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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>?  
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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>?  
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== Solution ==
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== Solution 1==
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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>.  
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It is evident that <math>k</math> has only 2s and 3s in its prime factorization, or <math>k = 2^a3^b</math>.  
  
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*<math>\displaystyle 6^6 = 2^6\cdot3^6</math>
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*<math>6^6 = 2^6\cdot3^6</math>
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*<math>\displaystyle 8^8 = 2^24</math>
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*<math>8^8 = 2^{24}</math>
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*<math>\displaystyle 12^{12} = 2^{24}\cdot3^{12}</math>
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*<math>12^{12} = 2^{24}\cdot3^{12}</math>
  
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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. 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>.
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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>.
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== Solution 2==
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We want the number of <math>k</math> such that <math>\operatorname{lcm}(6^6,8^8,k)=12^{12}</math>.
 +
 
 +
Using <math>\operatorname{lcm}</math> properties, this is <math>\operatorname{lcm}(\operatorname{lcm}(2^6\cdot 3^6,2^{24}),k)=2^{24} \cdot 3^{12}</math>, or <math>\operatorname{lcm}(2^{24}\cdot 3^6,k)=2^{24}\cdot 3^{12}</math>.
 +
 
 +
At this point, we realize that <math>k=2^a\cdot3^b</math>, as any other prime factors would be included in the <math>\operatorname{lcm}</math>.
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 +
Also, <math>b=12</math> (or the power of <math>3</math> in the <math>\operatorname{lcm}</math> wouldn't be <math>12</math>) and <math>0\le a\le 24</math> (or the power of <math>2</math> in the <math>\operatorname{lcm}</math> would be <math>a</math> and not <math>24</math>).
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Therefore, <math>a</math> can be any integer from <math>0</math> to <math>24</math>, for a total of <math>25</math> values of <math>a</math> and <math>\boxed{025}</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 ==
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{{AIME box|year=1986|before=First question|num-a=2}}
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{{AIME box|year=1998|before=First question|num-a=2}}
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 +
[[Category:Intermediate Number Theory Problems]]
 +
{{MAA Notice}}

Latest revision as of 21:50, 23 April 2024

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 1

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 $k$.

Solution 2

We want the number of $k$ such that $\operatorname{lcm}(6^6,8^8,k)=12^{12}$.

Using $\operatorname{lcm}$ properties, this is $\operatorname{lcm}(\operatorname{lcm}(2^6\cdot 3^6,2^{24}),k)=2^{24} \cdot 3^{12}$, or $\operatorname{lcm}(2^{24}\cdot 3^6,k)=2^{24}\cdot 3^{12}$.

At this point, we realize that $k=2^a\cdot3^b$, as any other prime factors would be included in the $\operatorname{lcm}$.

Also, $b=12$ (or the power of $3$ in the $\operatorname{lcm}$ wouldn't be $12$) and $0\le a\le 24$ (or the power of $2$ in the $\operatorname{lcm}$ would be $a$ and not $24$).

Therefore, $a$ can be any integer from $0$ to $24$, for a total of $25$ values of $a$ and $\boxed{025}$ 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
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All AIME Problems and Solutions

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