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Re: divisible by 8? [#permalink]
15 Jun 2014, 10:42

Expert's post

mahendru1992 wrote:

Bunuel wrote:

Is \(x^2 - y^2\) divisible by 8?

(1) \(x\) and \(y\) are even integers. Clearly insufficient, consider \(x=y=0\) for an YES answer and \(x=2\) and \(y=0\) for a NO answer.

(2) \(x + y\) is divisible by \(8\) --> \(x^2 - y^2=(x+y)(x-y)\), if one of the multiples is divisible by 8 then so is the product: true for integers, but we are not told that \(x\) and \(y\) are integers. If \(x=4.8\) and \(y=3.2\), \(x+y\) is divisible by \(8\), BUT \(x^2 - y^2\) is not. Not sufficient.

(1)+(2) \(x\) and \(y\) integers. \(x+y\) divisible by 8. Hence \((x+y)(x-y)\) is divisible by \(8\). Sufficient.

Answer: C.

I don't know, I think I'm missing something here. If x=4.8 and y=3.2, Still the product would be divisible by 8 IMO. (x+y)(x-y)/8 (4.8+3.2)(4.8-3.2)/8 8(1.6)/8 =1.6 What am I doing wrong here?

1.6 is not an integer. Divisible means that you get an integer as a result of division. _________________

Re: divisible by 8? [#permalink]
16 Jun 2014, 06:34

mahendru1992 wrote:

Bunuel wrote:

Is \(x^2 - y^2\) divisible by 8?

(1) \(x\) and \(y\) are even integers. Clearly insufficient, consider \(x=y=0\) for an YES answer and \(x=2\) and \(y=0\) for a NO answer.

(2) \(x + y\) is divisible by \(8\) --> \(x^2 - y^2=(x+y)(x-y)\), if one of the multiples is divisible by 8 then so is the product: true for integers, but we are not told that \(x\) and \(y\) are integers. If \(x=4.8\) and \(y=3.2\), \(x+y\) is divisible by \(8\), BUT \(x^2 - y^2\) is not. Not sufficient.

(1)+(2) \(x\) and \(y\) integers. \(x+y\) divisible by 8. Hence \((x+y)(x-y)\) is divisible by \(8\). Sufficient.

Answer: C.

I don't know, I think I'm missing something here. If x=4.8 and y=3.2, Still the product would be divisible by 8 IMO. (x+y)(x-y)/8 (4.8+3.2)(4.8-3.2)/8 8(1.6)/8 =1.6 What am I doing wrong here?

Here is where you are wrong mate:

8(1.6)/8=1.6. This clearly shows that 8(1.6) is not divisible by 8 because the result is not an integer result. Revisit your definition of divisibility. If 8 is a divisor of a number 'a', it means that 8 can get into 'a' without leaving a remainder. In other words, the operation of dividing 'a' by 8 should give an integer result, which in your case above, it isn't.

Re: Is x^2 - y^2 divisible by 8? [#permalink]
18 Jun 2014, 22:42

mahendru1992 wrote:

Bunuel wrote:

Is \(x^2 - y^2\) divisible by 8?

(1) \(x\) and \(y\) are even integers. Clearly insufficient, consider \(x=y=0\) for an YES answer and \(x=2\) and \(y=0\) for a NO answer.

(2) \(x + y\) is divisible by \(8\) --> \(x^2 - y^2=(x+y)(x-y)\), if one of the multiples is divisible by 8 then so is the product: true for integers, but we are not told that \(x\) and \(y\) are integers. If \(x=4.8\) and \(y=3.2\), \(x+y\) is divisible by \(8\), BUT \(x^2 - y^2\) is not. Not sufficient.

(1)+(2) \(x\) and \(y\) integers. \(x+y\) divisible by 8. Hence \((x+y)(x-y)\) is divisible by \(8\). Sufficient.

Answer: C.

I don't know, I think I'm missing something here. If x=4.8 and y=3.2, Still the product would be divisible by 8 IMO. (x+y)(x-y)/8 (4.8+3.2)(4.8-3.2)/8 8(1.6)/8 =1.6 What am I doing wrong here?

We get the formula (x-y)(x+y) = x^2+xy-yx-y^2=x^2-y^2, so for xy and yx it is no matter whether x and y are integer or not, they eliminate each other. Problem with squares, if x and y integers their squares will be integer, if they are not integer squares will not be integers and, what is more, their difference will never be the integer or 0. So, it will not be divisible by integer

Re: Is x^2 - y^2 divisible by 8? [#permalink]
20 Jun 2014, 06:13

Bunuel wrote:

Is \(x^2 - y^2\) divisible by 8?

(1) \(x\) and \(y\) are even integers. Clearly insufficient, consider \(x=y=0\) for an YES answer and \(x=2\) and \(y=0\) for a NO answer.

(2) \(x + y\) is divisible by \(8\) --> \(x^2 - y^2=(x+y)(x-y)\), if one of the multiples is divisible by 8 then so is the product: true for integers, but we are not told that \(x\) and \(y\) are integers. If \(x=4.8\) and \(y=3.2\), \(x+y\) is divisible by \(8\), BUT \(x^2 - y^2\) is not. Not sufficient.

(1)+(2) \(x\) and \(y\) integers. \(x+y\) divisible by 8. Hence \((x+y)(x-y)\) is divisible by \(8\). Sufficient.

Answer: C.

Hi Bunuel, I agree with the answer but pls tell me " if one of the multiples is divisible by 8 then so is the product: true for integers, but we are not told that x and y are integers. If x=4.8 and y=3.2, x+y is divisible by 8". Sometimes we take variable as decimal and sometimes not , i have seen questions which solve by considering x and y as integers only.

Re: Is x^2 - y^2 divisible by 8? [#permalink]
20 Jun 2014, 06:36

Expert's post

nidhi12 wrote:

Bunuel wrote:

Is \(x^2 - y^2\) divisible by 8?

(1) \(x\) and \(y\) are even integers. Clearly insufficient, consider \(x=y=0\) for an YES answer and \(x=2\) and \(y=0\) for a NO answer.

(2) \(x + y\) is divisible by \(8\) --> \(x^2 - y^2=(x+y)(x-y)\), if one of the multiples is divisible by 8 then so is the product: true for integers, but we are not told that \(x\) and \(y\) are integers. If \(x=4.8\) and \(y=3.2\), \(x+y\) is divisible by \(8\), BUT \(x^2 - y^2\) is not. Not sufficient.

(1)+(2) \(x\) and \(y\) integers. \(x+y\) divisible by 8. Hence \((x+y)(x-y)\) is divisible by \(8\). Sufficient.

Answer: C.

Hi Bunuel, I agree with the answer but pls tell me " if one of the multiples is divisible by 8 then so is the product: true for integers, but we are not told that x and y are integers. If x=4.8 and y=3.2, x+y is divisible by 8". Sometimes we take variable as decimal and sometimes not , i have seen questions which solve by considering x and y as integers only.

When we are told that variables are integers, then we should consider integers only. When we are not told that, then we should not assume that the variables are integers only.

By the way this question is not a good example of GMAT questions because every GMAT divisibility question will tell you in advance that any unknowns represent positive integers (ALL GMAT divisibility questions are limited to positive integers only). _________________

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