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Intern  Joined: 30 Aug 2007
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If k and t are integers and k^2 – t^2 is an odd integer  [#permalink]

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Question Stats: 77% (01:45) correct 23% (01:55) wrong based on 416 sessions

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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
Senior Manager  Joined: 08 Jun 2007
Posts: 478
Re: GMAT Prep 2: PS Q25  [#permalink]

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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

For k^2 – t^2 to be odd , I guess either k or t must be odd .
So lets assume k= even t = odd

I. k + t + 2
even + odd + 2 = even + odd = odd

II. k^2 + 2kt + t^2
even + even + odd = even + odd = odd

III. k^2 + t^2
even + odd = odd .

So I guess NONE
Senior Manager  Joined: 18 Jul 2006
Posts: 407

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none..i.e E

take any example.. 3 & 4 and plug it in each option.
Intern  Joined: 30 Aug 2007
Posts: 37

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OA is A i.e. (None).

Thanks
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Re: Q25) If k and t are integers and k^2 t^2 is an odd  [#permalink]

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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.
Math Expert V
Joined: 02 Sep 2009
Posts: 58402
Re: Q25) If k and t are integers and k^2 t^2 is an odd  [#permalink]

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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.

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Re: Q25) If k and t are integers and k^2 t^2 is an odd  [#permalink]

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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.

Thanks, i misinterpreted the option (III),rather i solved it for k^2 - t^2.
Intern  Joined: 14 Aug 2012
Posts: 7
Re: If k and t are integers and k^2 – t^2 is an odd integer  [#permalink]

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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?
Math Expert V
Joined: 02 Sep 2009
Posts: 58402
Re: If k and t are integers and k^2 – t^2 is an odd integer  [#permalink]

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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.

Check here: if-k-and-t-are-integers-and-k-2-t-2-is-an-odd-integer-53248.html#p1082758

Hope it helps.
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Re: If k and t are integers and k^2 – t^2 is an odd integer  [#permalink]

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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 = odd
so,if k is even,t will be odd.

I. k+t+2 = even+odd+2 = odd
II. k^2 + 2kt+t^2 = even + even + odd = odd
III. k^2 + t^2 = even + odd = odd

Hence answer will be option A.
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Re: If k and t are integers and k^2 – t^2 is an odd integer  [#permalink]

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2
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.

Hope that helped. Manager  B
Joined: 31 Mar 2013
Posts: 60
Location: India
GPA: 3.02
Re: Q25) If k and t are integers and k^2 t^2 is an odd  [#permalink]

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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.

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?
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Joined: 02 Sep 2009
Posts: 58402
Re: Q25) If k and t are integers and k^2 t^2 is an odd  [#permalink]

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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.

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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Re: If k and t are integers and k^2 – t^2 is an odd integer  [#permalink]

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Thank you, Bunuel.
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Re: If k and t are integers and k^2 – t^2 is an odd integer  [#permalink]

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avenger 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

Plug in some no's and check

k = 7 , t = 4

I. k + t + 2 = 7 + 4 + 2 = 13 ( Odd )
II. k^2 + 2kt + t^2 = 7^2 + 2*7*4 + 4^2 = 121 ( Odd )
III. k^2 + t^2 = 7^2 + 4^2 = 65 ( Odd )

k = 6 , t = 3

I. k + t + 2 = 6 + 3 + 2 = 11 ( Odd )
II. k^2 + 2kt + t^2 = 6^2 + 2*6*3 + 3^2 = 81 ( Odd )
III. k^2 + t^2 = 6^2 + 3^2 = 45 ( Odd )

Check in all the cases the answer will be ODD, hence we will not get even number..

Answer will be (A) None...

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If k and t are integers and k^2 – t^2 is an odd integer  [#permalink]

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avenger 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

Here is a methodical approach to solve this question:

Given Info:
$$k^2 – t^2$$ is odd.

Inferences:
$$k^2 – t^2$$ can be written as $$(k – t)*(k + t)$$
Since the product is odd, both $$(k – t)$$ and $$(k + t)$$ must be odd.

So, $$(k – t)$$ is odd ……. (1)
$$(k + t)$$ is odd …… (2)

The above also implies that exactly one of {$$k$$, $$t$$} is odd and the other is even …… (3)

Approach:
We’ll use the above inferences to identify the even-odd nature of each of the given expressions.

Working Out:

(I) $$k + t + 2$$
We already determined that $$(k + t)$$is odd.
So, adding an even number ($$2$$) to $$(k + t)$$ won’t change its even-odd nature.

Think: $$3$$is odd. $$3+2 = 5$$ is also odd.

(II) $$k^2 + 2kt + t^2$$
This is simply $$(k + t)^2$$

Since $$(k + t)$$ is odd, its square is also odd.

Think: $$3$$ is odd. $$3^2 = 9$$ is odd).

(You can also look at this case as: product of two odd integers is always odd).

(III) $$k^2 + t^2$$

$$k^2 + t^2$$ will have the same even-odd nature as$$k^2 – t^2$$.
So, $$k^2 + t^2$$ is also odd.

Think: $$(a + b)$$ will have the same even-odd nature as$$(a – b)$$.
Eg: $$5-2 = 3$$ (odd) $$5 + 2 = 7$$ (odd)

Since none of the given expressions are even, the correct answer is option A.

Hope this helps. Cheers,
Krishna
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Re: If k and t are integers and k^2 – t^2 is an odd integer  [#permalink]

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avenger 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

We can see that k^2 - t^2 = odd or (k - t)(k + t) = odd. Thus, (k - t) is odd, and (k + t) is odd.

Let’s consider each Roman numeral.

I. Since(k + t) is odd, k + t + 2 = odd + 2 = odd.

II. Since k^2 + 2kt + t^2 = (k + t)^2 = (odd)^2 = odd, k^2 + 2kt + t^2 is odd.

III. Since k^2 – t^2 is odd, k^2 + t^2 is also odd.

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Re: If k and t are integers and k^2 – t^2 is an odd integer  [#permalink]

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