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

Latest revision as of 22:50, 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

@@ Line 11: / Line 11: @@
 == Solution 2==
-We want the number of <math>k</math> such that <math>\operatorname{lcm}(6^6,8^8,k)=12^12</math>.
+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>. 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>).
+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>.
+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>).
 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>.

1998 AIME (Problems • Answer Key • Resources)
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