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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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21 Mar 2018, 04:15
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Difficulty:

95% (hard)

Question Stats:

20% (01:36) correct 80% (01:08) wrong based on 143 sessions

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FRESH GMAT CLUB TESTS QUESTION

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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21 Mar 2018, 04:58
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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21 Mar 2018, 09:30
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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21 Mar 2018, 11:38
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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21 Mar 2018, 15:55
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.

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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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21 Mar 2018, 16:07
2
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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.

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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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21 Mar 2018, 18:11
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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27 Mar 2018, 02:07
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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31 Mar 2018, 05:51
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.  [#permalink]

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24 Dec 2018, 01:34
Bunuel wrote:

FRESH GMAT CLUB TESTS QUESTION

Is k an even integer?

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

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Joined: 06 Nov 2018
Posts: 1
Re: Is k an even integer? (1) 5k + 2 is even. (2) k^2 + 1 is odd.  [#permalink]

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26 Dec 2018, 02:50
Hello guys,

For the combination, what if k = 0 ? (Both statements are satisfied but it is not even)
0 is considered as non-even and non-odd, right?

Thanks

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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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26 Dec 2018, 17:36
choufik wrote:
Hello guys,

For the combination, what if k = 0 ? (Both statements are satisfied but it is not even)
0 is considered as non-even and non-odd, right?

Thanks

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Nope 0 is considered to be even (as it's divisible by 2). You may be thinking of the rule about positives and negatives: 0 is neither positive nor negative.
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Re: Is k an even integer? (1) 5k + 2 is even. (2) k^2 + 1 is odd.   [#permalink] 26 Dec 2018, 17:36
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