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Difference between revisions of "1985 AHSME Problems/Problem 24"

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==Problem==
 
==Problem==
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A non-zero [[digit]] is chosen in such a way that the probability of choosing digit <math> d </math> is <math> \log_{10}{(d+1)}-\log_{10}{d} </math>. The probability that the digit <math> 2 </math> is chosen is exactly <math> \frac{1}{2} </math> the probability that the digit chosen is in the set
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A non-zero digit is chosen in such a way that the probability of choosing digit <math>d</math> is <math>\log_{10}{(d+1)}-\log_{10}{d}</math>. The probability that the digit <math>2</math> is chosen is exactly <math>1/2</math> the probability that the digit chosen is in the set
  
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<math> \mathrm{(A)\ } \{2, 3\} \qquad \mathrm{(B) \ }\{3, 4\} \qquad \mathrm{(C) \  } \{4, 5, 6, 7, 8\} \qquad \mathrm{(D) \  } \{5, 6, 7, 8, 9\} \qquad \mathrm{(E) \  }\{4, 5, 6, 7, 8, 9\} </math>
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<math> \mathrm{(A)\ } \{2,3\} \qquad \mathrm{(B) \ }\{3,4\} \qquad \mathrm{(C) \  } \{4,5,6,7,8\} \qquad \mathrm{(D) \  } \{5,6,7,8,9\} \qquad \mathrm{(E) \  }\{4,5,6,7,8,9\} </math>
  
 
==Solution==
 
==Solution==
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Notice that <math> \log_{10}{(d+1)}-\log_{10}{d}=\log_{10}{\frac{d+1}{d}} </math>. Therefore, the probability of choosing <math> 2 </math> is <math> \log_{10}{\frac{3}{2}} </math>. The probability that the digit is chosen out of the set is twice that, <math> 2\log_{10}{\frac{3}{2}}=\log_{10}{\left(\frac{3}{2}\right)^2}=\log_{10}{\frac{9}{4}} </math>.
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We have <math>\log_{10}{(d+1)}-\log_{10}{d} = \log_{10}{\left(\frac{d+1}{d}\right)}</math>, so the probability of choosing <math>2</math> is <math>\log_{10}{\left(\frac{3}{2}\right)}</math>. The probability that the digit chosen is in the set must therefore be <cmath>\begin{align*}2\log_{10}{\left(\frac{3}{2}\right)} = &\log_{10}{\left(\left(\frac{3}{2}\right)^2\right)} \\ = &\log_{10}{\left(\frac{9}{4}\right)} \\ = &\log_{10}{9}-\log_{10}{4} \\ = &\left(\log_{10}{9}-\log_{10}{8}\right)+\left(\log_{10}{8}-\log_{10}{7}\right)+\left(\log_{10}{7}-\log_{10}{6}\right) \\ &+\left(\log_{10}{6}-\log_{10}{5}\right)+\left(\log_{10}{5}-\log_{10}{4}\right),\end{align*}</cmath>
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which, by definition, is the probability that the digit chosen is in the set <math>\boxed{\text{(C)} \ \{4,5,6,7,8\}}</math>.
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<math> \log_{10}{\frac{9}{4}}=\log_{10}{9}-\log_{10}{4} </math>
 
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<math> =(\log_{10}{9}-\log_{10}{8})+(\log_{10}{8}-\log_{10}{7})+(\log_{10}{7}-\log_{10}{6})+(\log_{10}{6}-\log_{10}{5})+(\log_{10}{5}-\log_{10}{4}) </math>
 
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which is the probability that the digit is from the set <math> \{4, 5, 6, 7, 8\}, \boxed{\text{C}} </math>.
 
  
 
==See Also==
 
==See Also==
 
{{AHSME box|year=1985|num-b=23|num-a=25}}
 
{{AHSME box|year=1985|num-b=23|num-a=25}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 00:03, 20 March 2024

Problem

A non-zero digit is chosen in such a way that the probability of choosing digit $d$ is $\log_{10}{(d+1)}-\log_{10}{d}$. The probability that the digit $2$ is chosen is exactly $1/2$ the probability that the digit chosen is in the set

$\mathrm{(A)\ } \{2,3\} \qquad \mathrm{(B) \ }\{3,4\} \qquad \mathrm{(C) \ } \{4,5,6,7,8\} \qquad \mathrm{(D) \ } \{5,6,7,8,9\} \qquad \mathrm{(E) \ }\{4,5,6,7,8,9\}$

Solution

We have $\log_{10}{(d+1)}-\log_{10}{d} = \log_{10}{\left(\frac{d+1}{d}\right)}$, so the probability of choosing $2$ is $\log_{10}{\left(\frac{3}{2}\right)}$. The probability that the digit chosen is in the set must therefore be \begin{align*}2\log_{10}{\left(\frac{3}{2}\right)} = &\log_{10}{\left(\left(\frac{3}{2}\right)^2\right)} \\ = &\log_{10}{\left(\frac{9}{4}\right)} \\ = &\log_{10}{9}-\log_{10}{4} \\ = &\left(\log_{10}{9}-\log_{10}{8}\right)+\left(\log_{10}{8}-\log_{10}{7}\right)+\left(\log_{10}{7}-\log_{10}{6}\right) \\ &+\left(\log_{10}{6}-\log_{10}{5}\right)+\left(\log_{10}{5}-\log_{10}{4}\right),\end{align*} which, by definition, is the probability that the digit chosen is in the set $\boxed{\text{(C)} \ \{4,5,6,7,8\}}$.

See Also

1985 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 23 		Followed by
Problem 25
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 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

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