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Difference between revisions of "1998 AIME Problems/Problem 1"

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== Problem ==
 
== Problem ==
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For how many values of <math>\displaystyle k</math> is <math>\displaystyle 12^{12}</math> the [[least common multiple]] of the positive integers <math>6^6</math> and <math>8^8</math>?  
+
For how many values of <math>k</math> is <math>12^{12}</math> the [[least common multiple]] of the positive integers <math>6^6</math>, <math>8^8</math>, and <math>k</math>?  
 
== Solution ==
 
== Solution ==
−
It is evident that <math>\displaystyle k</math> has only 2s and 3s in its prime factorization, or <math>\displaystyle k = 2^a3^b</math>.  
+
It is evident that <math>k</math> has only 2s and 3s in its prime factorization, or <math>k = 2^a3^b</math>.  
  
−
*<math>\displaystyle 6^6 = 2^6\cdot3^6</math>
+
*<math>6^6 = 2^6\cdot3^6</math>
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*<math>\displaystyle 8^8 = 2^24</math>
+
*<math>8^8 = 2^{24}</math>
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*<math>\displaystyle 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. The <math>\displaystyle lcm(6^6,8^8) = 2^{24}3^6</math>. Therefore <math>\displaystyle 12^{12} = 2^{24}\cdot3^{12} = lcm(2^{24}3^6,2^a3^b) = 2^{max(24,a)}3^{max(6,b)}</math>, and <math>b \displaystyle  = 12</math>.  Since <math>0 \le \displaystyle a \le 24</math>, there are <math>\displaystyle 025</math> values of <math>\displaystyle 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>025</math> values of <math>k</math>.
  
 
== See also ==
 
== See also ==

Revision as of 13:17, 19 November 2007

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 $025$ values of $k$.

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
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