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 \boxed{\mathrm{E}}</math>. | 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>. | ||
| + | |||
| + | == Video Solution by Power Solve == | ||
| + | https://youtu.be/p3bwMuK1RUM?si=KWRsbwzVa2Z8MN69&t=157 | ||
== 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}} | ||
Latest revision as of 17:40, 25 August 2025
Problem
How many positive integers 
have the property that 
is a positive integer?
Solution
If 
, then 
. Since 
, must be 
to some factor of 6. Thus, there are four (3, 9, 27, 729) possible values of 
.
Video Solution by Power Solve
https://youtu.be/p3bwMuK1RUM?si=KWRsbwzVa2Z8MN69&t=157
See also
| 2000 AMC 12 (Problems • Answer Key • Resources) | |
| 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 | |
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
