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

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== Solution ==
 
== Solution ==
To be a three digit number in base-10: 
 
  
<math>10^{2} \leq n \leq 10^{3}-1</math>
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The smallest base-11 number that has 3 digits in base-10 is <math>100_{11}</math> which is <math>121_{10}</math>.
  
<math>100 \leq n \leq 999</math>  
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The largest number in base-9 that has 3 digits in base-10 is <math>8\cdot9^2+8\cdot9^1+8\cdot9^0=888_{9}=728_{10}</math>
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Alternatively, you can do <math>9^3-1</math>
  
Thus there are <math>900</math> three-digit numbers in base-10
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The smallest number in base-9 that has 3 digits in base-10 is <math>1\cdot9^2+2\cdot9^1+1\cdot9^0=121_{9}=100_{10}</math>
  
To be a three-digit number in base-9
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Hence, all numbers that will have 3 digits in base-9, 10, and 11 will be between <math>728_{10}</math> and <math>121_{10}</math>, thus the total amount of numbers that will have 3 digits in base-9, 10, and 11 is <math>728-121+1=608</math>
  
<math>9^{2} \leq n \leq 9^{3}-1</math>
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There are 900 possible 3 digit numbers in base 10.
  
<math>81 \leq n \leq 728</math>
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Hence, the answer is <math>\frac{608}{900}\approx .675 \approx \boxed{0.7}</math>
  
To be a three-digit number in base-11: 
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==Video Solution==
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https://youtu.be/YaV5oanhAlU
  
<math>11^{2} \leq n \leq 11^{3}-1</math>
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~IceMatrix
 
 
<math>121 \leq n \leq 1330</math>
 
 
 
So, <math>121 \leq n \leq 728</math>
 
 
 
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>.
 
  
 
== See Also ==
 
== See Also ==
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[[Category:Introductory Number Theory Problems]]
 
[[Category:Introductory Number Theory Problems]]
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{{MAA Notice}}

Revision as of 19:19, 27 January 2020

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

The smallest base-11 number that has 3 digits in base-10 is $100_{11}$ which is $121_{10}$.

The largest number in base-9 that has 3 digits in base-10 is $8\cdot9^2+8\cdot9^1+8\cdot9^0=888_{9}=728_{10}$ Alternatively, you can do $9^3-1$

The smallest number in base-9 that has 3 digits in base-10 is $1\cdot9^2+2\cdot9^1+1\cdot9^0=121_{9}=100_{10}$

Hence, all numbers that will have 3 digits in base-9, 10, and 11 will be between $728_{10}$ and $121_{10}$, thus the total amount of numbers that will have 3 digits in base-9, 10, and 11 is $728-121+1=608$

There are 900 possible 3 digit numbers in base 10.

Hence, the answer is $\frac{608}{900}\approx .675 \approx \boxed{0.7}$

Video Solution

https://youtu.be/YaV5oanhAlU

~IceMatrix

See Also

2003 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 19
Followed by
Problem 21
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All AMC 10 Problems and Solutions

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