#include<stdio.h>

#include<conio.h>

void main()

{

int i,n;

float j=0;

clrscr();

printf("enter the no");

scanf("%d",&n);

for(i=n;i>2;i=i-3)

{

j=j+1;

if(i==4)

{

j=j+1.333333;

}

if(i==5)

{

j=j+1.666666;

}

}

printf("%f",j);

getch();

}

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The number is N, the decimal places to which the division will be done is D.

int quot=0,remainder=0,rem=0;

int x=0;

Void DivisionBySubtraction(N,D)

{

while(N>=3)

{

N-=3;

++quot;

}

if (N==0)

cout<<"quotient="<<quot;

else

{

while(x<D)

{

N*=10;

while(N>=3)

{ N-=3;

rem++;

}

remainder = remainder*10+rem;

if (N==0)

break;

++x;

}

}

cout<<"remainder="<<remainder;

}

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Can we use logarithm."logarithm of a number-logarithm of 3".Taking antilog gives the result

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void dvdby3(int n)

{

int count=0,rem;

while(n>0)

{

rem=n;//reminder

n=n-3;

count++;

}

}

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quo=-1;rem=0;

//n the number entered

for(i=0;i<=n;i+=3)

{

quo+=1;

}

rem=n-i+3;

printf("%d is the quotient n%d is the remainder",quo,rem);

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int quoitent=-1, rem=0;

int number;

while(number>=0)

{

rem=number;

quoitent +=1; //quoitent= quoitent+1;

number -=3; //number=number-3;

}

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**Alok Mishra**
Answered On : Aug 4th, 2011

{geshibot language="c"}#include
#include
void main()
{
int i,n;
float m,j=0;
clrscr();
printf("enter the no");
scanf("%d",&n);
if(n==1)
{
m=0.3333;
printf("%f",m);
}
if(n==2)
{
m=0.6666;
printf("%f",m);
}
if(n >= 3)
{
for(i=n;i>2;i=i-3)
{
j=j+1;
if(i==4)
j=j+1.333333;
if(i==5)
j=j+1.666666;
}
printf("%f",j);
}
getch();
}
}{/geshibot}

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