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

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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

@@ Line 31: / Line 31: @@
 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>
 By the hockey-stick identity, it is <math>2\binom{11}{3} = 330</math>.
+~rayfish
 == Solution 2 ==

2013 AMC 12A (Problems • Answer Key • Resources)
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