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

(Another Solution)
(Another Solution)
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==Another Solution ==  
 
==Another Solution ==  
  
It is known that the palindromes can be expressed as: <math>1000x+100y+10y+x </math>(as it is a four digit palindrome it must be of the form xyyx, where x and y are positive integers from [0,9].  
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It is known that the palindromes can be expressed as: <math>1000x+100y+10y+x </math> (as it is a four digit palindrome it must be of the form <math>xyyx</math> , where x and y are positive integers from [0,9].  
Using the divisibility rules of 7, 100x+10y+y-2x = <math>98x+11y \equiv 0 \pmod 7</math>
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Using the divisibility rules of 7, <math>100x+10y+y-2x</math> = <math>98x+11y \equiv 0 \pmod 7</math>
  
 
The 98x is now irrelelvant
 
The 98x is now irrelelvant

Revision as of 23:38, 21 January 2014

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

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 $\frac{18}{90}=\boxed{\textbf{(E)}\ \frac15}$

Another Solution

It is known that the palindromes can be expressed as: $1000x+100y+10y+x$ (as it is a four digit palindrome it must be of the form $xyyx$ , where x and y are positive integers from [0,9]. Using the divisibility rules of 7, $100x+10y+y-2x$ = $98x+11y \equiv 0 \pmod 7$

The 98x is now irrelelvant

Thus we solve:

$11y \equiv 0 \pmod 7$

Which has two solutions: 0 and 7

There are thus, two options for y out of the 10, so 2/10 = 1/5.

See also

2010 AMC 10B (ProblemsAnswer KeyResources)
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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