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

Revision as of 15:57, 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 $\displaystyle k$ .

@@ Line 11: / Line 11: @@
 == See also ==
-{{AIME box|year=1988|before=First question|num-a=2}}
+{{AIME box|year=1998|before=First question|num-a=2}}
 [[Category:Intermediate Number Theory Problems]]

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

Revision as of 15:57, 7 September 2007

Problem

Solution

See also