Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution 1)
 
(22 intermediate revisions by 6 users not shown)
Line 9: Line 9:
 
By working backwards, we can multiply 5-digit palindromes <math>ABCBA</math> by <math>11</math>, giving a 6-digit palindrome:
 
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>
+
<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
 
Note that if <math>A + B >= 10</math> or <math>B + C >= 10</math>, then the symmetry will be broken by carried 1s
Line 23: Line 23:
 
<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>
+
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>
  
−
== Solution 2 ==
+
===Note===
 +
 
 +
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 (using the answer choices) ==
 
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>.
 +
 +
== 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 <math>ABCBA</math>, where the digits could be the same. If we multiply by <math>11</math> 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 --> <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> 
 +
 +
330/900 = <math>\boxed{\textbf{(E)} \ \frac{11}{30}}</math>
 +
 +
~Solution by Daily Dose of Math (Thesmartgreekmathdude)
 +
 +
==Video Solution by Richard Rusczyk==
 +
https://artofproblemsolving.com/videos/amc/2013amc12a/361
 +
 +
~dolphin7
 +
 
== 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}}

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2013_AMC_12A_Problems/Problem_22&oldid=224830"