2013 AMC 12A Problems/Problem 22

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 (using the answer choices)

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 $\boxed{\textbf{(E)} \ \frac{11}{30}}$ .

Solution 3 (extremely simple)

A six-digit palindrome, that when divided by 11, has to, if another palindrome, equal a five-digit palindrome. It cannot be a four-digit palindrome, otherwise, it would be too small. We now have a five-digit palindrome $ABCBA$ , where the digits could be the same. If we multiply by $11$ we are adding like this:

ABCBA0

+ ABCBA=

A, A+B, B+C, B+C, A+B, A

Commas separating the digits, We don't have to worry about A because it can be any digit, however, we have to do casework on B, as it affects all the digits.

B=0 A=1-9 C=0-9 B=1 A=1-8 C=0-8 B=2 A=1-7 C=0-7 B=3 A=1-6 C=0-6 B=4 A=1-5 C=0-5 B=5 A=1-4 C=0-4 B=6 A=1-3 C=0-3 B=7 A=1-2 C=0-2 B=8 A=1 C=0-1 B=9 does not work because A's only option is equal to 0 which makes it an invalid number (four-digit)

These solutions work as we ensure that all the digits do not carry into the following digits (e.g., The tens cannot carry into the hundreds or it would not be a palindrome).

For probability, as Richard Rusczyk says the total possibilities over the total successful outcomes Possibilities--> which are A=1-9, B=0-9, and C=0-9 --> 9 x 10 x 10 = 900

Successful Outcomes--> which are from the casework above --> (9x10) + (8x9) + (7x8) + (6x7) + (5x6) + (4x5) + (3x4) + (2x3) + (1x2) = 90 + 72 + 56 + 42 + 30 + 20 + 12 + 6 + 2 = 330

$\boxed{\textbf{(E)} \ \frac{330}{900}}$ = $\boxed{\textbf{(E)} \ \frac{11}{30}}$

~Solution by Daily Dose of Math (Thesmartgreekmathdude)

Video Solution by Richard Rusczyk

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

~dolphin7

2013 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 21	Followed by Problem 23
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