Check out divisibility concept here:
https://www.veritasprep.com/blog/2011/0 ... unraveled/

When n is split into groups of 6, 3 is left over.
If you remove 3 from n, there will be only groups of 6 leftover. So (n - 3) will be divisible by 6.
If you add 3 to n, the remaining will also form a complete group of 6 and hence there will be only groups of 6. So (n + 3) will be divisible by 6.

When you double n, 3 left over will become 6 and so we will have only groups of 6. Hence 2n will be divisible by 6.
If 2n is divisible by 6, 4n must be divisible by 6 too.

When you triple n, 3 will still be left over. So 3n will not be divisible by 6.

Answer (D)
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Since 6q+3=n where q is the quotient, n-3 is a multiple of 6.

Now, 6q+6=n+3, so n+3 is also a multiple.

2 x (6q-3)=2n is also a multiple, so it 4n

We can see that only 3n is not a multiple.

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The integer n is divided by 6, the remainder is 3

This meabs that \(n = 6q+3\)

If n is divided by even number and leaves remainder it means that N is ODD number

to prove this just plug any numbers for q

n=9= 6(1)+3

now test 9 in aswer choices

(A) 9 – 3 = 6 (MULTIPLE OF 6)
(B) 9 + 3=12 (MULTIPLE OF 6)
(C) 2(9)=18 (MULTIPLE OF 6)
(D) 3(9)= 27 ( 27 IS NOT MULTIPLE OF 6 )
(E) 4 (9) =36 (MULTPLE OF 6)

IMO: D

n=6q+3→
3n=18q+9
18q is a multiple of 6
9 is not a multiple of 6
so 18q+9 is not a multiple of 6
3n
D

What about 20? if 20/6, it still leaves a remainder but 20 ain't odd. Can anyone explain?

n is divided by 6, the remainder is 3 => n = 3 [3 = 0 * 1 + 3]

=> 3 - 3 = 0 and 3 + 3 = 6 is multiple of 6. (Option A and B)

=> 2 * 3 = 6 and 4 * 3 = 12 is multiple of 6. (Option C and D)

=> 3 * 3 = 9 is not multiple of 6 - (Option D)

Answer D
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let us plug in and check
let n=21
18 is dvisible by 6 therefore A is out
24 is divisible therefore B is out
63 is bot divisible by 6
84 is divisible by 6 therefore out

hence IMO D

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