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Difference between revisions of "2002 AMC 12P Problems/Problem 15"

(Created page with "== Problem == How many positive integers <math>b</math> have the property that <math>\log_{b} 729</math> is a positive integer? <math> \mathrm{(A) \ 0 } \qquad \mathrm{(B...")
 
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{{duplicate|[[2002 AMC 12P Problems|2002 AMC 12P #15]] and [[2002 AMC 10P Problems|2002 AMC 10P #17]]}}
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== Problem ==
 
== Problem ==
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How many positive [[integer]]s <math>b</math> have the property that <math>\log_{b} 729</math> is a positive integer?
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There are <math>1001</math> red marbles and <math>1001</math> black marbles in a box. Let <math>P_s</math> be the probability that two marbles drawn at random from the box are the same color, and let <math>P_d</math> be the probability that they are different colors. Find <math>|P_s-P_d|.</math>
  
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<math> \mathrm{(A) \ 0 } \qquad \mathrm{(B) \ 1 } \qquad \mathrm{(C) \ 2 } \qquad \mathrm{(D) \ 3 } \qquad \mathrm{(E) \ 4 } </math>
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<math>
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\text{(A) }0
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\qquad
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\text{(B) }\frac{1}{2002}
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\qquad
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\text{(C) }\frac{1}{2001}
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\qquad
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\text{(D) }\frac {2}{2001}
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\qquad
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\text{(E) }\frac{1}{1000}
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</math>
  
 
== Solution ==
 
== Solution ==
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If <math>\log_{b} 729 = n</math>, then <math>b^n = 729</math>. Since <math>729 = 3^6</math>, <math>b</math> must be <math>3</math> to some [[factor]] of 6. Thus, there are four (3, 9, 27, 729) possible values of <math>b \Longrightarrow \boxed{\mathrm{E}}</math>.
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First we find the value of <math>P_s</math>. Note that whatever color we choose on our first marble, there are exactly <math>1000</math> of <math>2001</math> marbles remaining that match that color. Therefore, <math>P_s = \frac {1000}{2001}</math>.
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Now we find the value of <math>P_d</math>. Again, the actual color of the first marble does not matter, since there are always exactly <math>1001</math> of <math>2001</math> marbles remaining that don't match that color. Therefore, <math>P_d = \frac{1001}{2001}</math>.
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The value of <math>|P_s - P_d|</math> is therefore <math>\frac {|1000-1001|}{2001} = \boxed {\text{(C) }\frac{1}{2001}}</math>.
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~Minor edits from [[User:Astro2010|Astro2010]]~
  
 
== See also ==
 
== See also ==
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{{AMC12 box|year=2000|num-b=6|num-a=8}}
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{{AMC10 box|year=2002|ab=P|num-b=16|num-a=18}}
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{{AMC12 box|year=2002|ab=P|num-b=14|num-a=16}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 15:59, 14 August 2024

The following problem is from both the 2002 AMC 12P #15 and 2002 AMC 10P #17, so both problems redirect to this page.

Problem

There are $1001$ red marbles and $1001$ black marbles in a box. Let $P_s$ be the probability that two marbles drawn at random from the box are the same color, and let $P_d$ be the probability that they are different colors. Find $|P_s-P_d|.$

$\text{(A) }0 \qquad \text{(B) }\frac{1}{2002} \qquad \text{(C) }\frac{1}{2001} \qquad \text{(D) }\frac {2}{2001} \qquad \text{(E) }\frac{1}{1000}$

Solution

First we find the value of $P_s$. Note that whatever color we choose on our first marble, there are exactly $1000$ of $2001$ marbles remaining that match that color. Therefore, $P_s = \frac {1000}{2001}$.

Now we find the value of $P_d$. Again, the actual color of the first marble does not matter, since there are always exactly $1001$ of $2001$ marbles remaining that don't match that color. Therefore, $P_d = \frac{1001}{2001}$.

The value of $|P_s - P_d|$ is therefore $\frac {|1000-1001|}{2001} = \boxed {\text{(C) }\frac{1}{2001}}$.


~Minor edits from Astro2010~

See also

2002 AMC 10P (Problems • Answer Key • Resources)
Preceded by
Problem 16 		Followed by
Problem 18
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
2002 AMC 12P (Problems • Answer Key • Resources)
Preceded by
Problem 14 		Followed by
Problem 16
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 12 Problems and Solutions

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