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

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Revision as of 15:43, 7 September 2007

Problem

For how many values of $\displaystyle k$ is $\displaystyle 12^{12}$ the least common multiple of the positive integers $6^6$ and $8^8$?

Solution

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

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

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

See also

1986 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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