Exponents PS

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vishubn: Legendary Member; Posts: 574; Joined: Sun Jun 01, 2008 8:48 am; Location: Bangalore; Thanked: 28 times

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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Source: Beat The GMAT — Problem Solving | Fri Oct 03, 2008 10:14 pm

krazy800: Master | Next Rank: 500 Posts; Posts: 136; Joined: Sun Sep 28, 2008 8:20 pm; Location: fssf; Thanked: 17 times

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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vishubn: Legendary Member; Posts: 574; Joined: Sun Jun 01, 2008 8:48 am; Location: Bangalore; Thanked: 28 times

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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cramya: Legendary Member; Posts: 2467; Joined: Thu Aug 28, 2008 6:14 pmFollowed by: 11 members; Thanked: 331 times

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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stop@800: Legendary Member; Posts: 871; Joined: Wed Aug 13, 2008 7:48 am; Thanked: 48 times

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.

www.interrogationpoint.com

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