Difference between revisions of "2003 AMC 10A Problems/Problem 20"

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Thus, there are <math>608</math> base-10 three-digit numbers that are three digit numbers in base-9 and base-11.  
 
Thus, there are <math>608</math> base-10 three-digit numbers that are three digit numbers in base-9 and base-11.  
  
Therefore the desired probability is <math>\frac{608}{900}\approx 0.7 \Rightarrow E</math>.  
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Therefore the desired probability is <math>\frac{608}{900}\approx 0.7 \Rightarrow\boxed{\mathrm{(E)}\ 0.7}</math>.
  
 
== See Also ==
 
== See Also ==

Revision as of 23:50, 16 July 2014

Problem 20

A base-10 three digit number $n$ is selected at random. Which of the following is closest to the probability that the base-9 representation and the base-11 representation of $n$ are both three-digit numerals?

$\mathrm{(A) \ } 0.3\qquad \mathrm{(B) \ } 0.4\qquad \mathrm{(C) \ } 0.5\qquad \mathrm{(D) \ } 0.6\qquad \mathrm{(E) \ } 0.7$

Solution

To be a three digit number in base-10:

$10^{2} \leq n \leq 10^{3}-1$

$100 \leq n \leq 999$

Thus there are $900$ three-digit numbers in base-10

To be a three-digit number in base-9:

$9^{2} \leq n \leq 9^{3}-1$

$81 \leq n \leq 728$

To be a three-digit number in base-11:

$11^{2} \leq n \leq 11^{3}-1$

$121 \leq n \leq 1330$

So, $121 \leq n \leq 728$

Thus, there are $608$ base-10 three-digit numbers that are three digit numbers in base-9 and base-11.

Therefore the desired probability is $\frac{608}{900}\approx 0.7 \Rightarrow\boxed{\mathrm{(E)}\ 0.7}$.

See Also

2003 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 19
Followed by
Problem 21
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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