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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.

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.

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

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05 Aug 2013, 21:04

Please help explain: 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 05 Aug 2013, 23:32, edited 1 time in total.

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

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05 Aug 2013, 21:11

smartyman wrote:

Please help explain: 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, 21:49

smartyman wrote:

Please help explain: 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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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, 11:38

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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.

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