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If x and y are positive integers, is x2 + x + y + 1 an even number?

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If x and y are positive integers, is x2 + x + y + 1 an even number?  [#permalink]

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New post 19 Mar 2015, 22:49
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If x and y are positive integers, is x^2 + x + y + 1 an even number?

1. x is equal to half the sum of the HCF of 12, 26, and 48 and LCM of 25, 35, and 50.
2. y^4/16 is an even number.

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Re: If x and y are positive integers, is x2 + x + y + 1 an even number?  [#permalink]

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New post 20 Mar 2015, 02:48
1) is in the form of x(x+1) + y +1. No matter what the value of x is, we will still need to know whether y is even or odd to judge whether the whole thing is even or odd.
Also x(x+1) is always even
2) No. is divisible by 16, hence it is even. Therefore according to the above rule, the whole expression in odd

Hence option B
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Re: If x and y are positive integers, is x2 + x + y + 1 an even number?  [#permalink]

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New post 04 Apr 2015, 08:22
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samichange wrote:
If x and y are positive integers, is x^2 + x + y + 1 an even number?

1. x is equal to half the sum of the HCF of 12, 26, and 48 and LCM of 25, 35, and 50.
2. y^4/16 is an even number.

A. Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
B. Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
C. BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
D. EACH statement ALONE is sufficient.
E. Statements (1) and (2) TOGETHER are NOT sufficient.


X2 will be odd when x is odd , will be even when X is even .

4 possible scenarios for F(XY)=x^2 + x + y + 1

X=Odd , Y= Even , F(XY) = ODD
X=Even , Y= Even , F(XY) = ODD

X=Odd, Y=Odd, F(XY) = EVEN
X=Even, Y=Odd F(XY)=EVEN


1. x is equal to half the sum of the HCF of 12, 26, and 48 and LCM of 25, 35, and 50.
not sufficient as we do not know about Y (it may be odd or even) .
2. y^4/16 is an even number.
clearly Y is even , let look at above scenarios.
we see that whenever Y is EVEN F(XY) is ODD.
so this statement is sufficient.

not that hard question IMO.
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Re: If x and y are positive integers, is x2 + x + y + 1 an even number?  [#permalink]

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New post 04 Apr 2015, 09:38
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samichange wrote:
If x and y are positive integers, is x^2 + x + y + 1 an even number?

1. x is equal to half the sum of the HCF of 12, 26, and 48 and LCM of 25, 35, and 50.
2. y^4/16 is an even number.

A. Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
B. Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
C. BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
D. EACH statement ALONE is sufficient.
E. Statements (1) and (2) TOGETHER are NOT sufficient.



At first we should analyze question: exponents doesn't matter in the tasks about parity, so we have x + x = 2x and this part will be even, so for solving this task we should only know about parity of y

1) in this statement we don't have information about y
insufficient
2) exponents doesn't matter so we have y / 16 = a so y = 16a and we know that y is even
sufficient

and answer is B
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Re: If x and y are positive integers, is x2 + x + y + 1 an even number?  [#permalink]

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New post 09 Jun 2016, 15:37
samichange wrote:
If x and y are positive integers, is x^2 + x + y + 1 an even number?

1. x is equal to half the sum of the HCF of 12, 26, and 48 and LCM of 25, 35, and 50.
2. y^4/16 is an even number.

A. Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
B. Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
C. BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
D. EACH statement ALONE is sufficient.
E. Statements (1) and (2) TOGETHER are NOT sufficient.



According to question , \(x^2\)+x=even number (no matter what x could be even or odd integer,)

So basically the Scope of the answer lies on y (if y is odd,\(x^2\)+x+y+1=even,But if y is even \(x^2\)+x+y+1=odd)

Statement 1. didn't say anything about y,Insufficient

Statement 2. \(\frac{y^4}{16}\)=even,So y=even number.Sufficient

Correct Answer B
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Re: If x and y are positive integers, is x2 + x + y + 1 an even number?  [#permalink]

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New post 21 Nov 2016, 05:43
This is a tremendous Question Testing our even/odd understanding
We are asked if x^2+x+y+1 is even
here let V=x^2+x+y+1
V=x(x+1)+y+1
V=Even+y+odd
V=Odd+y
If y is odd => V is even
If y is even => V is odd
So the Question is asking "IS Y ODD"
Statement 1
x is some value
Dont stress on this statement as no clue of y is mentioned
hence insufficient
Statement 2
y^4/16 => even
hence y^4 must be even too
so y*y*y*y is even => This can happen only if y is even
So as y is even => x^2+x+y+1=> odd
Hence sufficient

Hence B
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Re: If x and y are positive integers, is x2 + x + y + 1 an even number?  [#permalink]

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New post 19 Sep 2017, 11:04
sandeep4488 wrote:
1) is in the form of x(x+1) + y +1. No matter what the value of x is, we will still need to know whether y is even or odd to judge whether the whole thing is even or odd.
Also x(x+1) is always even
2) No. is divisible by 16, hence it is even. Therefore according to the above rule, the whole expression in odd

Hence option B


how is the whole expression odd?
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Re: If x and y are positive integers, is x2 + x + y + 1 an even number?  [#permalink]

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New post 23 Sep 2017, 21:13
rever08 wrote:
sandeep4488 wrote:
1) is in the form of x(x+1) + y +1. No matter what the value of x is, we will still need to know whether y is even or odd to judge whether the whole thing is even or odd.
Also x(x+1) is always even
2) No. is divisible by 16, hence it is even. Therefore according to the above rule, the whole expression in odd

Hence option B


how is the whole expression odd?



hi

here you can see, x(x+1) is a product of 2 consecutive numbers, and hence the product will always be an even number. So the odd- even nature of the whole expression will rest on the odd-even nature of y only...
statement 1 says nothing about y, not sufficient
statement 2 says y is even, sufficient
because, even + even + odd = odd

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Re: If x and y are positive integers, is x2 + x + y + 1 an even number? &nbs [#permalink] 23 Sep 2017, 21:13
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