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Difference between revisions of "2013 AMC 12A Problems/Problem 22"

(Created page with "Working backwards, we can multiply 5-digit palindromes <math>ABCBA</math> by <math>11</math>, giving a 6-digit palindrome: <math>A (A+B) (B+C) (B+C) (A+B) A</math> Note that if...")
 
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== Problem==
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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 <math>n</math> is chosen uniformly at random. What is the probability that <math>\frac{n}{11}</math> is also a palindrome?
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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>
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==Solution==
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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>
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== See also ==
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{{AMC12 box|year=2013|ab=A|num-b=21|num-a=23}}

Revision as of 17:54, 22 February 2013

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 $n$ is chosen uniformly at random. What is the probability that $\frac{n}{11}$ is also a palindrome?

$\textbf{(A)} \ \frac{8}{25} \qquad \textbf{(B)} \ \frac{33}{100} \qquad \textbf{(C)} \ \frac{7}{20} \qquad \textbf{(D)} \ \frac{9}{25} \qquad \textbf{(E)} \ \frac{11}{30}$

Solution

Working backwards, we can multiply 5-digit palindromes $ABCBA$ by $11$, giving a 6-digit palindrome:

$A (A+B) (B+C) (B+C) (A+B) A$

Note that if $A + B > 10$ or $B + C > 10$, then the symmetry will be broken by carried 1s

Simply count the combinations of $(A, B, C)$ for which $A + B < 10$ and $B + C < 10$

$A = 1$ implies $9$ possible $B$ (0 through 8), for each of which there are $10, 9, 8, 7, 6, 5, 4, 3, 2$ possible C, respectively. There are $54$ valid palindromes when $A = 1$

$A = 2$ implies $8$ possible $B$ (0 through 7), for each of which there are $10, 9, 8, 7, 6, 5, 4, 3$ possible C, respectively. There are $52$ valid palindromes when $A = 2$

Following this pattern, the total is

$54 + 52 + 49 + 45 + 40 + 34 + 27 + 19 + 10 = 330$

6-digit palindromes are of the form $XYZZYX$, and the first digit cannot be a zero, so there are $9 * 10 * 10 = 900$ combinations of $(X, Y, Z)$

So, the probability is $\frac{330}{900} = \frac{11}{30}$

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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