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Magoosh tricky question (or error)

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Magoosh tricky question (or error)  [#permalink]

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New post 15 May 2017, 19:52
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Question is

Is positive integer n divisible by 4 ?

1)n^2 is divisible by 8
2) sqrt n is an even integer.

Now answer states both options alone are enough to answer the question but my question is for second option

you have sqrt 16 = 4 (divisible by 4 and also even integer) but then you got sqrt 100 = 10 (even integer but not divisible by 4)

What do you guys think ?

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Magoosh tricky question (or error)  [#permalink]

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New post 15 May 2017, 20:50
Hi mbsingh
The answer is absolutely correct.


We need to see if n is divisible by 4 or not.

Statment 1 ->
n^2=8k(for some integer k)=> 2^3*k

As n is an integer => n^2 must be a perfect square => All powers of its primes must be even.
Hence n^2=2^4*some integer.

Hence n must be divisible by 4.
Sufficient

Statment 2 ==>
√n=even=2m (say)
Squaring on both sides
n=even*even => 4m^2

Hence n must be divisible by 4.
Sufficient.
Hence D.

Smash that D.

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Re: Magoosh tricky question (or error)  [#permalink]

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New post 15 May 2017, 22:33
stonecold wrote:
Hi mbsingh
The answer is absolutely correct.


We need to see if n is divisible by 4 or not.

Statment 1 ->
n^2=8k(for some integer k)=> 2^3*k

As n is an integer => n^2 must be a perfect square => All powers of its primes must be even.
Hence n^2=2^4*some integer.

Hence n must be divisible by 4.
Sufficient

Statment 2 ==>
√n=even=2m (say)
Squaring on both sides
n=even*even => 4m^2

Hence n must be divisible by 4.
Sufficient.
Hence D.

Smash that D.


How exactly is second statement sufficient ?

Square root of 100 is 10 , its even but not divisible by 4
Square root of 16 is 4, its even and divisible by 4.

Both 100 and 16 are even, so how can you surely say its sufficient ?
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Re: Magoosh tricky question (or error)  [#permalink]

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New post 15 May 2017, 22:40
mbsingh wrote:
stonecold wrote:
Hi mbsingh
The answer is absolutely correct.


We need to see if n is divisible by 4 or not.

Statment 1 ->
n^2=8k(for some integer k)=> 2^3*k

As n is an integer => n^2 must be a perfect square => All powers of its primes must be even.
Hence n^2=2^4*some integer.

Hence n must be divisible by 4.
Sufficient

Statment 2 ==>
√n=even=2m (say)
Squaring on both sides
n=even*even => 4m^2

Hence n must be divisible by 4.
Sufficient.
Hence D.

Smash that D.


How exactly is second statement sufficient ?

Square root of 100 is 10 , its even but not divisible by 4
Square root of 16 is 4, its even and divisible by 4.

Both 100 and 16 are even, so how can you surely say its sufficient ?


Here is your mistake -->

Square root of 100 is 10 , its even but not divisible by 4==> 100 is divisible by 4.
Square root of 16 is 4, its even and divisible by 4.==> 16 is divisible by 4.


You are just proving it sufficient yourself mate

This actually isn't a fruitful way to solve this up.

Think of it like => Let the number be n

√n = even
So n = even * even

Every even must have atleast one 2.
Hence a product of two even must be divisible by 4.

Hence sufficient.

I hope i am being clear now :)


Regards
Stone Cold

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Re: Magoosh tricky question (or error)  [#permalink]

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New post 15 May 2017, 22:44
fudge my life lol. Thanks stone cold now i see the issue. For whatever reason i kept on thinking that 10 should be divisible by 4. I forgot its 100 not 10 that needs to be divisible by 4.

Thanks a lot mate, i should probably catch up on my sleep. argggh
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Re: Magoosh tricky question (or error)  [#permalink]

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New post 16 May 2017, 04:41
mbsingh wrote:
Question is

Is positive integer n divisible by 4 ?

1)n^2 is divisible by 8
2) sqrt n is an even integer.

Now answer states both options alone are enough to answer the question but my question is for second option

you have sqrt 16 = 4 (divisible by 4 and also even integer) but then you got sqrt 100 = 10 (even integer but not divisible by 4)

What do you guys think ?


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Re: Magoosh tricky question (or error) &nbs [#permalink] 16 May 2017, 04:41
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