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

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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 xyyx, where x and y are positive integers from [0,9].  
 
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].  
−
Using the divisibility rules of 7, 100x+10y+y-2x = 98x+11y \equiv 0 \mod{7}
+
Using the divisibility rules of 7, 100x+10y+y-2x = 98x+11y  
  
 
The 98x is now irrelelvant
 
The 98x is now irrelelvant
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Thus we solve:  
 
Thus we solve:  
  
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11y \equiv 0 \mod{7}
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11y  
  
 
Which has two solutions: 0 and 7
 
Which has two solutions: 0 and 7

Revision as of 22:36, 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

The 98x is now irrelelvant

Thus we solve:

11y

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 (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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