When positive integer k is divided by 6 the remainder is 3

Question

When positive integer k is divided by 6 the remainder is 3. Which of the following CANNOT be an even integer?

a. k + 1 
b. k -11 
c. 4k + 2
d. (k-3)/3 +2 
e. k/3

BrentGMATPrepNow · Accepted Answer

Another approach is to TEST values. 
 
 When positive integer k is divided by 6 the remainder is 3  
So, k could equal 3, 9, 15, 21, etc

let's TEST k =  3

a.  3  + 1 = 4 (EVEN) 
b.  3  -11 = -8 (EVEN) 
c. 4( 3 ) + 2 = 14 (EVEN)
d. ( 3 -3)/3 +2 = 2 (EVEN)

At this point, we can already see the answer must be E. 
Let's check E for "fun"
e.  3 /3 = 1 (ODD)
Great!
Answer: E

Cheers,
Brent

davedekoos · Answer

First, because k/6 leaves a remainder of 3, we know k must be odd.
(k=6x+3 = 3(2x+1) = odd*odd = odd)

Knowing k is odd we can examine the answer choices:

a. k + 1 --> odd + 1 -->  must be even 
b. k -11 --> odd - odd -->  must be even 
c. 4k + 1 --> 4*odd + 1 = even + 1 -->  must be odd 
d. (k-3)/3 +2 --> (6x+3-3)/3 + 2 = 2x + 2 -->  must be even 
e. k/3 --> (6x+3)/3 = 2x+1 =  Must be odd

It seems we have a problem with the answer choices... Both C and E CANNOT result in an even integer.

paisaj87  - are you sure that these are the correct answer choices? What is the source of the problem?

paisaj87 · Answer

Thnak you for your observation, answer choice c had a typo.

paisaj87 · Answer

Thnak you for your observation, answer choice c had a typo.

adkikani · Answer

Is there any better approach than plugging numbers here?
I spent quite a few time after arriving to below step:

k = 6* Quotient + 3 (remainder)

AkshdeepS · Answer

First step should be to find the value of k.

So from question we know that k = 3,9,15,21 .....

Plug k = 15 in answer choices

You will get :

A. 15+1 = 16

B. 15-11 = 4

C. 4*15+2 = 62

D. (15 - 3)/3 + 2 = 6

E. 15/3 = 5

Hence E.

GMATBusters · Answer

I feel the approach by davedekoos is nice one without plugging numbers... (PS: while relying to your post, I mistakenly edited it

.)

niks18 · Answer

hi  adkikani

once you have arrived at k=6*Q+3, you need to realize that k is odd because 6Q is even and 3 is odd, Hence even+odd=odd

Now Odd divided by Odd is ODD / Non Even Integer.

Hence straight option E

BrentGMATPrepNow · Answer

Here's a different approach:

GIVEN: When positive integer k is divided by 6 the remainder is 3

So, we can say that  k = 6n + 3  for some integer n

Now go to the answer choices and replace k with  6n+3 ....

a. k + 1 =  6n+3  + 1 = 6n + 4 =  2 (3n + 2). We can see that this expression is a multiple of  2 . ELIMINATE A
b. k -11 =  6n+3  - 11 = 6n - 8 =  2 (3n - 4). We can see that this expression is a multiple of  2 . ELIMINATE B
c. 4k + 2 = 4( 6n+3 ) + 2 = 24n + 14 =  2 (12n + 7). We can see that this expression is a multiple of  2 . ELIMINATE C)
d. (k-3)/3 +2 = ( 6n+3  - 3)/3 + 2 = 6n/3 + 2 = 2n + 2 =  2 (n + 1). We can see that this expression is a multiple of  2 . ELIMINATE D
e. k/3 = ( 6n+3 )/3 = 2n + 1. Since we know that 2n is EVEN, it must be the case that 2n+1 is ODD

Answer: E

Cheers,
Brent

Abhishek009 · Answer

Possible values of k are : 3 , 9 , 15 , 21.................

a. k + 1 , if k = 3 and is divided by 6 result will be even 
b. k -11 ,  if k = 15 and is divided by 6 result will be even 
c. 4k + 2 ,  if k = 3 and is divided by 6 result will be even 
d. (k-3)/3 +2 ,  if k = 3 and is divided by 6 result will be even 
 e. k/3, (Any value of K will be completely divisible by 3 and will not produce even Integer...

Answer must be (E)

ScottTargetTestPrep · Answer

k can be values such as:

3, 9, 15, 21…

When we scan the answer choices, we see that all of these values are odd when they are divided by 3, so k/3 cannot be even.

Answer: E

BrushMyQuant · Answer

When positive integer k is divided by 6 the remainder is 3

Theory: Dividend = Divisor*Quotient + Remainder

k -> Dividend
6 -> Divisor
a -> Quotient (Assume)
3 -> Remainders
=> k = 6*a + 3 = 6a + 3

Which of the following CANNOT be an even integer

Let's take each option and evaluate if it CANNOT be an even integer

a. k + 1  
k + 1 = 6a + 3 + 1 = 6a + 4 =>   EVEN

b. k -11  
k - 11 = 6a + 3 - 11 = 6a - 8 =>   EVEN

c. 4k + 2  
4k + 2 = 4*(6a + 3) + 2 = 24a + 20 =>   EVEN

d.  \frac{(k-3)}{3}  + 2  
 \frac{(k-3)}{3}  + 2 =  \frac{(6a + 3 -3)}{3}  + 2 = 2a + 2 =>   EVEN

e.  \frac{k}{3}  
 \frac{k}{3}  =  \frac{6a + 3}{3}  = 2a + 1 =>   ODD

So,  Answer will be E 
Hope it helps!

Watch the following video to learn the Basics of Remainders

https://www.youtube.com/watch?v=P0S6j2uTaj4

bumpbot · Answer

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When positive integer k is divided by 6 the remainder is 3

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When positive integer k is divided by 6 the remainder is 3

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