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Difference between revisions of "2010 AMC 10B Problems/Problem 21"

(→‎Solution)
(→‎Solution: Simplified the solution and made it more rigorous)
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==Solution==  
 
==Solution==  
 +
The palindromes can be expressed as: <math>1000x+100y+10y+x </math> (since it is a four digit palindrome, it must be of the form <math>xyyx</math> , where x and y are integers from <math>[1, 9]</math> and <math>[0, 9]</math>, respectively.)
  
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It is known that the palindromes can be expressed as: <math>1000x+100y+10y+x </math> (since it is a four digit palindrome, it must be of the form <math>xyyx</math> , where x and y are integers from [1,9] and [0,9], respectively.)
 
−
Using the divisibility rules of 7, <math>100x+10y+y-2x</math> = <math>98x+11y \equiv 0 \pmod 7</math>
 
  
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Since <math>98 \equiv 0 \pmod 7</math>, The <math>98x</math> is now irrelelvant.
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We simplify this to <math>1001x+110y</math>.
 +
 
 +
Because the question asks for it to be divisible by 7,
 +
 
 +
We express it as <math>1001x+110y \equiv 0 \pmod 7</math>.
 +
 
 +
 
 +
Because <math>1001 \equiv 0 \pmod 7</math>,
 +
 
 +
We can substitute <math>1001</math> for <math>0</math>
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 +
We are left with <math>110y \equiv 0 \pmod 7</math>
 +
 
 +
 
 +
Since <math>110 \equiv 5 \pmod 7</math> we can simplify the <math>110</math> in the expression to
 +
 
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<math>5y \equiv 0 \pmod 7</math>.
 +
 
 +
 
 +
In order for this to be true, <math>y \equiv 0 \pmod 7</math> must also be true.  
 +
 
  
 
Thus we solve:  
 
Thus we solve:  
  
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<math>11y \equiv 0 \pmod 7</math>
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<math>y \equiv 0 \pmod 7</math>
  
 
Which has two solutions: <math>0</math> and <math>7</math>
 
Which has two solutions: <math>0</math> and <math>7</math>
  
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There are thus two options for <math>y</math> out of the 10, so <math>2/10 = \boxed{\textbf{(E)}\ \frac15}</math>
+
There are thus two options for <math>y</math> out of the 10, so the answer is <math>2/10 = \boxed{\textbf{(E)}\ \frac15}</math>
  
 
== See also ==
 
== See also ==
 
{{AMC10 box|year=2010|ab=B|num-b=20|num-a=22}}
 
{{AMC10 box|year=2010|ab=B|num-b=20|num-a=22}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 15:40, 30 April 2016

Problem 21

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

The palindromes can be expressed as: $1000x+100y+10y+x$ (since it is a four digit palindrome, it must be of the form $xyyx$ , where x and y are integers from $[1, 9]$ and $[0, 9]$, respectively.)


We simplify this to $1001x+110y$.

Because the question asks for it to be divisible by 7,

We express it as $1001x+110y \equiv 0 \pmod 7$.


Because $1001 \equiv 0 \pmod 7$,

We can substitute $1001$ for $0$

We are left with $110y \equiv 0 \pmod 7$


Since $110 \equiv 5 \pmod 7$ we can simplify the $110$ in the expression to

$5y \equiv 0 \pmod 7$.


In order for this to be true, $y \equiv 0 \pmod 7$ must also be true.


Thus we solve:

$y \equiv 0 \pmod 7$

Which has two solutions: $0$ and $7$

There are thus two options for $y$ out of the 10, so the answer is $2/10 = \boxed{\textbf{(E)}\ \frac15}$

See also

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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