If s, u, and v are positive integers and 2s = 2u + 2v, which of the fo

Option D.
s,u,v can't be equal to 0
And s=u+v
So s>v definitely.
II statement could be or couldn't be true.
And I statement can never be true because if s=u,then v=0 which is not possible as v is +ve integer.

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I got this one wrong at first but figured it out after. Here's what helped me.

I set up each equation,

S = U + V
V = U + S
U = V - S

I. From above we can plug zero in for V, and tell right away that S and U do not equal one another. FALSE
II. From above we can plug zero in for S, and in that instance it would tell us that U can equal V. FALSE
III. From the equation S = U + V, we can incur that S is a product of V + something else. So S is always greater than V. TRUE

Answer is (D)

Hi,

why we can't consider 0 as a value for U - as 0 is a positive integer. If U = 0 then S = V thus option 3 will not hold true.

Following post from tells that we can consider 0 as non-negative integer.

Bunuel Math Expert

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follow send pm Re: PS-which of the following must be true [#permalink] 22 May 2012, 01:23 Expert's post Joy111 wrote:
Bunuel wrote:
LM wrote:
If x and y are integers such that ,and z is non negative integer then which of the following must be true?

A)

B)

C)

D)

E)


Note that we are asked "which of the following MUST be true, not COULD be true. For such kind of questions if you can prove that a statement is NOT true for one particular set of numbers, it will mean that this statement is not always true and hence not a correct answer.

Given: and .

Evaluate each option:
A) --> not necessarily true, for example: and ;

B) --> not necessarily true, for example: and ;

C) --> not necessarily true, if then ;

D) --> not necessarily true, it's true only for ;

E) --> as then and as then --> always true.

Answer: E.



amazing ! couldn't figure out how option 3 was not necessarily true , forgot that non negative could mean that 0 is possible ,folks : non negative does not mean only positive integers , it could be 0 as well

Hypothetically speaking, Bunuel so if a question says, non positive numbers can we consider 0 as well , rather than only negative numbers.

Set of Non positive numbers { 0,-1,-5,-9 }
Set of Non negative numbers { 0,1, 4, 7, }

is this correct?


Yes, a set of non-positive numbers consists of zero and negative numbers.

0 is neither positive nor negative integer.

Also, sets of positive integers {1, 2, 3, 4, ....} and non-negative integers {0, 1, 2, 3, ...} are NOT the same.

Hi Bunuel,

Thanks very much for your quick reply. I understood my mistake.

Also I am going through the posts in the quantitative forum for solving the questions and noticed one thing that in almost every post you have provided solution with detailed explanation, which are very helpful. This is a great job considering the huge number of posts we have in the quantitative forum.

2S = 2U + 2V
Take out commons
2S = 2 ( U + V )
(2/2) S = U + V
S = U + V
1, S = U may or may not
2, U # V may or may not
3, S > V Must be true
D

We are given that 2s = 2u + 2v; thus, s = u + v. We need to determine which of the Roman numerals MUST BE true.

I. s = u

We see that s cannot equal u. Since s, u, and v are positive integers, s, the sum of u and v, will always be greater than u. Roman numeral one is NOT true.

II. u ≠ vu ≠ v

Roman numeral two does not have to be true. For example, if u = 1 and v = 1, then we have u = uv = v.

III. s > v

Since s = u + v, and s, u, and v are positive integers, s will ALWAYS be greater than v. Roman numeral three IS true.

Answer: D

Key word: If s, u, and v are positive integers, cant be 0 and -ive integers

Only one statement satisfies this 2s =2(u+v)
s = u + v

III, s > v

D

Given:
1. s, u, and v are positive integers
2. 2s = 2u + 2v
s = u + v
s, u ,v > 0

Asked: Which of the following must be true?

I. s = u
Since v > 0
s > u
NOT TRUE

II. \(u\neq{v}\)
Since there is no restriction on u & v
u may or may not be equal to v
COULD BE TRUE

III. s > v
Since u > 0
s = u + v > v
MUST BE TRUE

IMO D

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