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

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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>
  
−
<math> \mathrm{(A) \ 0 } \qquad \mathrm{(B) \ 1 } \qquad \mathrm{(C) \ 2 } \qquad \mathrm{(D) \ 3 } \qquad \mathrm{(E) \ 4 } </math>
+
<math>
 +
\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}
 +
</math>
  
 
== Solution ==
 
== Solution ==

Revision as of 23:51, 29 December 2023

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

If $\log_{b} 729 = n$, then $b^n = 729$. Since $729 = 3^6$, $b$ must be $3$ to some factor of 6. Thus, there are four (3, 9, 27, 729) possible values of $b \Longrightarrow \boxed{\mathrm{E}}$.

See also

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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