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Master | Next Rank: 500 Posts
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by gmatrant » Sat Oct 27, 2007 9:59 am
if s, u, v are positive integers and
2^s = 2^u + 2^v, which of the following must be true?
i. s=u
ii. u <> v
iii s>v

A)None
B) 1 only
C) II only
D) III only
E) II and III

OA is D

I was able to solve this by POE, but mathematically how to solve this.Is there a way to prove s>v mathematically rather than by substitution.
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by ri2007 » Sat Oct 27, 2007 10:42 am
I did not understand what the II option was. It is easy to eliminate no I and III has to be right.

It is given that S is a positive integer. Since S is a positive integer and 2^u + 2^V is = 2^s, S has to be greater than V. The smallest value of 2 ^V is 2. So the greatest value of v is 2^s - 2.

Not sure if I am explaining this right. But hope it makes some sense.
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by gmatrant » Sat Oct 27, 2007 5:16 pm
ri2007 wrote:I did not understand what the II option was. It is easy to eliminate no I and III has to be right.

It is given that S is a positive integer. Since S is a positive integer and 2^u + 2^V is = 2^s, S has to be greater than V. The smallest value of 2 ^V is 2. So the greatest value of v is 2^s - 2.

sorry... couldnt get u.. in the second condition states <> means not equal to
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