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

Question

FRESH GMAT CLUB TESTS QUESTION

Is k an even integer?

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

M36-79

Bunuel · Accepted Answer

Official Solution:

Is  k  an even integer? 
 
(1)  5k + 2  is an even integer.
 
If  k  is an integer, then from  5k + 2=even  it follows that  5k=even  and in this case  k  must be even.
 
If  k  is NOT an integer, then from  5k + 2=even  it follows that  k =\frac{even}{5}  and in this case  k  must be some fraction like:  ..., \ -\frac{4}{5}, \ -\frac{2}{5}; \ \frac{2}{5},\ \frac{4}{5}, ... , so NOT an even intger.
 
Not sufficient.
 
(2)  k^2 + 1  is an odd integer. 
 
If  k  is an integer, then from  k^2+1=odd  it follows that  k^2=even  and in this case  k  must be even.
 
If  k  is NOT an integer, then from  k^2+1=odd  it follows that  k =\sqrt{even}  and in this case  k  must some irrational number like:  ..., \ -\sqrt{6}, \ -\sqrt{2}; \ \sqrt{2},\ \sqrt{6}, ... , so NOT an even intger.
 
Not sufficient.
 
(1)+(2) No irrational number satisfices (1):  5k + 2=5*irrational + integer=irrational +integer=irrational . So, we are left with the first case from (2):  k  is an even integer. Sufficient.

Answer: C

Hatakekakashi · Answer

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

VincentJongen · Answer

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

Hatakekakashi · Answer

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

aserghe1 · Answer

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

aserghe1 · Answer

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

rahul16singh28 · Answer

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.

Shobhit7 · Answer

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

Posted from my mobile device

AkshdeepS · Answer

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)

Bunuel · Answer

Par of  GMAT CLUB'S New Year's Quantitative Challenge Set

choufik · Answer

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

Posted from my mobile device

ccooley · Answer

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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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.

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