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Math Expert V
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If x and y are positive integers, is xy even?  [#permalink]

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Difficulty:   25% (medium)

Question Stats: 75% (01:40) correct 25% (01:51) wrong based on 1424 sessions

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If x and y are positive integers, is xy even?

(1) x^2 + y^2 − 1 is divisible by 4.
(2) x + y is odd.

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Re: If x and y are positive integers, is xy even?  [#permalink]

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4
Bunuel wrote:
If x and y are positive integers, is xy even?

(1) x^2 + y^2 − 1 is divisible by 4.
(2) x + y is odd.

We need to determine whether the product of x and y is even.

Statement One Alone:

x^2 + y^2 − 1 is divisible by 4.

Since 4 is an even number, we need x^2 + y^2 − 1 to be even. In order for x^2 + y^2 − 1 to be even, we need x^2 + y^2 to be odd. If the sum of two squares is odd, one of them must be odd and the other must be even. This means that either x = odd and y = even OR x = even and y = odd. In either case, the product of x and y will be even. Statement one alone is sufficient to answer the question.

Statement Two Alone:

x + y is odd.

Since x + y = odd, either x = odd and y = even OR x = even and y = odd. In either case, the product of x and y will be even. Statement two alone is sufficient to answer the question.

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Re: If x and y are positive integers, is xy even?  [#permalink]

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5
x^2 + y^2 - 1 has to be even to be divisible by 4.
Hence x^2 + y^2 is odd.
This means either x or y has to be even. Statement 1 is sufficient.
Similarly for x+y to be odd, either x or y has to be even. Hence product xy is even.
Statement 2 is also sufficient.

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Re: If x and y are positive integers, is xy even?  [#permalink]

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1
Statement 1
X^2 +Y^2 - 1 is divisible by 4. can be 3^2 + 5^2 - 1 = 24 or 4^2 + 1^2 - 1 = 16 S
Statement 2. X+Y = odd. it has to be one even and one odd number. Suff

Ans D

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Originally posted by Ejiroosa on 26 Jun 2017, 01:50.
Last edited by Ejiroosa on 25 Jul 2017, 03:36, edited 1 time in total.
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If x and y are positive integers, is xy even?  [#permalink]

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Bunuel wrote:
If x and y are positive integers, is xy even?

(1) x^2 + y^2 − 1 is divisible by 4.
(2) x + y is odd.

(1) $$x^2 + y^2 − 1$$ is divisible by 4.

Let ($$x^2 + y^2$$ ) be $$z$$.

$$z - 1$$ is divisible by $$4$$.

Only even number is divisible by $$4$$. Hence $$z - 1$$ should be even.

Odd - Odd = Even.

Therefore $$(x^2 - y^2)$$ should be Odd. ($$Odd^2$$ will be Odd. $$Even^2$$ will be Even)

Even - Odd = Odd

Therefore either $$x$$ or $$y$$ should be even. Therefore $$xy$$ will be even. I is Sufficient.

(2) $$x + y$$ is odd.

Even + Odd = Odd.

Therefore either $$x$$ or $$y$$ should be even. Therefore $$xy$$ will be even. II is Sufficient.

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Re: If x and y are positive integers, is xy even?  [#permalink]

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1
If x and y are positive integers, is $$xy$$even?

(1) $$x^2 + y^2 − 1$$ is divisible by 4

This means that either x or y has to be odd.

As you know ODD * EVEN = EVEN

Question - Is xy EVEN ? is TRUE =====> Eq. (1) SUFFICIENT

(2) $$x + y$$ is odd

As we know, ODD + EVEN = ODD

And ODD * EVEN = EVEN

Question - Is xy EVEN ? is TRUE =====> Eq. (2) SUFFICIENT

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Re: If x and y are positive integers, is xy even?  [#permalink]

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Ans is D used 5 instead of 4....

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Re: If x and y are positive integers, is xy even?  [#permalink]

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Bunuel wrote:
If x and y are positive integers, is xy even?

(1) x^2 + y^2 − 1 is divisible by 4.
(2) x + y is odd.

St 1

(x^2 + y^2 − 1) /4 = some integer - therefore some odd number minus 1 is divisible by 4

x^2 + y^2= some odd number... in order for this to be true either X and Y must be some even and odd mix

(1)^2 +(2)^2= 5 odd

knowing x and y must be different ( even and odd) x and y must odd

St 2

Even + Odd = Odd so

Suff

D
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If x and y are positive integers, is xy even?  [#permalink]

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Bunuel pushpitkc niks18 Hatakekakashi
amanvermagmat

$$x^2$$ + $$y^2$$ - 1 = 4 *(m) where m is an integer since there is no remainder (given)

or

$$x^2$$ + $$y^2$$ = odd

or

x + y = odd (a positive odd / even no when squared will give odd and even values respectively)

This is only possible when either of x or y is odd.

Does that help to make St 1 suff?
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Re: If x and y are positive integers, is xy even?  [#permalink]

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Bunuel pushpitkc niks18 Hatakekakashi
amanvermagmat

$$x^2$$ + $$y^2$$ - 1 = 4 *(m) where m is an integer since there is no remainder (given)

or

$$x^2$$ + $$y^2$$ = odd

or

x + y = odd (a positive odd / even no when squared will give odd and even values respectively)

This is only possible when either of x or y is odd.

Does that help to make St 1 suff?

Hello

yes i think this approach is correct
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Re: If x and y are positive integers, is xy even?  [#permalink]

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Bunuel pushpitkc niks18 Hatakekakashi
amanvermagmat

$$x^2$$ + $$y^2$$ - 1 = 4 *(m) where m is an integer since there is no remainder (given)

or

$$x^2$$ + $$y^2$$ = odd

or

x + y = odd (a positive odd / even no when squared will give odd and even values respectively)

This is only possible when either of x or y is odd.

Does that help to make St 1 suff?

this approach is correct good usage of basic concepts (Y)
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Re: If x and y are positive integers, is xy even?  [#permalink]

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Bunuel wrote:
If x and y are positive integers, is xy even?

(1) x^2 + y^2 − 1 is divisible by 4.
(2) x + y is odd.

1) x^2 + y^2 is odd. So, one of x and y will be even which will make xy even Sufficient
2) same as 1. Sufficient.
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Re: If x and y are positive integers, is xy even?  [#permalink]

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1
Bunuel wrote:
If x and y are positive integers, is xy even?

(1) x^2 + y^2 − 1 is divisible by 4.
(2) x + y is odd.

Question: Is x*y even?

STatement 1: x^2 + y^2 − 1 = 4a

i.e. x^2 + y^2 = 4a + 1 = ODD
i.e. one of x and y must be even and other must be odd

i.e. x*y = even

SUFFICIENT

Statement 2: x + y = odd

i.e. one of x and y must be even and other must be odd

SUFFICIENT

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Re: If x and y are positive integers, is xy even?  [#permalink]

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Bunuel wrote:
If x and y are positive integers, is xy even?

(1) x^2 + y^2 − 1 is divisible by 4.
(2) x + y is odd.

****************
x,y>0 , is xy even?

xy can be even when

x and y both are even or
either x or y is even

(1) x^2 + y^2 − 1 is divisible by 4, therefore, x^2 + y^2 − 1=4I (I is an integer)

x^2 + y^2 =4I+ 1 (can you correlate it with 2n+1) for x^2 + y^2 to be odd, both x and y should be odd.
Odd*Odd=Odd
Sufficient.

(2) x + y is odd (Refer above).

Sufficient.

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************************************* Re: If x and y are positive integers, is xy even?   [#permalink] 21 Apr 2020, 03:28

# If x and y are positive integers, is xy even?   