Divisibility by 8
Method 1:
 If the number formed by the last three digits at the hundreds ,tens and unit place is divisible by 8 then the number is divisible by 8.
Method 2:

 The digit at the hundreds place is multiplied by 4.
 The digit at the tens place is multiplied by 2.
 The result of step 1 and 2 is added to the digit at the unit place.
 Repeat the process if you get a large number.
 The result is divided by 8. If the remainder is zero then the number is divisible else the remainder is the remainder of the original number.
Proof:
Let the number be abcde.
abcde can be written as
a×10000 + b×1000 + c×100 + d×10 + e
which can also be written as
a×8×1250 + b×8×125 + c×100 + d×10 + e
Every power of 10 i.e. 10^{x} is divisible by 8 if x>2 and 2 is a positive integer.
So a×8×1250 + b×8×125 is divisible by 8 as it has 8 as factor or common factor.
The number is divisible by 8 if 100×c+10×d+e or cde is divisible by 8.
The left is 100×c+10×d+e
Which can be written as
c(8×12+4)+ d(8+2) + e = c×8×12 + 4c + 8d + 2d + e
The above is divisible by 8 if 4c+2d+e is divisible by 8.
Remainder
 The number formed by the last three digits when divided by 8 gives a remainder and this is the remainder of the original number.
 If the number formed by the last three digits is abc. Then the remainder is the remainder of the number 4a+2b+c when divided by 8.
Examples
 Is 348 divisible by 8.
(3×4+4×2+8)=28. The remainder when 28 is divided by 8 is 4. Hence the remainder is 4.
 Is 1344 divisible by 8.
The number formed by the last three digits is 344 and it is divisible by 8. Hence, the number is divisible by 8.