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Re: Q25) If k and t are integers and k^2 t^2 is an odd [#permalink]
07 May 2012, 07:42

avenger wrote:

Q25) If k and t are integers and k^2 – t^2 is an odd integer, which of the following must be an even integer?

I. k + t + 2 II. k^2 + 2kt + t^2 III. k^2 + t^2

(a) None (b) I only (c) II only (d) III only (e) I, II, and III

I have a doubt regarding option(III) of the stem,below is my explanation: Given : K^2-t^2--> odd it means (k+t)(k-t) both are odd. take option 3 we have to chek whether k^2+t^2 is odd or even. k^2+t^2=(k+t)^2-(k-t)^2 = k^2+t^2+2kt-k^2-t^2+2kt =4kt Here as 4 is an even number, and any odd number multiplied by an even results in an even number.

Please let me know whether this is correct as i had interpreted or not. and provide a suitable explanation.

Re: Q25) If k and t are integers and k^2 t^2 is an odd [#permalink]
07 May 2012, 08:23

Expert's post

piyushksharma wrote:

avenger wrote:

Q25) If k and t are integers and k^2 – t^2 is an odd integer, which of the following must be an even integer?

I. k + t + 2 II. k^2 + 2kt + t^2 III. k^2 + t^2

(a) None (b) I only (c) II only (d) III only (e) I, II, and III

I have a doubt regarding option(III) of the stem,below is my explanation: Given : K^2-t^2--> odd it means (k+t)(k-t) both are odd. take option 3 we have to chek whether k^2+t^2 is odd or even. k^2+t^2=(k+t)^2-(k-t)^2 = k^2+t^2+2kt-k^2-t^2+2kt =4kt Here as 4 is an even number, and any odd number multiplied by an even results in an even number.

Please let me know whether this is correct as i had interpreted or not. and provide a suitable explanation.

The red part is not correct. k^2+t^2 does not equal to 4kt.

If k and t are integers and k^2 – t^2 is an odd integer, which of the following must be an even integer?

I. k + t + 2 II. k^2 + 2kt + t^2 III. k^2 + t^2

A. None B. I only C. II only D. III only E. I, II, and III

k^2–t^2 is an odd integer means that either k is even and t is odd or k is odd and t is even.

Check all options: I. k + t + 2 --> even+odd+even=odd or odd+even+even=odd. Discard; II. k^2 + 2kt + t^2 --> even+even+odd=odd or odd+even+even=odd. Discard; III. k^2 + t^2 --> even+odd=odd or odd+even=odd. Discard.

Re: Q25) If k and t are integers and k^2 t^2 is an odd [#permalink]
07 May 2012, 09:54

Bunuel wrote:

piyushksharma wrote:

avenger wrote:

Q25) If k and t are integers and k^2 – t^2 is an odd integer, which of the following must be an even integer?

I. k + t + 2 II. k^2 + 2kt + t^2 III. k^2 + t^2

(a) None (b) I only (c) II only (d) III only (e) I, II, and III

I have a doubt regarding option(III) of the stem,below is my explanation: Given : K^2-t^2--> odd it means (k+t)(k-t) both are odd. take option 3 we have to chek whether k^2+t^2 is odd or even. k^2+t^2=(k+t)^2-(k-t)^2 = k^2+t^2+2kt-k^2-t^2+2kt =4kt Here as 4 is an even number, and any odd number multiplied by an even results in an even number.

Please let me know whether this is correct as i had interpreted or not. and provide a suitable explanation.

The red part is not correct. k^2+t^2 does not equal to 4kt.

If k and t are integers and k^2 – t^2 is an odd integer, which of the following must be an even integer?

I. k + t + 2 II. k^2 + 2kt + t^2 III. k^2 + t^2

A. None B. I only C. II only D. III only E. I, II, and III

k^2–t^2 is an odd integer means that either k is even and t is odd or k is odd and t is even.

Check all options: I. k + t + 2 --> even+odd+even=odd or odd+even+even=odd. Discard; II. k^2 + 2kt + t^2 --> even+even+odd=odd or odd+even+even=odd. Discard; III. k^2 + t^2 --> even+odd=odd or odd+even=odd. Discard.

Answer: A.

Thanks, i misinterpreted the option (III),rather i solved it for k^2 - t^2.

Re: If k and t are integers and k^2 – t^2 is an odd integer [#permalink]
27 Mar 2013, 03:51

i have chosen to solve using fact that since k^2 - t^2 is odd, both K+T and K-T should be odd. Making this choice, i get that both options 1 and 2 are odd 1) k+t+2 means odd number + 2 = odd number 2) (k+t)^2 means (odd)^2 = odd number 3) k^2 + t^2 = ((k+t)^2 + (k-t)^2)/2 => (odd + odd)/2 = even

so, result is (1) and (2) are odd while (3) is even, since this combination is not part of any answer, chose NONE.

Re: If k and t are integers and k^2 – t^2 is an odd integer [#permalink]
27 Mar 2013, 05:36

Expert's post

mamathak wrote:

i have chosen to solve using fact that since k^2 - t^2 is odd, both K+T and K-T should be odd. Making this choice, i get that both options 1 and 2 are odd 1) k+t+2 means odd number + 2 = odd number 2) (k+t)^2 means (odd)^2 = odd number 3) k^2 + t^2 = ((k+t)^2 + (k-t)^2)/2 => (odd + odd)/2 = even

so, result is (1) and (2) are odd while (3) is even, since this combination is not part of any answer, chose NONE.

Is this approach correct?

No, that's NOT correct.

k^2–t^2 is an odd integer means that either k is even and t is odd or k is odd and t is even.

Re: If k and t are integers and k^2 – t^2 is an odd integer [#permalink]
27 Mar 2013, 07:28

1

This post received KUDOS

This problem mobilizes addition, substraction and multiplication rules of odd and even integers so knowing your rules will be the key to helping you solve this correctly.

First of all, \(k^2 - t^2\) can be rewritten as \((k-t)*(k+t)\)

If you're unsure about the notation, just develop \((k-t)*(k+t)\).

Now, assuming that \(k^2 - t^2\) = \((k-t)*(k+t)\) is odd, then according to the following rule :

odd * odd = odd (1)

We'll get \(k-t\) is odd and \(k+t\) is odd, which is extremely helpful since, if you notice the answer choices, all of them revolve around \(k+t\). So let's go through them one by one :

I. \(k+t+2\).

Using parenthesis to isolate \(k+t\), we get \((k+t)+2\) which is a sum involving an odd number and an even number. So, according to the following rule :

odd + even = odd (2)

Which means that \(k+t+2\) is odd. So answer I is not possible. (Since we're looking for an even result)

II. \(k^2 + 2kt + t^2\)

Now this answer choice may seem intimidating, but it actually isn't. Since \(k^2 + 2kt + t^2\) is equal to \((k+t)^2\). And since \(k+t\) is odd, then its square will be odd as well (rule 1). So answer II is also not possible.

III.\(k^2 + t^2\)

Once again, this answer choice may seem intimidating since you have no data on k nor t. But, looking at answer choice II., \(k^2+t^2\) is actually equal to \((k+t)^2 - 2kt\). This is a difference between an odd number \((k+t)^2\) and an even number \(2kt\), so according to rule 2, the result will be odd. So answer III. is also not possible.

As such, the only correct answer to this question is answer A.

Re: Q25) If k and t are integers and k^2 t^2 is an odd [#permalink]
04 Sep 2013, 03:18

Bunuel wrote:

piyushksharma wrote:

avenger wrote:

Q25) If k and t are integers and k^2 – t^2 is an odd integer, which of the following must be an even integer?

I. k + t + 2 II. k^2 + 2kt + t^2 III. k^2 + t^2

(a) None (b) I only (c) II only (d) III only (e) I, II, and III

I have a doubt regarding option(III) of the stem,below is my explanation: Given : K^2-t^2--> odd it means (k+t)(k-t) both are odd. take option 3 we have to chek whether k^2+t^2 is odd or even. k^2+t^2=(k+t)^2-(k-t)^2 = k^2+t^2+2kt-k^2-t^2+2kt =4kt Here as 4 is an even number, and any odd number multiplied by an even results in an even number.

Please let me know whether this is correct as i had interpreted or not. and provide a suitable explanation.

The red part is not correct. k^2+t^2 does not equal to 4kt.

If k and t are integers and k^2 – t^2 is an odd integer, which of the following must be an even integer?

I. k + t + 2 II. k^2 + 2kt + t^2 III. k^2 + t^2

A. None B. I only C. II only D. III only E. I, II, and III

k^2–t^2 is an odd integer means that either k is even and t is odd or k is odd and t is even.

Check all options: I. k + t + 2 --> even+odd+even=odd or odd+even+even=odd. Discard; II. k^2 + 2kt + t^2 --> even+even+odd=odd or odd+even+even=odd. Discard; III. k^2 + t^2 --> even+odd=odd or odd+even=odd. Discard.

Answer: A.

Hi Bunuel, You've mentioned that if k^2–t^2 is an odd integer means that either k is even and t is odd or k is odd and t is even.

Isn't a 3rd case also possible where K is odd and T is 0?

Re: Q25) If k and t are integers and k^2 t^2 is an odd [#permalink]
04 Sep 2013, 03:26

Expert's post

emailmkarthik wrote:

Bunuel wrote:

piyushksharma wrote:

If k and t are integers and k^2 – t^2 is an odd integer, which of the following must be an even integer?

I. k + t + 2 II. k^2 + 2kt + t^2 III. k^2 + t^2

(a) None (b) I only (c) II only (d) III only (e) I, II, and III

I have a doubt regarding option(III) of the stem,below is my explanation: Given : K^2-t^2--> odd it means (k+t)(k-t) both are odd. take option 3 we have to chek whether k^2+t^2 is odd or even. k^2+t^2=(k+t)^2-(k-t)^2 = k^2+t^2+2kt-k^2-t^2+2kt =4kt Here as 4 is an even number, and any odd number multiplied by an even results in an even number.

Please let me know whether this is correct as i had interpreted or not. and provide a suitable explanation.

The red part is not correct. k^2+t^2 does not equal to 4kt.

If k and t are integers and k^2 – t^2 is an odd integer, which of the following must be an even integer?

I. k + t + 2 II. k^2 + 2kt + t^2 III. k^2 + t^2

A. None B. I only C. II only D. III only E. I, II, and III

k^2–t^2 is an odd integer means that either k is even and t is odd or k is odd and t is even.

Check all options: I. k + t + 2 --> even+odd+even=odd or odd+even+even=odd. Discard; II. k^2 + 2kt + t^2 --> even+even+odd=odd or odd+even+even=odd. Discard; III. k^2 + t^2 --> even+odd=odd or odd+even=odd. Discard.

Answer: A.

Hi Bunuel, You've mentioned that if k^2–t^2 is an odd integer means that either k is even and t is odd or k is odd and t is even.

Isn't a 3rd case also possible where K is odd and T is 0?

Well, since 0 is an even number, then this scenario falls into the case when k=odd and t=even. _________________

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