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

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<math> \mathrm{(A) \ } 0.3\qquad \mathrm{(B) \ } 0.4\qquad \mathrm{(C) \ } 0.5\qquad \mathrm{(D) \ } 0.6\qquad \mathrm{(E) \ } 0.7 </math>
 
<math> \mathrm{(A) \ } 0.3\qquad \mathrm{(B) \ } 0.4\qquad \mathrm{(C) \ } 0.5\qquad \mathrm{(D) \ } 0.6\qquad \mathrm{(E) \ } 0.7 </math>
  
== Solution ==
+
it's A lmaooooo xd lel
To be a three digit number in base-10: 
 
 
 
<math>10^{2} \leq n \leq 10^{3}-1</math>
 
 
 
<math>100 \leq n \leq 999</math>
 
 
 
Thus there are <math>900</math> three-digit numbers in base-10
 
 
 
To be a three-digit number in base-9: 
 
 
 
<math>9^{2} \leq n \leq 9^{3}-1</math>
 
 
 
<math>81 \leq n \leq 728</math>
 
 
 
To be a three-digit number in base-11: 
 
 
 
<math>11^{2} \leq n \leq 11^{3}-1</math>
 
 
 
<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\boxed{\mathrm{(E)}\ 0.7}</math>.
 
  
 
== See Also ==
 
== See Also ==

Revision as of 20:59, 5 June 2019

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?

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

it's A lmaooooo xd lel

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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