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 $k$ is $12^{12}$ the least common multiple of the positive integers $6^6$, $8^8$, and $k$?

Solution

It is evident that $k$ has only 2s and 3s in its prime factorization, or $k = 2^a3^b$.

  • $6^6 = 2^6\cdot3^6$
  • $8^8 = 2^{24}$
  • $12^{12} = 2^{24}\cdot3^{12}$

The LCM of any numbers an be found by writing out their factorizations and taking the greatest power for each factor. $[6^6,8^8] = 2^{24}3^6$. Therefore $12^{12} = 2^{24}\cdot3^{12} = [2^{24}3^6,2^a3^b] = 2^{\max(24,a)}3^{\max(6,b)}$, and $b = 12$. Since $0 \le a \le 24$, there are $\boxed{025}$ values of $k$.

See also

1998 AIME (ProblemsAnswer KeyResources)
Preceded by
First question
Followed by
Problem 2
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All AIME Problems and Solutions
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