Determine whether a number n is divisible by both 3 and 5. The algorithm for the problem is represented in a traditional flowchart

A number divisible by 3 has individual digits that sum to a multiple of 3.

Numbers divisible by 5 with a final digit of 0 or 5.

==>Numbers divisible by both 3 and 5 with individual digits summing to a multiple of 3 and a final digit of 0 or 5.

Or more simply, divisible by 15.

Algorithm:

1. Determine if it's a multiple of 3

---- split the n

---. --- Add each number and divide by 3

---- remainder == 0?

- --- Yes, go to the next step; No, exit no

2. Determine whether the last digit is 0 or 5

--- --- last digit == 0 or == 5?

----Yes, exit yes; No, exit no

Or

1, to determine whether it is a multiple of 15

----n take the remainder of 15

---- remainder == 0?

----Yes, exit yes; No, exit no

Program:

#include

void

main()

{

int

i,m,k=10,s=0;

int

a[10];

printf("Input a positive integer:");

scanf("%d",&m);

a[0]=0;

for(i=1;i& lt;10;i++)

{a[i]=(m-s)%k;k*=10;s+=a[i];}

if(s%3==0)

{if(a[1]==0||a[1]==5)printf("\nm can be divisible by both 3 and 5\n");

else

printf("\nm cannot be divisible by both 3 and 5\n");

}

else

printf("\nm cannot be divisible by both 3 and 5\n");

}