Difference between revisions of "1998 AIME Problems/Problem 1"
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*<math>12^{12} = 2^{24}\cdot3^{12}</math> | *<math>12^{12} = 2^{24}\cdot3^{12}</math> | ||
| − | The [[LCM]] of any numbers an be found by writing out their factorizations and taking the greatest power for each factor. <math>[6^6,8^8] = 2^{24}3^6</math>. Therefore <math>12^{12} = 2^{24}\cdot3^{12} = [2^{24}3^6,2^a3^b] = 2^{\max(24,a)}3^{\max(6,b)}</math>, and <math>b = 12</math>. Since <math>0 \le a \le 24</math>, there are <math>025</math> values of <math>k</math>. | + | The [[LCM]] of any numbers an be found by writing out their factorizations and taking the greatest power for each factor. <math>[6^6,8^8] = 2^{24}3^6</math>. Therefore <math>12^{12} = 2^{24}\cdot3^{12} = [2^{24}3^6,2^a3^b] = 2^{\max(24,a)}3^{\max(6,b)}</math>, and <math>b = 12</math>. Since <math>0 \le a \le 24</math>, there are <math>\boxed{025}</math> values of <math>k</math>. |
== See also == | == See also == | ||
Revision as of 05:17, 29 June 2011
Problem
For how many values of 
is 
the least common multiple of the positive integers 
, 
, and ?
Solution
It is evident that has only 2s and 3s in its prime factorization, or 
.
The LCM of any numbers an be found by writing out their factorizations and taking the greatest power for each factor. ![$[6^6,8^8] = 2^{24}3^6$](http://latex.artofproblemsolving.com/a/3/7/a3784d45cf90b4f699775de824f0cb19215f08f0.png)
. Therefore ![$12^{12} = 2^{24}\cdot3^{12} = [2^{24}3^6,2^a3^b] = 2^{\max(24,a)}3^{\max(6,b)}$](http://latex.artofproblemsolving.com/1/a/7/1a76da6ba7634a57a636d73a6428f2f9713b4c47.png)
, and 
. Since 
, there are 
values of .
See also
| 1998 AIME (Problems • Answer Key • Resources) | ||
| Preceded by First question |
Followed by Problem 2 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
