Why isn't $6$ necessarily a divisor of $\ \ \ \cdots 336$? I have to prove that if a given positive integer n ends with digits 336, then n has to be divisible by 2, by 4 and by 8, but its not necessarily divisible by 6.
I know that a positive integer is divisible by 2 if the last digit is 0, 2, 4, 6, and 8. And because 4, 6, and are multiples of 2 they should also be divisible. However I can't find why it wouldn't be divisible by 6?
Here's my proof so far for divisibility by 2:
10 divided by 2 has a remainder of 0.
So 10 ≡ 0 (mod 2). Then 10k ≡ 0
k ≡ 0 (mod 2) for k = 1, 2, 3, . . .. 
so
x ≡ a0 + a1 · 0 + a2 · 0 + a3 · 0 + a4 · 0 + · · · + am · 0
≡ a0 (mod 2).
Therefore x is divisible by 2 if and only if its last digit a0 is divisible by 2, which
happens if and only if the last digit is one of 0, 2, 4, 6, 8.
a0 is the digit in the one's place, a1 in the ten's place etc.
 A: If the digits before $336$ do not form a number divisble by $3$, the number is not divisible by $3$ and hence not divisible by $6$, the smallest (counter-)example is $1336$.
A: Usually for this type of question you just need to give a counterexample.
1336 ends with 336, but is it divisible by 6?
A: Hint:
A number is divisible by $6$ if it is divisible by $2$ and by $3$. 
Add to $...336$ some digit such that the sum of the digits is not a multiple of $3$.
A: In order to show that a set of numbers are not all divisible by 6, all you need to do is show an example where it is not (as well as, for completeness, one that is).
The lowest positive integers ending with $336$ are $336$ itself ($= 6 × 56$) and $1336$ ($= 6 × 222 + 4$). These should do nicely.
The reason why the same logic used for 2, 4 and 8 cannot be used for 6, is that 2, 4 and 8 all divide 1000 (which itself divides the difference between two numbers ending in the last 3 digits), 6 does not - so if 6 divides $n$, it does not divide $n+1000$.
A: Let $n$ be a positive integer ending in digits $336$.  Observe that $336$ is divisible by $8$.  Since $1000 = 125 \times 8$ is divisible by $8$ and $n$ is of the form $n = m \times 1000 + 336$, $n$ is also divisible by $8$. In particular, $n$ is also divisible by $2$ and $4$.  
$n$ is divisible by $6$ if and only if it is divisible by $2$ and $3$.  But a number ending in digits $336$ need not be divisible by $3$ (because the sum of the digit of $n$ need not be divisible by $3$; for example, take $n=1336$).
