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

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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{25}</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{25}</math> values of <math>k</math>.
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== Video Solution by OmegaLearn ==
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https://youtu.be/HISL2-N5NVg?t=2899
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 +
~ pi_is_3.14
  
 
== See also ==
 
== See also ==

Revision as of 03:08, 16 January 2023

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 $\boxed{25}$ values of $k$.

Video Solution by OmegaLearn

https://youtu.be/HISL2-N5NVg?t=2899

~ pi_is_3.14

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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