Question

Implementing a recursive function that calculates the

remainder of dividing a very large integer by three

Implementing a recursive function that calculates the

remainder of dividing a very large integer by three

1 .) Remainder of Dividing by Three

The easiest way to find the remainder of a given integer like 235 when dividing by three is to

calculate the remainder of 2 + 3 + 5 instead. Use this idea to implement a recursive function

that calculates the remainder of any given non-negative integer when dividing by three.

Your function must obey the following signature:

int nMOD3 (int n[ ], int size)

where array n specifies the input integer by storing each digit in one cell and size determines

how many digits the input integer has. For example, nMOD3 ({2, 3, 5}, 3) is a valid function

call. The remainder of n when dividing by three has to be returned by the nMOD3 function

as its output.

Answer 1

int div3(int p) //Recursive function name

{

if(p>=3)

{

p=p/n;

div3(p); //Recursive calls

}

}