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# If x, y and z are integers and xyz is divisible by 8, is x e

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If x, y and z are integers and xyz is divisible by 8, is x e [#permalink]

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29 Jun 2011, 13:24
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85% (hard)

Question Stats:

39% (02:19) correct 61% (01:20) wrong based on 178 sessions

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If x, y and z are integers and xyz is divisible by 8, is x even?

(1) yz is divisible by 4.
(2) x,y and z are all NOT divisible by 4.

[Reveal] Spoiler: My doubt
I am struggling to understand this question. The approach that I am trying to use is :

Statemenen 1 alone: INSUFFICIENT. Why? Because it doesn't say anything about x. x could be 1 and yz could be 4. x could be 8 and yz could e 4. So x can be ODD and EVEN both. Therefore, not sufficient. Is my understanding and concept correct for declining Statement 1?

Statement 2: If x,y and z are not divisible by 4, but they are divisible by 8 i.e they are divisible by at least 2^3. I am stuck after that. Can someone please help to tell me how to approach this mathematically?

The correct answer is B i.e statement 2 alone is sufficient.
[Reveal] Spoiler: OA
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Re: Divisibility (Odds & Evens) [#permalink]

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29 Jun 2011, 15:47
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enigma123 wrote:
Hi Fluke and everyone,

If x,y and z are integers and xyz is divisible by 8, is x even?

1. yz is divisible by 4.
2. x,y and z are all NOT divisible by 4.

[Reveal] Spoiler: My doubt
I am struggling to understand this question. The approach that I am trying to use is :

Statemenen 1 alone: INSUFFICIENT. Why? Because it doesn't say anything about x. x could be 1 and yz could be 4. x could be 8 and yz could e 4. So x can be ODD and EVEN both. Therefore, not sufficient. Is my understanding and concept correct for declining Statement 1?

Statement 2: If x,y and z are not divisible by 4, but they are divisible by 8 i.e they are divisible by at least 2^3. I am stuck after that. Can someone please help to tell me how to approach this mathematically?

The correct answer is B i.e statement 2 alone is sufficient.

Just think in terms of prime factors;

xyz is divisible by 8.
means; x*y*z must have at least 3 2's, doesn't matter from where those 2 come; the 3 2's are there.

Then think what possible ways can I get those three 2's.
x=8, y=1,z=1
We got 3 2's; this time from x.

x=1,y=8,z=1
I got 3 2's; this time from y.

x=1,y=1,z=8
I got 3 2's; this time from z.

There are many such combinations.
x=2,y=2,z=2
x=4,y=2,z=1
x=1,y=2,z=4

Q: Is x=even?

1. yz is divisible by 4.
This means that
yz must have at least two 2's because 4 has two 2's.
Now,
if y=2, z=2; yz=4;
But from the stem we know xyz contains three 2's
Thus, the one additional 2 must come from x AND x becomes even. If "x" contains one 2, it becomes even.

But, we don't know whether y=2, z=2;
y may be 4 or 8 or 16
Then the odd/even for x or z doesn't make a difference. Because, y alone took care of both statements.

If y=8; y has 3 2's
Now even if x=1; z=1; xyz will be divisible by 8 AND yz will be divisible by 4.
Thus, we saw two different scenario where x may be an even or an odd.
Not sufficient.

2. x,y and z are all NOT divisible by 4.
This statement tells us that neither of x, y or z is divisible by 4. What does it mean?

It means;
x does not contain two 2's. It may contain 0 2's or 1 2's.
y does not contain two 2's. It may contain 0 2's or 1 2's.
z does not contain two 2's. It may contain 0 2's or 1 2's.

However, we already know that xyz contain 3 2's.
Combining both the stem condition and statement 2, we can conclude that each variable has one AND only one 2 in its factors.

x=2
y=2
z=2

OR

x=2*3*7*11*13*13*13
y=2*5*43
z=2

The point is x will contain exactly one 2 in its factor. And if x contains one 2 in its factors, it must be even.
Sufficient.

Ans: "B"
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Re: Divisibility (Odds & Evens) [#permalink]

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29 Jun 2011, 17:39
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1. Not sufficient

yz is divisible by 4 = > yz is factor of 4

x may or may not be even.

x y z
2 4 1 => x is even
3 4 2 => x is odd

2 . Sufficient

We need total of 3 2's in x,y and z. But we know each number is not divisible by 4.

=> each number has to have exactly one 2 .

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Re: Divisibility (Odds & Evens) [#permalink]

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29 Jun 2011, 18:12
B..good one fluke..
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Re: Divisibility (Odds & Evens) [#permalink]

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29 Jun 2011, 20:17
As always - you are spot on fluke. Many thanks.
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If X, Y and z are integers and xyz is divisible by 8, is x [#permalink]

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05 Aug 2013, 22:04
If x,y, and z are integers and xyz is divisible by 8, is x even?
1. yz is divisible by 4.
2. x,y, and z are all NOT divisible by 4.

I could not comprehend the explanation given by MGMAT for the second part of the statement. Please further explain. Thanks. Please ignore the difficulty tag.

Last edited by mau5 on 06 Aug 2013, 00:32, edited 1 time in total.
Edited the Q.
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Re: Special Case of Divisibilty (odds and evens) [#permalink]

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05 Aug 2013, 22:11
smartyman wrote:
If x,y, and z are integers and xyz is divisible by 8, is x even?
1. yz is divisible by 4.
2. x,y, and z are all NOT divisible by 8.

I could not comprehend the explanation given by MGMAT for the second part of the statement. Please further explain. Thanks. Please ignore the difficulty tag.

there is a typo in second statement.
question is as follows:

If X, Y and z are integers and xyz is divisible by 8, isx even?

(1) yz is divisible by 4
(2) X, Y, and z are all NOTdivisible by 4

solution:

(1) yz is divisible by 4
If the value of yz = 4, then x should be even
If the value of yz = 8, then x can be odd or even
Insufficient!

Quote:
(2) X, Y, and z are all NOT divisible by 4
If x , y , z are multiples of 2, then xyz is divisible by 8 and x is even. This is the only condition which satisfies the conditions xyz divisible by 8 and x,y,z are all not divisible by 4
Sufficient!

hence B
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Re: Special Case of Divisibilty (odds and evens) [#permalink]

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05 Aug 2013, 22:49
smartyman wrote:
If x,y, and z are integers and xyz is divisible by 8, is x even?
1. yz is divisible by 4.
2. x,y, and z are all NOT divisible by 8.

I could not comprehend the explanation given by MGMAT for the second part of the statement. Please further explain. Thanks. Please ignore the difficulty tag.

In which book of Manhattans' did you find this problem can you tell me?
did you write the question correctly? check it once more..plz
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Asif vai.....

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Re: If X, Y and z are integers and xyz is divisible by 8, is x [#permalink]

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06 Aug 2013, 00:23
Sorry,

I misread the question; statement 2 should be
x,y, and z are all NOT divisible by 4.
then it makes sense.
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Re: If X, Y and z are integers and xyz is divisible by 8, is x [#permalink]

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06 Aug 2013, 00:37
smartyman wrote:
Sorry,

I misread the question; statement 2 should be
x,y, and z are all NOT divisible by 4.
then it makes sense.

thats best. sometimes it happens.
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Re: If x, y and z are integers and xyz is divisible by 8, is x e [#permalink]

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23 Jul 2015, 12:38
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Re: If x, y and z are integers and xyz is divisible by 8, is x e [#permalink]

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04 Aug 2016, 08:47
enigma123 wrote:
If x, y and z are integers and xyz is divisible by 8, is x even?

(1) yz is divisible by 4.
(2) x,y and z are all NOT divisible by 4.

[Reveal] Spoiler: My doubt
I am struggling to understand this question. The approach that I am trying to use is :

Statemenen 1 alone: INSUFFICIENT. Why? Because it doesn't say anything about x. x could be 1 and yz could be 4. x could be 8 and yz could e 4. So x can be ODD and EVEN both. Therefore, not sufficient. Is my understanding and concept correct for declining Statement 1?

Statement 2: If x,y and z are not divisible by 4, but they are divisible by 8 i.e they are divisible by at least 2^3. I am stuck after that. Can someone please help to tell me how to approach this mathematically?

The correct answer is B i.e statement 2 alone is sufficient.

Hi,

I am a bit confused with the second statement in the question.

The statement states - x,y and z are all NOT divisible by 4. I am confused because "all" is mentioned in the statement.
Because of this I inferred that "Not all x, y,z are divisible by 4"- implying some can be divisible by 4. Hence I marked "E" as my answer.

Can anyone please clarify my doubt....

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Re: If x, y and z are integers and xyz is divisible by 8, is x e [#permalink]

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21 Aug 2016, 05:42
I marked the wrong choice
Lesson learnt => XYZ are not all divisible by 4 ≠ XYZ are all not divisible by 4
Damn..!!!
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Re: If x, y and z are integers and xyz is divisible by 8, is x e   [#permalink] 21 Aug 2016, 05:42
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