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

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== Problem20 ==
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== Problem 20 ==
 
A base-10 three digit number <math>n</math> is selected at random. Which of the following is closest to the probability that the base-9 representation and the base-11 representation of <math>n</math> are both three-digit numerals?  
 
A base-10 three digit number <math>n</math> is selected at random. Which of the following is closest to the probability that the base-9 representation and the base-11 representation of <math>n</math> are both three-digit numerals?  
  
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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.  
  
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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>.
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== Video Solution by OmegaLearn ==
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https://youtu.be/SCGzEOOICr4?t=596
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~ pi_is_3.14
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==Video Solution==
 +
https://youtu.be/YaV5oanhAlU
 +
 
 +
~IceMatrix
 +
 
 +
==Video Solution by WhyMath==
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https://youtu.be/-ei6Ni-jnlc
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 +
~savannahsolver
  
 
== See Also ==
 
== See Also ==
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[[Category:Introductory Number Theory Problems]]
 
[[Category:Introductory Number Theory Problems]]
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{{MAA Notice}}

Latest revision as of 00:04, 1 August 2023

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 $\frac{608}{900}\approx 0.7 \Rightarrow\boxed{\mathrm{(E)}\ 0.7}$.

Video Solution by OmegaLearn

https://youtu.be/SCGzEOOICr4?t=596

~ pi_is_3.14

Video Solution

https://youtu.be/YaV5oanhAlU

~IceMatrix

Video Solution by WhyMath

https://youtu.be/-ei6Ni-jnlc

~savannahsolver

See Also

2003 AMC 10A (Problems • Answer Key • Resources)
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Problem 19 		Followed by
Problem 21
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All AMC 10 Problems and Solutions

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