When the positive integer x is divided by 6, the remainder is 4. Each

Question

When the positive integer x is divided by 6, the remainder is 4. Each of the following could also be an integer EXCEPT

(A)  \frac{x}{2}

(B)  \frac{x}{3}

(C)  \frac{x}{7}

(D)  \frac{x}{11}

(E)  \frac{x}{17}

KSBGC · Accepted Answer

Note : If any number is not divisible by 3 it's not divisible by 6.

X is not divisible by 6. So, x is not divisible by 3.

The best answer is B.

souvonik2k · Answer

x/6 leaves a remainder of 4 i.e. the number could be 4, 10, 16, 22, 28, 34, 40....
Let us check the options.
A) x/2 = Integer, if x=4,10,16...--Eliminate
B) x/3 = Not integer, HOLD
C) x/7 = Integer, if x=28 --Eliminate
D) x/11 = Integer, if x=22 --Eliminate
E) x/17 = Integer, if x=34 --Eliminate.
Answer B.

gracie · Answer

x/6=q+4→
x=6q+4→
x/3=2q+4/3
if q is integer, 2q+4/3 cannot be
B

Abhishek009 · Answer

Possibles values of x are { 4 , 10 , 16 , 22, 28, 34......................}

(A) Possible as all the numbers are divisible by 2
(B) Not possible as none of the number is divisible by 3
(C) Possible, 28 is divisible by 7
(D) Possible , 22 is divisible by 11
(E) Possible, 34 is divisible by 7

Thus, Answer must be (B)

JeffTargetTestPrep · Answer

Since when x is divided by 6, the remainder is 4, x can be the following:

4, 10, 16, 22, 28, 34, ...

So we see that x will never be a multiple of 3.

Answer: B

MathRevolution · Answer

x = 6p + 4 {for p = 0, 1, 2, 3 we get x = 4, 10, 16, 22, 28, 34....}

'x' is divisible by (2, 7, 11 and 17 but not 3)

Answer B

Fdambro294 · Answer

Two rules:

Multiple of N (+/-) another Multiple of N = Multiple of N

Multiple of N (+/-) NON-Multiple of N = will always be a NON-Multiple of N

X divided by 6 leaves a remainder of 4

X/6 = a + (4/6)

X = 6(a) + 4

Where a = non-negative integer quotient

6(a), regardless of the value or a, will be a Multiple of 3

4 is NOT a Multiple of 3

X = the addition of a Multiple of 3 and a NON Multiple of 3

Thus X can never be divisible by 3

Logically:

We can think of X as the following

There is some unknown A value of balls on the floor. We are able to break up those A balls into groups of 6 ———-> with 4 balls left over

For every one of the A groups of 6 balls, we can sub-divide those groups into two groups of 3 balls each

However, the 4 balls that remain, there will always be 1 left over when we try to create another group of 3 balls

(B)
X/3 can never be an integer

Posted from my mobile device

Budhaditya_Saha · Answer

IMO, this is a very risky assumption. Although it is true that if a number is divisible by 6 it is definitely divisible by 3, the reverse is not true always. 15 is not divisle by 6 but is perfectly divisible by 3.

Correct me if I'm wrong.

Gmatnk5259 · Answer

no bcz 21 can be divisible by 3 but not by 6.

bumpbot · Answer

Automated notice from GMAT Club BumpBot: 

A member just gave Kudos to this thread, showing it’s still useful. I’ve bumped it to the top so more people can benefit. Feel free to add your own questions or solutions.

 This post was generated automatically.

When the positive integer x is divided by 6, the remainder is 4. Each

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards

GMAT Problem Solving (PS) Questions

When the positive integer x is divided by 6, the remainder is 4. Each

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards