Question

if 6^y is a factor of (2^14)*(3^24) , what is the greatest possible value for y:
a)7
b)8
c)14
d)18
e)36

Answer 1

The way I see it..
\[
2^{14} \cdot 3^{24} = 2_1 \cdot 2_2 \cdot 2_3 \cdot .... \cdot 2_{14} \cdot 3_1 \cdot 3_2 \cdot ... \cdot 3_{24} \\
\]
While 6 is basically \(2 \cdot 3\). This means that
\[
6^y = 2^y \cdot 3^y = 2_1 \cdot 2_2 \cdot .... \cdot 2_y \cdot 3_1 \cdot 3_2 \cdot ...\cdot3_{y}
\]
\(2^{14} \cdot 3^{24}\) is constructed from 14 such pairs and 10 more 3's so like:
\[
2^{14} \cdot 3^{14} \cdot 3^{10} = 6^{14} \cdot 3^{10}
\]
Then we don't have more than 14 pairs of such components required for the '6'.
Means that the largets part we can convert of this expression would go into \(6^{14}\) at max.
Hopefully I'm not doing an extremely dumb mistakes, but that's how I get it.
I would go with 14

Answer 2

sounds good..

Answer 3

The ".." tells differently, but that's good enough for me. =]

Answer 4

hhhahahah. no, I believe you. Can't believe it's really that simple.
you are so gooD1

Answer 5

ty ty =]