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

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==Solution 1==
 
==Solution 1==
  
Working backwards, we can multiply 5-digit palindromes <math>ABCBA</math> by <math>11</math>, giving a 6-digit palindrome:
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By 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>
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<math>\overline{A (A+B) (B+C) (B+C) (A+B) A}</math>
  
Note that if <math>A + B > 10</math> or <math>B + C > 10</math>, then the symmetry will be broken by carried 1s
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Note that if <math>A + B >= 10</math> or <math>B + C >= 10</math>, then the symmetry will be broken by carried 1s
  
 
Simply count the combinations of <math>(A, B, C)</math> for which <math>A + B < 10</math> and <math>B + C < 10</math>
 
Simply count the combinations of <math>(A, B, C)</math> for which <math>A + B < 10</math> and <math>B + C < 10</math>
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<math>54 + 52 + 49 + 45 + 40 + 34 + 27 + 19 + 10 = 330</math>
 
<math>54 + 52 + 49 + 45 + 40 + 34 + 27 + 19 + 10 = 330</math>
  
6-digit palindromes are of the form <math>XYZZYX</math>, and the first digit cannot be a zero, so there are <math>9 * 10 * 10 = 900</math> combinations of <math>(X, Y, Z)</math>
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6-digit palindromes are of the form <math>XYZZYX</math>, and the first digit cannot be a zero, so there are <math>9 \cdot 10 \cdot 10 = 900</math> combinations of <math>(X, Y, Z)</math>
  
 
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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===Note===
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You can more easily count the number of triples <math>(A, B, C)</math> by noticing that there are <math>9 - B</math> possible values for <math>A</math> and <math>10 - B</math> possible values for <math>C</math> once <math>B</math> is chosen. Summing over all <math>B</math>, the number is <cmath>9\cdot 10 + 8\cdot 9 + \ldots + 1\cdot 2 = 2\left(\binom{2}{2} + \binom{3}{2} + \ldots + \binom{10}{2}\right).</cmath>
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By the hockey-stick identity, it is <math>2\binom{11}{3} = 330</math>.
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~rayfish
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== Solution 2 ==
 
== 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>.
 
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>.
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==Video Solution by Richard Rusczyk==
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https://artofproblemsolving.com/videos/amc/2013amc12a/361
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 +
~dolphin7
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== 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 15:51, 28 August 2021

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 1

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

$\overline{A (A+B) (B+C) (B+C) (A+B) A}$

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 $9 \cdot 10  \cdot 10 = 900$ combinations of $(X, Y, Z)$

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

Note

You can more easily count the number of triples $(A, B, C)$ by noticing that there are $9 - B$ possible values for $A$ and $10 - B$ possible values for $C$ once $B$ is chosen. Summing over all $B$, the number is \[9\cdot 10 + 8\cdot 9 + \ldots + 1\cdot 2 = 2\left(\binom{2}{2} + \binom{3}{2} + \ldots + \binom{10}{2}\right).\] By the hockey-stick identity, it is $2\binom{11}{3} = 330$.

~rayfish

Solution 2

Let the palindrome be the form in the previous solution which is $XYZZYX$. It doesn't matter what $Z$ is because it only affects the middle digit. There are $90$ ways to pick $X$ and $Y$, and the only answer choice with denominator a factor of $90$ is $\boxed{\textbf{(E)} \ \frac{11}{30}}$.

Video Solution by Richard Rusczyk

https://artofproblemsolving.com/videos/amc/2013amc12a/361

~dolphin7

See also

2013 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 21
Followed by
Problem 23
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All AMC 12 Problems and Solutions

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