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convert int to char* and add all digits.. continue this recursively until the sum reaches 1 digit. if the sum is 3/6/9, then it is divisible by 3

@ak
what we we have to find whether the number is divisible by 3 or we want to implement this number /3.

for checking divisibility solution given by "kiran" is good.

Otherwise. (n>>1) -1

Assuming it is a positive integer and you are looking for integer (i.e. 5/3 = 1 and not 1.666666) then we can use binary search.

You double your estimate(x) for n/3 till you hit one for which x+x+x>n. Now you can binary search the range based on the previous value and current value of x.

n/3 = pow(2, log2(n) - log2(3))

geeksforgeeks.org/?p=511

The simple way of doing that will be , convert the given number into string using itoa, then add get the sum of the given number, check whether sum is 3 6 or 9 , if yes then its divisible :)