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If s, u, and v

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by ektamatta » Thu May 29, 2008 7:53 pm
If s, u, and v are positive integers and , which of the following must be true? vus222+=
I. s = u
II. vu≠
III. s > v
(A) None
(B) Ι only
(C) ΙI only
(D) ΙII only
(E) ΙΙ and ΙΙΙ

OA is D
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by keepsmilinyaar » Fri May 30, 2008 6:52 am
you sure you got the question correct??
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by cheryl3007 » Thu Dec 30, 2010 1:40 am
ektamatta wrote:If s, u, and v are positive integers and , which of the following must be true? vus222+=
I. s = u
II. vu≠
III. s > v
(A) None
(B) Ι only
(C) ΙI only
(D) ΙII only
(E) ΙΙ and ΙΙΙ

OA is D
Question should be
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. u # (is different from) v
III. s > v
Last edited by cheryl3007 on Thu Dec 30, 2010 2:13 am, edited 1 time in total.
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by RACHVIK » Thu Dec 30, 2010 2:07 am
If s, u, and v are positive integers and 2^s = s^u + 2^v, which of the following must be true?
I. s = u
II. u # (is different from) v
III. s > v

I think the answer should be NONE.

Lets take each case

Case 1: s=u

ie. 2^s - 2^v = s^s

s=2, 4 - 2^v = 4 or 2^v = 0 which is not possible

Case 2: u# v

Let say U=V

i.e 2^s = s^v + 2^v

let V=1, 2^s = s + 2 or s = 2^s -2

which is possible for S = 2

Case 3: s>v

let say s=3 & v=1

2^3 - 2^1 = 3^u

or 3^u = 6 not possible.
Rachvik
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by stormier » Thu Dec 30, 2010 6:47 am
Question should be
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. u # (is different from) v
III. s > v
note that each of the quantities 2^s, 2^u and 2^v is > 0 for all values of s, u and v.

I. For condition 1 to be true, 2^v must be equal to zero, which is not possible.

II. u could be same as v, as well as different from v. So its not a must true condition. Let's say u = v, then the original equation becomes 2^s = 2^u +2^u = 2.2^u=2^(u+1): This would be possible if s = u+1, when u=v, as an example.

III. since each of the quantities are individually greater than zero

2^s = 2^v + a quantitiy greater than zero

=> 2^s > 2^v

=> 2^(s-v)>1

=> s-v>0
=> s>v must be true.

Hence III must be true. Choose D
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by cheryl3007 » Thu Dec 30, 2010 7:56 am
Nice explanation, Stormier. Indeed, the answer is D
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