middle digit. There are
ways to pick
and
, and the only answer choice with denominator a factor of
is
.
See also
2013 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 21 |
Followed by Problem 23 |
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 12 Problems and Solutions |
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