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

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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

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

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

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