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Exponents PS

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Exponents PS

by vishubn » Fri Oct 03, 2008 10:14 pm
I am not sure if its posted before if it is then i am sorry But found the problem Nice

Which of the following is equal to (2^12 – 2^6) / (2^6 – 2^3)?
A. 2^6 + 2^3
B. 2^6 - 2^3
C. 2^9
D. 2^3
E. 2
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by krazy800 » Fri Oct 03, 2008 10:23 pm
The answer is A

Taking 2^6 common in the numerator and 2^3 common in the denominator we get

2^6(2^6 -1)/2^3(2^3-1) = 72 = 2^6+2^3

Hope this helps!
Aiming High
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by vishubn » Fri Oct 03, 2008 10:31 pm
Ya i got it s welll ! and the OA is indeed A !
Thanks krazy800 ....

p.s. i don't know why i felt this was anice problem ( thats the reason i posted it )

Vishu
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by cramya » Sat Oct 04, 2008 12:44 am
I think we can do a^2-b^2 = (a+b) (a-b)

2^12 - 2 ^ 6 = (2^6+2^3) (2^6-2^3)

So (2^6-2^3) cancels out and the ans is (2^6+2^3)

Thoughts??
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by stop@800 » Sat Oct 04, 2008 2:51 am
cramya wrote:I think we can do a^2-b^2 = (a+b) (a-b)

2^12 - 2 ^ 6 = (2^6+2^3) (2^6-2^3)

So (2^6-2^3) cancels out and the ans is (2^6+2^3)

Thoughts??
Absolutely correct
Perhaps this is what was expected from this Qm.
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