Difference between revisions of "2000 AMC 12 Problems/Problem 7"

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== Solution ==
 
== Solution ==
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 \mathrm{E}</math>.
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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>.
  
 
== See also ==
 
== See also ==
 
{{AMC12 box|year=2000|num-b=6|num-a=8}}
 
{{AMC12 box|year=2000|num-b=6|num-a=8}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 02:05, 30 December 2020

Problem

How many positive integers $\displaystyle b$ have the property that $\displaystyle \log_{b} 729$ is a positive integer?

$\mathrm{(A) \ 0 } \qquad \mathrm{(B) \ 1 } \qquad \mathrm{(C) \ 2 } \qquad \mathrm{(D) \ 3 } \qquad \mathrm{(E) \ 4 }$

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

2000 AMC 12 (ProblemsAnswer KeyResources)
Preceded by
Problem 6
Followed by
Problem 8
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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