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Difference between revisions of "2010 AMC 12B Problems/Problem 11"
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== Problem 11 ==
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{{duplicate|[[2010 AMC 12B Problems|2010 AMC 12B #11]] and [[2010 AMC 10B Problems|2010 AMC 10B #21]]}}
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== Problem ==
 
A palindrome between <math>1000</math> and <math>10,000</math> is chosen at random. What is the probability that it is divisible by <math>7</math>?
 
A palindrome between <math>1000</math> and <math>10,000</math> is chosen at random. What is the probability that it is divisible by <math>7</math>?
  
 
<math>\textbf{(A)}\ \dfrac{1}{10} \qquad \textbf{(B)}\ \dfrac{1}{9} \qquad \textbf{(C)}\ \dfrac{1}{7} \qquad \textbf{(D)}\ \dfrac{1}{6} \qquad \textbf{(E)}\ \dfrac{1}{5}</math>
 
<math>\textbf{(A)}\ \dfrac{1}{10} \qquad \textbf{(B)}\ \dfrac{1}{9} \qquad \textbf{(C)}\ \dfrac{1}{7} \qquad \textbf{(D)}\ \dfrac{1}{6} \qquad \textbf{(E)}\ \dfrac{1}{5}</math>
  
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== Solution ==
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== Solution 1==
 
View the palindrome as some number with form (decimal representation):
 
View the palindrome as some number with form (decimal representation):
 
<math>a_3 \cdot 10^3 + a_2 \cdot 10^2 + a_1 \cdot 10 + a_0</math>. But because the number is a palindrome, <math>a_3 = a_0, a_2 = a_1</math>. Recombining this yields <math>1001a_3 + 110a_2</math>. 1001 is divisible by 7, which means that as long as <math>a_2 = 0</math>, the palindrome will be divisible by 7. This yields 9 palindromes out of 90 (<math>9 \cdot 10</math>) possibilities for palindromes. However, if <math>a_2 = 7</math>, then this gives another case in which the palindrome is divisible by 7. This adds another 9 palindromes to the list, bringing our total to <math>18/90 = \boxed {\frac{1}{5} } = \boxed {E}</math>
 
<math>a_3 \cdot 10^3 + a_2 \cdot 10^2 + a_1 \cdot 10 + a_0</math>. But because the number is a palindrome, <math>a_3 = a_0, a_2 = a_1</math>. Recombining this yields <math>1001a_3 + 110a_2</math>. 1001 is divisible by 7, which means that as long as <math>a_2 = 0</math>, the palindrome will be divisible by 7. This yields 9 palindromes out of 90 (<math>9 \cdot 10</math>) possibilities for palindromes. However, if <math>a_2 = 7</math>, then this gives another case in which the palindrome is divisible by 7. This adds another 9 palindromes to the list, bringing our total to <math>18/90 = \boxed {\frac{1}{5} } = \boxed {E}</math>
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== Solution (Divisibility Rules) ==
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We can notice the palindrome is of the form <math>\overline{abba}</math>. Then, by the divisibility rule of <math>7</math>, <math>7</math> must divide <cmath>100a+11b-2a = 98a+11b.</cmath> This nicely simplifies to the fact that <math>7 \mid 4b,</math> so <math>b</math> is clearly <math>0</math> or <math>7</math>. This gives us <math>9 \cdot 2</math> total choices for the palindrome divisible by <math>7</math>, divided by <math>9 \cdot 10</math> total choices for <math>\overline{abba}</math>, giving us an answer of <math>\boxed{\text{(E)}} \ \dfrac{1}{5}</math>.
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~icecreamrolls8
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== Addendum (Alternate) ==
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<math>7\mid 1001a^3+110b^2</math> and <math>1001 \equiv 0 \pmod 7</math>. Knowing that <math>a</math> does not factor (pun intended) into the problem, note 110's prime factorization and <math>7\mid b</math>. There are only 10 possible digits for <math>b</math>, 0 through 9, but <math>7\mid b</math> only holds if <math>b=0, 7</math>. This is 2 of the 10 digits, so <math>\frac{2}{10}=\boxed{\textbf{(E) }\frac{1}{5}}</math>
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~BJHHar
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==Video Solution==
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https://youtu.be/ZfnxbpdFKjU
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~IceMatrix
  
 
== See also ==
 
== See also ==
 
{{AMC12 box|year=2010|num-b=10|num-a=12|ab=B}}
 
{{AMC12 box|year=2010|num-b=10|num-a=12|ab=B}}
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{{AMC10 box|year=2010|num-b=20|num-a=22|ab=B}}
  
 
[[Category:Introductory Combinatorics Problems]]
 
[[Category:Introductory Combinatorics Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 17:58, 22 June 2024

The following problem is from both the 2010 AMC 12B #11 and 2010 AMC 10B #21, so both problems redirect to this page.

Problem

A palindrome between $1000$ and $10,000$ is chosen at random. What is the probability that it is divisible by $7$?

$\textbf{(A)}\ \dfrac{1}{10} \qquad \textbf{(B)}\ \dfrac{1}{9} \qquad \textbf{(C)}\ \dfrac{1}{7} \qquad \textbf{(D)}\ \dfrac{1}{6} \qquad \textbf{(E)}\ \dfrac{1}{5}$

Solution 1

View the palindrome as some number with form (decimal representation): $a_3 \cdot 10^3 + a_2 \cdot 10^2 + a_1 \cdot 10 + a_0$. But because the number is a palindrome, $a_3 = a_0, a_2 = a_1$. Recombining this yields $1001a_3 + 110a_2$. 1001 is divisible by 7, which means that as long as $a_2 = 0$, the palindrome will be divisible by 7. This yields 9 palindromes out of 90 ($9 \cdot 10$) possibilities for palindromes. However, if $a_2 = 7$, then this gives another case in which the palindrome is divisible by 7. This adds another 9 palindromes to the list, bringing our total to $18/90 = \boxed {\frac{1}{5} } = \boxed {E}$

Solution (Divisibility Rules)

We can notice the palindrome is of the form $\overline{abba}$. Then, by the divisibility rule of $7$, $7$ must divide \[100a+11b-2a = 98a+11b.\] This nicely simplifies to the fact that $7 \mid 4b,$ so $b$ is clearly $0$ or $7$. This gives us $9 \cdot 2$ total choices for the palindrome divisible by $7$, divided by $9 \cdot 10$ total choices for $\overline{abba}$, giving us an answer of $\boxed{\text{(E)}} \ \dfrac{1}{5}$.

~icecreamrolls8

Addendum (Alternate)

$7\mid 1001a^3+110b^2$ and $1001 \equiv 0 \pmod 7$. Knowing that $a$ does not factor (pun intended) into the problem, note 110's prime factorization and $7\mid b$. There are only 10 possible digits for $b$, 0 through 9, but $7\mid b$ only holds if $b=0, 7$. This is 2 of the 10 digits, so $\frac{2}{10}=\boxed{\textbf{(E) }\frac{1}{5}}$

~BJHHar

Video Solution

https://youtu.be/ZfnxbpdFKjU

~IceMatrix

See also

2010 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 10 		Followed by
Problem 12
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
2010 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 20 		Followed by
Problem 22
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 10 Problems and Solutions

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