Difference between revisions of "2013 AMC 12A Problems/Problem 22"
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<math> \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}</math> | <math> \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}</math> | ||
− | ==Solution== | + | ==Solution 1== |
− | + | 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 |
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> | ||
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 | + | 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> | ||
+ | |||
+ | ===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>. | ||
+ | |||
+ | ==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 00:52, 16 April 2023
Contents
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 is chosen uniformly at random. What is the probability that
is also a palindrome?
Solution 1
By working backwards, we can multiply 5-digit palindromes by
, giving a 6-digit palindrome:
Note that if or
, then the symmetry will be broken by carried 1s
Simply count the combinations of for which
and
implies
possible
(0 through 8), for each of which there are
possible C, respectively. There are
valid palindromes when
implies
possible
(0 through 7), for each of which there are
possible C, respectively. There are
valid palindromes when
Following this pattern, the total is
6-digit palindromes are of the form , and the first digit cannot be a zero, so there are
combinations of
So, the probability is
Note
You can more easily count the number of triples by noticing that there are
possible values for
and
possible values for
once
is chosen. Summing over all
, the number is
By the hockey-stick identity, it is
.
~rayfish
Solution 2 (using the answer choices)
Let the palindrome be the form in the previous solution which is . It doesn't matter what
is because it only affects the middle digit. There are
ways to pick
and
, and the only answer choice with denominator a factor of
is
.
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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