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If s, u, and v are positive integers and 2^s = 2^u + 2^v,

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If s, u, and v are positive integers and 2^s = 2^u + 2^v, [#permalink] New post 29 Sep 2008, 21:12
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If s, u, and v are positive integers and 2^s = 2^u + 2^v, which of the following must be true?

I. s = u
II. v≠ u
III. s > v

(A) None
(B) Ι only
(C) ΙI only
(D) ΙII only
(E) ΙΙ and ΙΙΙ

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Re: PS: Exponents [#permalink] New post 29 Sep 2008, 21:24
D for me.

The original equation can be translated into 2^(u-s) + 2^(v-s) = 1.

In the above equation, u cannot be equal to s as in that case one of the terms will become 1 and since none of terms can be equal to zero, u <>s and v <>s.

Similarly, u and v can be the same and in that case 2*2^(u-s) = 1 => u-s = -1

Now, looking at the original equation itself, since none of the terms can be equal to zero, s will at least be more than both u and v.
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Re: PS: Exponents [#permalink] New post 29 Sep 2008, 21:29
amitdgr wrote:
If s, u, and v are positive integers and 2^s = 2^u + 2^v, which of the following must be true?

I. s = u
II. v≠ u
III. s > v

(A) None
(B) Ι only
(C) ΙI only
(D) ΙII only
(E) ΙΙ and ΙΙΙ


Will go with (D).

2^4=2^3+2^3

I. s = u => Not true
II. v≠ u => Not true
III. s > v => True

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Last edited by leonidas on 29 Sep 2008, 21:39, edited 1 time in total.
Re: PS: Exponents   [#permalink] 29 Sep 2008, 21:29
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If s, u, and v are positive integers and 2^s = 2^u + 2^v,

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