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Difference between revisions of "2013 AMC 12A Problems/Problem 22"
 
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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.
 
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=0 A=1-9 C=0-9,
−
B=1 A=1-8 C=0-8
+
B=1 A=1-8 C=0-8,
−
B=2 A=1-7 C=0-7
+
B=2 A=1-7 C=0-7,
−
B=3 A=1-6 C=0-6
+
B=3 A=1-6 C=0-6,
−
B=4 A=1-5 C=0-5
+
B=4 A=1-5 C=0-5,
−
B=5 A=1-4 C=0-4
+
B=5 A=1-4 C=0-4,
−
B=6 A=1-3 C=0-3
+
B=6 A=1-3 C=0-3,
−
B=7 A=1-2 C=0-2
+
B=7 A=1-2 C=0-2,
−
B=8 A=1  C=0-1
+
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)
+
 
 +
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).
 
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).
Line 64: Line 65:
 
Possibilities--> which are A=1-9, B=0-9, and C=0-9 --> 9 x 10 x 10 = 900
 
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
+
Successful Outcomes--> which are from the casework above --> <cmath> (9 \cdot 10) + (8 \cdot 9) + (7 \cdot 8) + (6 \cdot 7) + (5 \cdot 6) + (4 \cdot 5) + (3 \cdot 4) + (2 \cdot 3) + (1 \cdot 2)</cmath> = <math>90 + 72 + 56 + 42 + 30 + 20 + 12 + 6 + 2 = 330</math> 
  
−
<math>\boxed{\textbf{(E)} \ \frac{330}{900}}</math> = <math>\boxed{\textbf{(E)} \ \frac{11}{30}}</math>
+
330/900 = <math>\boxed{\textbf{(E)} \ \frac{11}{30}}</math>
  
 
~Solution by Daily Dose of Math (Thesmartgreekmathdude)
 
~Solution by Daily Dose of Math (Thesmartgreekmathdude)

Latest revision as of 14:00, 9 August 2024

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 --> \[(9 \cdot 10) + (8 \cdot 9) + (7 \cdot 8) + (6 \cdot 7) + (5 \cdot 6) + (4 \cdot 5) + (3 \cdot 4) + (2 \cdot 3) + (1 \cdot 2)\] = $90 + 72 + 56 + 42 + 30 + 20 + 12 + 6 + 2 = 330$

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

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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