If x and y are positive odd integers, then which of the following must

Question

If x and y are positive odd integers, then which of the following must also be an odd integer?

I.  x^{(y+1)}

II.  x(y+1)

III.  (y+1)^{(x-1)} + 1

A. I only
B. II only
C. III only
D. I and III
E. None of the above

Bunuel · Accepted Answer

MAGOOSH OFFICIAL SOLUTION: must_be_odd.png

PareshGmat · Answer

Answer = A. I only

odd^{(odd+1)} = odd^{even} = odd

odd * (odd+1) = odd * even = even

(odd+1)^{(odd-1)} + 1 = even^{even} +1 = even + 1 = odd

However, this contradicts for x = 3, y = 1

(3+1)^{1-1} + 1 = 1 + 1 = 2

Question asked is "must be odd", so answer = A

vik09 · Answer

I think the answer to this question is A only because anything to the power of odd will always be odd as odd*odd = odd. In case of II y+1 will be even as odd+odd = even and from even*odd = even In case of III anything to the power of even will churn out even only. Sorry for the clumsy explanation

ynaikavde · Answer

I. x^(y+1)>> (odd)^even=odd
II. x(y+1)>>(odd)*even=even
III. (y+1)^(x-1) + 1>>(even)^even+1>>even+1>> odd if x is not equal to 1 and if x=1 then its (even)^0+1=1+!=2 even.

answer A

pacifist85 · Answer

I also say A.

I used 1 as an odd, positive number to test:
I. x^(y+1) = 1^(1+1) = 1^2=1, and a second effort: 3(1+1) = 3^2 = 9. So, yes.
II. x(y+1) = 1(1+1) = 1*2 = 2. So, no.
III. (y+1)^(x-1) + 1 = (1+1)^(1-1) +1 = 2^0 + 1 = 1+1 = 2. So, no.

ANS A

OptimusPrepJanielle · Answer

The best way to do this is to assume values and put it in the equation. Let x = 1 and y = 3 Option I: x^(y+1) = 1^(4+1) = 1^5 = 1 (ODD) Option II: x(y+1) = 1(3 + 1) = 1(4) = 4 (EVEN) Option III: (y+1)^(x-1) + 1 = (3+1)^(1-1) + 1 = 4^0 + 1 = 1 + 1 (EVEN) But since we've got a power of 0, let us reconfirm with another value of x. Let x = 3. So, option III: (y+1)^(x-1) + 1 = (3+1)^(3-1) + 1 = 4^2 + 1 = 17 (ODD). So option III doesn't give a unique answer. So option I MUST always be odd. Hence option A. -- Optimus Prep's GMAT On Demand course for only $299 covers all verbal and quant. concepts in detail. Visit the following link to get your 7 days free trial account: https://www.optimus-prep.com/gmat-on-demand-course

mvictor · Answer

tricky one...
we can eliminate B, C, and E right away.

1 is true, since an odd number raised to a positive number power will always be odd.
the third one, if x>1, then it will be odd, but if x=1, then everything is equal to 2, and not odd.

thus, 1 alone is correct.

mikeyg51 · Answer

I tested the three statements choosing positive, odd values for x and y. I used x=1 and y=3 first, then x=3 y=5 next. I realized about halfway through that number properties could've done it even quicker!

RR88 · Answer

Option A

x,y are positive Odd integers. Odd: O, Even: E & Fraction: F

I. x^(y+1) = O^(O + 1) = O^E = O
II. x(y+1) = O(O + 1) = O(E) = E
III. (y+1)^(x-1) + 1 = (O + 1)^(O - 1) + 1 = (E)^(E) + 1 = E + 1 = O || Spl. Case, if x = 1 then (y+1)^(x-1) + 1 = (y+1)^(0) + 1 = 1 + 1 = 2 =E

sahilvijay · Answer

If x and y are positive odd integers, then which of the following must also be an odd integer?

I. x^(y+1)
II. x(y+1)
III. (y+1)^(x-1) + 1

i) always odd as odd raise to anything which is even or odd= odd
ii) x(y+1) , since y is odd so y+1 will be even and => even x odd = even
iii) (y+1)^(x11) + 1
let y =1 , x= 1 both odd => 2^0 +1 = 2 even

Hence i) is always odd.

ANSWER IS A

shashankism · Answer

x is odd, so (x-1) is even.
y is odd, so (y+1) is even.

I. x^(y+1) = {odd} ^ {even} = odd 
II.  x(y+1) = odd * even = even 
III.  (y+1)^(x-1) + 1 = {even}^{even} + 1  OR  {even}^0 +1  = odd OR even

So, Only I is correct.

Answer A

gps5441 · Answer

No calculations required.

I.  x^{(y+1)} = odd^{(odd+1)} -any odd no raised to any odd or even power will always be odd.

II. x(y+1)=odd(odd+1)=odd(even)-any number multiplyed with even no will always be even

III.  (y+1)^{x-1}  + 1=

Let x=3(odd)

(odd+1)^{(odd-1)} +1= (even)^{(3-1)} +1=odd
 
 or

Let x=1(odd)

even^{(1-1)} +1= even^{(0)} +1=1+1=2(any integer raised to zero is 1) even

Hence 3rd option can be odd or even.

Answer-1

ThatDudeKnows · Answer

I. Odd raised to even. Odd times odd is odd. Times odd is odd. Times odd is odd. Times odd is odd. Times odd is odd. YES. B, C, and E are out. We only need to try III.

III. If x=0, we will have 1+1=2, which is not odd. NO. D is out.

Answer choice A.

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If x and y are positive odd integers, then which of the following must

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If x and y are positive odd integers, then which of the following must

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