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Is k an even integer? (1) 5k + 2 is even. (2) k^2 + 1 is odd.

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Is k an even integer? (1) 5k + 2 is even. (2) k^2 + 1 is odd.  [#permalink]

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New post 21 Mar 2018, 05:15
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A
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C
D
E

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  95% (hard)

Question Stats:

25% (01:34) correct 75% (00:58) wrong based on 75 sessions

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Is k an even integer? (1) 5k + 2 is even. (2) k^2 + 1 is odd.  [#permalink]

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New post 21 Mar 2018, 05:58
Bunuel wrote:

FRESH GMAT CLUB TESTS QUESTION



Is k an even integer?

(1) 5k + 2 is even.
(2) k^2 + 1 is odd.



5k +2 = even

2 is even
5k is even

hence 5k= even
k has to be even
but if k=2/5
a insufficient


k^2 + 1 = odd

even+ odd = odd

odd+odd = even

so k^2 = even

but if k=\sqrt{2}
the equation is still satisified

(c) imo
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Re: Is k an even integer? (1) 5k + 2 is even. (2) k^2 + 1 is odd.  [#permalink]

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New post 21 Mar 2018, 10:30
1)

O x E = E + E = E Thus, K = even
O x O = O + E = O K = odd does not satisfy the statement. Thus, k must be even = sufficient

2)

O x O = O + O = E Thus, K = odd does not satisfy statement
E x E = E + O = O K = even does satisfy statement, thus k must be even = sufficient

= D imo
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Re: Is k an even integer? (1) 5k + 2 is even. (2) k^2 + 1 is odd.  [#permalink]

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New post 21 Mar 2018, 12:38
VincentJongen wrote:
1)


O x O = O + O = E Thus, K = odd does not satisfy statement
E x E = E + O = O K = even does satisfy statement, thus k must be even = sufficient

= D imo


k^2 + 1 = odd

so k is even

what if k = \(\sqrt{2}\)

then we have k^2 + 1 = 3 as k =\(\sqrt{2}\)

but here k is \(\sqrt{2}\) , which is not an integer

so b wouldn't be sufficient imo
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Re: Is k an even integer? (1) 5k + 2 is even. (2) k^2 + 1 is odd.  [#permalink]

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New post 21 Mar 2018, 16:55
This is a tricky one. The question asks if \(k\) is an even integer.

Statement 1
If \(k = 2\), then the answer to the question is YES.
If \(k = \frac{2}{5}\), then the answer to the question is NO.
Not sufficient.

Statement 2
If \(k = 2\), then the answer to the question is YES.
If \(k = \sqrt{2}\), then the answer to the question is NO.
Not sufficient.

Combine Statements 1 & 2
There has to be a simpler approach (please show me). Here's what I did...

Statement 1 in equation form is:
\(5k + 2 = 2n\), where \(n\) is an integer.
Solve for k,
    \(5k + 2 = 2n\)
    \(5k = 2n - 2\)

    \(k = \frac{2(n - 1)}{5}\) .........Equation 1

Statement 2 in equation form is:
\(k^2 + 1 = 2m + 1\), where \(m\) is an integer.
Solve for k,
    \(k^2 + 1 = 2m + 1\)
    \(k^2 = 2m\)

    \(k = \sqrt{2m}\) .........Equation 2

Equate the two equations and simplify:
\(\frac{2(n - 1)}{5} = \sqrt{2m}\)

\(\frac{4(n - 1)^2}{25} = 2m\)

\(4(n - 1)^2 = 50m\)
\(2(n - 1)^2 = 25m\) .........Equation 3

Since \(n\) and \(m\) are integers, the LHS and RHS are integers. Also, the RHS is divisible by 25, so the LHS must be divisible by 25. Thus \((n - 1)^2\) must be divisible by 25. This means
\((n - 1)^2 = 25r\), where \(r\) is an integer.
\(n - 1 = 5\sqrt{r}\).
\(n - 1\) is an integer and thus \(5\sqrt{r}\) is an integer. And since \(r\) is an integer, \(\sqrt{r}\) must be an integer. Thus, \(n - 1\) is a multiple of 5. In other words,
\(n - 1 = 5s\) for some integer \(s\). .........Equation 4

Now we know that \(n\) can't be every single integer, but a subset of integers, e.g. {6,11,16,21,etc}. Let's substitute Equation 4 into Equation 1 and see what we get for \(k\):

\(k = \frac{2(n - 1)}{5} = \frac{2(5s)}{5} = 2s\)

This tells us that \(k\) must be an even integer. Both statements combined are sufficient.

Answer: C
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Re: Is k an even integer? (1) 5k + 2 is even. (2) k^2 + 1 is odd.  [#permalink]

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New post 21 Mar 2018, 17:07
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aserghe1 wrote:
This is a tricky one. The question asks if \(k\) is an even integer.

Statement 1
If \(k = 2\), then the answer to the question is YES.
If \(k = \frac{2}{5}\), then the answer to the question is NO.
Not sufficient.

Statement 2
If \(k = 2\), then the answer to the question is YES.
If \(k = \sqrt{2}\), then the answer to the question is NO.
Not sufficient.

Combine Statements 1 & 2
There has to be a simpler approach (please show me). Here's what I did...


Wow I think I just figured out the simpler approach...

Statement 1 says that 5k + 2 is even. Or more specifically (redundantly?), an even integer. So 5k must be an even integer. k can either be a fraction with 5 in the denominator and an even number in the numerator, or k can be an even integer.
Statement 2 says that k^2 + 1 is odd.... i.e. an odd integer. So k^2 must be an even integer. k can either be the square root of an even integer, or k can be an even integer.
Combining the 2 statements, the one thing in common is that k is an even integer.

Answer: C
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Re: Is k an even integer? (1) 5k + 2 is even. (2) k^2 + 1 is odd.  [#permalink]

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New post 21 Mar 2018, 19:11
1
Bunuel wrote:

FRESH GMAT CLUB TESTS QUESTION



Is k an even integer?

(1) 5k + 2 is even.
(2) k^2 + 1 is odd.


Statement I:

Consider \(K =\frac{2}{5}, 2\). Not Sufficient.

Statement II:

Consider \(K = \sqrt{2}, 2\). Not Sufficient.

Combining I & II:

K has to be even number.
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Re: Is k an even integer? (1) 5k + 2 is even. (2) k^2 + 1 is odd.  [#permalink]

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New post 27 Mar 2018, 03:07
Under root of a number can never be equal to a fraction of two integers.

Eg. √2 cannot be expressed as a/b where a & b are Integers.

Hence, when evaluating St.1+2:
Only option left for 'k' is to be an integer.

It is already clear from each statement that if k is an integer, then it's an even Integer.

So, imo, correct Ans is C

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Re: Is k an even integer? (1) 5k + 2 is even. (2) k^2 + 1 is odd.  [#permalink]

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New post 31 Mar 2018, 06:51
Bunuel wrote:

FRESH GMAT CLUB TESTS QUESTION



Is k an even integer?

(1) 5k + 2 is even.
(2) k^2 + 1 is odd.


Question : Is k even?

St 1: \(5k + 2\) even

If k = 2, answer to the QUESTION is YES

If k = 2/5, answer to the QUESTION is NO (Insufficient)

St 2: \(k^2 + 1\) is odd

If k = 2, answer to the QUESTION is YES

If k = \(\sqrt{2}\) or\(\sqrt{6}\), answer to the QUESTION is NO

Combining we get k = even

Hence (C)
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Re: Is k an even integer? (1) 5k + 2 is even. (2) k^2 + 1 is odd. &nbs [#permalink] 31 Mar 2018, 06:51
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