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devarshi9283
Bunuel
If p is an integer and p* = p^2 + 2, what is the value of k?

(1) When k + 2 is divided by 5 the remainder is 2.
(2) 18 < k* < 35



consider (1)...

When k + 2 is divided by 5 the remainder is 2.
So, basically ..k is divisible by 5...
now as k is integer...(positive, negative or zero)..following values of k are possible
k=......,-10, -5, 0,5,10... and so on...

Remember that 0 is divisible by all integers except 0 itself

So, (1) alone is not sufficient....remove option A and D.....

consider (2)...

18 < k* < 35

given that If p is an integer and p* = p^2 + 2..
so here...k*=k^2+2

So..
18 < k^2+2< 35
or
16 < k^2< 33

or k^2=25 for integer k...
but here again...
k= +5 or -5

(2) alone is not sufficient....remove option B .....

consider (1) and (2) simultaneously we get ..
k=......,-10, -5, 0,5,10...
k= +5 or -5

or
k= +5 or -5
hence (1) and (2) together not sufficient...So remove D

Correct choice=E

I think it should be C-How did you get remainder 2 when -3 is divided by 5?
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devarshi9283



consider (1)...

When k + 2 is divided by 5 the remainder is 2.
So, basically ..k is divisible by 5...
now as k is integer...(positive, negative or zero)..following values of k are possible
k=......,-10, -5, 0,5,10... and so on...

Remember that 0 is divisible by all integers except 0 itself

So, (1) alone is not sufficient....remove option A and D.....


k=......,-10, -5, 0,5,10.. and so on...

If we consider k = -5
than k+2 = -5+2 =-3
If we divide -3 by 5 reminder will be -3 and not 2. So here k cannot be -5

If we cant to work in negative number: we can say -7 / 5 = -1 quotient and reminder 2 => k+2 = -7 => k=-9
So for k+2 =5a+2
we can get -ve numbers for k = -4, -9, -14.... etc..
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devarshi9283



consider (1)...

When k + 2 is divided by 5 the remainder is 2.
So, basically ..k is divisible by 5...
now as k is integer...(positive, negative or zero)..following values of k are possible
k=......,-10, -5, 0,5,10... and so on...

Remember that 0 is divisible by all integers except 0 itself

So, (1) alone is not sufficient....remove option A and D.....





k=......,-10, -5, 0,5,10.. and so on...

If we consider k = -5
than k+2 = -5+2 =-3
If we divide -3 by 5 reminder will be -3 and not 2. So here k cannot be -

If we cant to work in negative number: we can say -7 / 5 = -1 quotient and reminder 2 => k+2 = -7 => k=-9
So for k+2 =5a+2
we can get -ve numbers for k = -4, -9, -14.... etc..



I did not get what you are saying.
Statement 2: K can be 5 or -5
Statement 1: K can be 10 etc etc but not negative. Using both, should the answer not be C i.e k=5?
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KS15



I did not get what you are saying.
Statement 2: K can be 5 or -5
Statement 1: K can be 10 etc etc but not negative. Using both, should the answer not be C i.e k=5?


Yes, i was just trying to tell " devarshi9283" that value of k =0,5, 10.... etc
By statement 1 we cannot say k = -10, -5...

So by 1 and 2... k =5 and Answer is C..
Same thing i wrote in my descriptive solution above.

:)
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Nikkb
devarshi9283



consider (1)...

When k + 2 is divided by 5 the remainder is 2.
So, basically ..k is divisible by 5...
now as k is integer...(positive, negative or zero)..following values of k are possible
k=......,-10, -5, 0,5,10... and so on...

Remember that 0 is divisible by all integers except 0 itself

So, (1) alone is not sufficient....remove option A and D.....


k=......,-10, -5, 0,5,10.. and so on...

If we consider k = -5
than k+2 = -5+2 =-3
If we divide -3 by 5 reminder will be -3 and not 2. So here k cannot be -

If we cant to work in negative number: we can say -7 / 5 = -1 quotient and reminder 2 => k+2 = -7 => k=-9
So for k+2 =5a+2
we can get -ve numbers for k = -4, -9, -14.... etc..



I did not get what you are saying.
Statement 2: K can be 5 or -5
Statement 1: K can be 10 etc etc but not negative. Using both, should the answer not be C i.e k=5?




Sorry guys...I did a calculation mistake...the possible negative values for k as negative integer would be -3,-8...and so on which will give a reminder 2 when divided by 5..
My bad...edited my previous post...correct answer would be C....but yes..I think we have to consider negative integer values of k also in eqn 1...I request Bunuel to shed more light onto this...
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Guys,

Please give me 1 kudo each :)
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St 1: K +2/5 = quotient + 2; so K is multiple of 5
But that can be any number. So NS

st 2: 18<k<35
I wrote down all the values that lies between them.
And then checked what value satisfies K square plus 2. Only 27 does (25 + 2)

so, 1) + 2) K is multiple of 5 with quotient 2; so value is k = 5 (root of 25). Ans C
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St 1: K +2/5 = quotient + 2; so K is multiple of 5
But that can be any number. So NS

st 2: 18<k<35
I wrote down all the values that lies between them.
And then checked what value satisfies K square plus 2. Only 27 does (25 + 2)

so, 1) + 2) K is multiple of 5 with quotient 2; so value is k = 5 (root of 25). Ans C
Hello Madhavi1990,

May you pls explain by cannot we take negative value of K ??

Imo we can consider the negative value of K as well and so the answer should be E.

What I understand is that if -3 is divided by 5, then the remainder is 2..

Thanks in advance..

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Madhavi1990
St 1: K +2/5 = quotient + 2; so K is multiple of 5
But that can be any number. So NS

st 2: 18<k<35
I wrote down all the values that lies between them.
And then checked what value satisfies K square plus 2. Only 27 does (25 + 2)

so, 1) + 2) K is multiple of 5 with quotient 2; so value is k = 5 (root of 25). Ans C
Hello Madhavi1990,

May you pls explain by cannot we take negative value of K ??

Imo we can consider the negative value of K as well and so the answer should be E.

What I understand is that if -3 is divided by 5, then the remainder is 2..

Thanks in advance..

Sent from my Lenovo TAB S8-50LC using GMAT Club Forum mobile app

Hi, I took a value which is multiple of 5 and added 2
So -15 +2 / 5 --> -3 + 2
or -20 + 2 --> -4 + 2
Either way, there are multiple values so A is insuff

When trying with 1) and 2) we know that it has to be positive (with the addition of st 2) and satisfying given condition. So C
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kumarparitosh123
Madhavi1990
St 1: K +2/5 = quotient + 2; so K is multiple of 5
But that can be any number. So NS

st 2: 18<k<35
I wrote down all the values that lies between them.
And then checked what value satisfies K square plus 2. Only 27 does (25 + 2)

so, 1) + 2) K is multiple of 5 with quotient 2; so value is k = 5 (root of 25). Ans C
Hello Madhavi1990,

May you pls explain by cannot we take negative value of K ??

Imo we can consider the negative value of K as well and so the answer should be E.

What I understand is that if -3 is divided by 5, then the remainder is 2..

Thanks in advance..

Sent from my Lenovo TAB S8-50LC using GMAT Club Forum mobile app

Hi, I took a value which is multiple of 5 and added 2
So -15 +2 / 5 --> -3 + 2
or -20 + 2 --> -4 + 2
Either way, there are multiple values so A is insuff

When trying with 1) and 2) we know that it has to be positive (with the addition of st 2) and satisfying given condition. So C
I feel it ha been mentioned in statement 2 that K* has to be positive and not K..
So even if we consider k=-5 , then also K * will be 27 and if k=5, then also it will be 27.

Moreover, the value of K* need to be between 18 and 35 and not K..


Note: I don't know where I m missing but this is what I have understood till now. May be some experts can shed some light on my misunderstanding.

Thanks in advance.

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value of integer k?

(1) Not Sufficient

(2) could be -5 or +5 when substituting -5/+5 in the equation . Not Sufficient

(c) on combining, only 5 is possible. (-5+2 divided by 5 gives remainder -3 ; we need remainder 2)

C
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value of integer k?

(1) Not Sufficient

(2) could be -5 or +5 when substituting -5/+5 in the equation . Not Sufficient

(c) on combining, only 5 is possible. (-5+2 divided by 5 gives remainder -3 ; we need remainder 2)

C
Hi
If I am not mistaken, if -3 is divided by 5 , the remainder will be 2 ...
It's something about negative remainder.

@bunuel.. @pushpitkc,... kindly correct me if I am wrong...

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k=5a+2
-3 = (5)(-1) +2
Quotient = -1 and remainder = 2
Hence the OA must be E

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I think OA is incorrect. The ans should be E.

1) k+2 is leaves a remainder of 2 ,when divided by 5. Hence k is a multiple of 5. As k is integer ( not positive , but any integer ) k can be -10, -5,0,5,10.

Not sufficient
Please note that there are no negative remainders. Hence when we divide -5+2 by 5. Remainder will still be 2.

2) 18<k*<35 and k*=K^2+2 , hence value of K would be +5 and -5.

Not sufficient.

Combining both still leaves us with +5 and -5.


Hence we don't have a unique value for k. Therefore both together are insufficient.

Ans should be E.

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