CareerCup

/*The code is as follows*/
int size(int n)
{
int c=1;
if(n/10!=0)
{
n=n/10;
c=c+1;
}
return c;
}
void main()
{
int n,i,l,k;
cin>>n;
m=size(n);
for(i=0;i<m;i++)
{
if(n%10!=0)
{
k=n%10;
n=n/10;
l=n-2k;
if(l%7==0)
{
cout<<"divisible";
break;
}
else
{
cout<<"Not Divisible";
break;
}
}
else
{
if(n>=0)
{
n=n/10;
}
else
{
cout<<"Not divisible";
break;
}
}

Is there any bit pattern based solution. like 'divisible by 2' problem ?

Convert the number to its octal notation, and sum up the digits. The sum is multiple of 7 if and only if the original number is.

Remove the last digit of the number, double that digit and subtract from the remainder. Continue this till left with one digit. If that digit is 7/-7/0, you have a winner.

e.g.
14
Remove 4 and double it i.e. 8
Subtract 8 from 1, we get -7 which is awesome.

35 => 3-10 = -7
154 => 15 - 8 = 7
147 => 14-14 = 0

You could subtract 7 in a loop until the number reaches either 0 or <1. If it reaches 0 then it is multiple of 7, else not divisible.

A number of the form 10x + y is divisible by 7 if and only if x − 2y is divisible by 7. In other words, subtract twice the last digit from the number formed by the remaining digits. Continue to do this until a number known to be divisible by 7 is obtained. The original number is divisible by 7 if and only if the number obtained using this procedure is divisible by 7. For example, the number 371: 37 − (2×1) = 37 − 2 = 35; 3 − (2 × 5) = 3 − 10 = −7; thus, since −7 is divisible by 7, 371 is divisible by 7.

Reference : en.wikipedia.org/wiki/Divisibility_rule#Divisibility_by_7

if(n==n&(n+1))
cout<<"yes Divisbile"<<endl;
else
cout<<"No"<<endl;