BECKONLAND2004

 Profile Answers by BECKONLAND2004 

• May 27th, 2008

 

Write a C-COde to blink the character

ravikth

 Profile Answers by ravikth 

• Oct 31st, 2008

 

Step1: We have to check the condition xy==3(x+y) -> xy/3 = x+y

The above statements unveil two essential points to write efficient program
1. x and y must be int value, not a real (float) value. so the xy will be
divisible by 3.
2. Maximum value of 3(x+y) is 54 if xy equals to 99 (max value of two digit).
so the xy will be less than 54.

Conclusion: we can get the result by checking the numbers which are divisible
by 3 and less than 54 instead of checking all the numbers.
Ans: 27

shreekanthss

 Profile Answers by shreekanthss 

• Feb 2nd, 2009

 

1  #include<stdio.h>
    #include<math.h>
    main()
{
   int x,y,r,m;
   printf("enter a numbern");
   scanf("%d",&x);
   y=x%10;
    x=x/10;
   r=x+y;
   m=3*r;
   printf("%d",m);
   getch();
}

RamLaw

 Profile Answers by RamLaw 

• Jun 16th, 2009

 

Answer for Q:5

#include<stdio.h>
int main(){
 int i = 32, n = 99, sq, val1, val2, val3;
 
 while(i <= n){
  sq = i * i;
  val1 = sq/100;
  val2 = sq%100;
  
  val3 = val1 + val2;
  
  if(val3 == i){
  printf("%d satisfy propertyn",i);
  }
 
  i++;
 }
 return 0;
}

milan.kadivar

 Profile Answers by milan.kadivar 

• Jun 16th, 2009

 

#include <stdio.h>
#include <conio.h>
void main()
{
 int i,x,y;
 clrscr();
 for(i=10;i<100;i++)
  {
    x=i/10;
    y=i%10;
    if(((x+y)*3)==i)
         printf("Peculiar Num is : %d",i);
  }
  getch();
}

jothejo2

 Profile Answers by jothejo2 

• Jun 26th, 2009

 

Q1.
This is what i've done but dere are still some bugs will fix them later
But this is the whole idea.

#include<stdio.h>

int main()
{
      for (i=10;i<=99;i++)
      { 
            j=i;
   //to do the splitting of the two digit value and find the sum of its digits
             while(j != 0)
               {    x = j%10;
                    y = y +x;
                   j = j/10;
                }
 // Because of the relation x +y = xy/3
         c = i/3;
         if (c == y)
           printf("Perculia vlue:  %d",i);
           //break;
          }
}

ranjini.np

 Profile Answers by ranjini.np 

• Aug 6th, 2009

 

/* Q1: Write a C program to find a peculiar two digit number which is three times the sum of its digits.*/


#include <stdio.h>
int main()
{
    char num;
    char second_pos, first_pos, total;
    for(num = 10; num < 100; num++)
    {
            second_pos = num / 10;
            first_pos = num % 10 ;
            total = second_pos + first_pos;

            printf("total = %d n", total);
            printf(" %d n", (total * 3 ));

            if(num == (3 * total))
            {
                printf("%d is the answern", num);
                return 0;
            }

    }
    return 0;
}

npoojasjce

 Profile Answers by npoojasjce 

• Oct 13th, 2009

 

#include<stdio.h>
void main()
{
   int i,a,b,sum;
   for(i=10;i<100;i++)
    {
        a=i/10;
        b=i%10;
       sum=a+b;
       if(b==(a-4) && (i/sum)==7)
          printf("%dn",i);
     }
}

meenu1992

 Profile Answers by meenu1992 

• Oct 23rd, 2009

 

5) Consider the following number 45*45=2025; 20+25=45.Write a program to generate number between 32 and 99 that satisfies the above property.

#include<stdio.h>
#include<conio.h>
void main()
{int i,n,a,b;
for(i=32;i<=99;i++)
{n=i*i;
a=n%100;
n=n100;
b=a+n;
{if(b==i)
printf("%dt",b);
}
}
getch();
}

abhishek tripathi

 Profile Answers by abhishek tripathi 

• Dec 15th, 2009

 

void main()
{
int i,x,y;
clrscr();
for(i=10;i<100;i++)
{
x=i/10;
y=i%10;
if(((x+y)*3)==i)
printf("Peculiar Num is : %d",i);
}
getch();
}

Ankur1989

 Profile Answers by Ankur1989 

• Jan 22nd, 2010

 

#include<stdio.h>
#include<conio.h>
void main()
{ clrscr();
  int n, x,y,z;
  printf("nEnter the number:");
  scanf("%d",&n);
  x=n%10;
  y=n/10;
  z=x+y;
  if(n==3*z)
{
  printf("nnThis is required number ....");
 }
   else
printf(""nnEnter Some other number");

}

Ankur1989

 Profile Answers by Ankur1989 

• Jan 22nd, 2010

 

//Program for numbert being three times its digits

#include<stdio.h>
#include<conio.h>

 void main()
{clrscr();
 
 int n, x,y,z;

 printf("nEnter the number you want to check for");
 
 scanf("%d",&n);

 y=n%10;
 x=n/10;
 z=x+y;

 if(n==3*z)
{
 printf("nThe number is three times some of its digits....");
}
 else

 printf("nnEnter some other number");
 
}

Ankur1989

 Profile Answers by Ankur1989 

• Jan 22nd, 2010

 

//Program to calculate bacteria on first day when bacteria on 10th day is 13, 312


 #include<stdio.h>
#include<conio.h>
 
 void main()
{
  clrscr();

 int n, i,x,
printf("nEnter the number of days:");
 scanf("%d",&n);
 n=10;

 printf("nEnter the bacteria's at end of 10th day...:");
 scanf("%d",&total);
 total=13,312;

 printf("nSo total bacteria's initially are.....:");
 x=total/pow(2,n-1);

 }

 

Ankur1989

 Profile Answers by Ankur1989 

• Jan 22nd, 2010

 

#include<stdio.h>
#include<conio.h>
#include<math.h>
 
 void main()
{clrscr();
 int n, sum,x;
 float result; 
 
  void main()
 printf("nEnter the number upto which prog is to be calculated :");
 scanf("%d",&n);

 int rec(int n)
{
 if(n==1)
 
 return 1;
 
 else
 

 return (n)*(n-1);

}

 printf("nEnter the value of X:");
 scanf("%d",&x);

 sum= 0;
 sum+=pow(x,2*(n-1))*pow(-1,n);

 result= sum/rec(n);

 printf("nnThe result is...:%f" result);
 }

dn007

 Profile Answers by dn007 

• Jan 27th, 2010

 

If (3(x+y)=10x+y

= no. is 27

(Answer: 27=3(2+7))

vkscool

 Profile Answers by vkscool 

• Sep 19th, 2010

 

programe to fine no where secound no is less then first by 4 and the no is divided by the sum of digit of the no then the quotient is 7

#include<stdio.h>
#include<conio.h>
void main()
{
 int num=20,i,j,k,l,m=0,arr[20];
 while(num<100)
 {
   j=num%10;
   k=num/10;
   if(k-j==4)
     {
       l=j+k;
       printf("n%d",num);
       if((num/l)==7)
         {
           arr[m]=num;
           m++;
         }
     }
     num++;
 }

 for(i=0;i<m;i++)
 {
  printf("n:- %d",arr[i]);
 }

 getch();

 }

Rohini Dasgupta

 Profile Answers by Rohini Dasgupta 

• Jun 10th, 2011

 

Q.2.
#include<stdio.h>
#include<conio.h>
int main()
{
int i,day,bact;
clrscr();
printf("Enter the no. of day and the total no. of bacterian");
scanf("%d %d",&day,&bact);
for(i=day;i>1;i--)
b=b/2;
printf("Number of bacteria present in the 1st day is=%dn",bact);
getch();
return 0;
}

Rohini Dasgupta

 Profile Answers by Rohini Dasgupta 

• Jun 10th, 2011

 

Q.1.
#include<stdio.h>
#include<conio.h>
int main()
{
int i,x,y;
clrscr();
for(i=10;i<100;i++)
{
x=i%10;
y=i/10;
if(i==3*(x+y))
printf("%dt",i);
}
getch();
return 0;
}

Rohini Dasgupta

 Profile Answers by Rohini Dasgupta 

• Jun 10th, 2011

 

Q.5.
#include<stdio.h>
#include<conio.h>
int main()
{
int i,p,x,y;
clrscr();
for(i=32;i<=99;i++)
{
p=i*i;
x=p%100;
y=p/100;
if(i==(x+y))
printf("%dt",i);
}
getch();
return 0;
}

Rohini Dasgupta

 Profile Answers by Rohini Dasgupta 

• Jun 10th, 2011

 

Q.6.
#include<stdio.h>
#include<conio.h>
int main()
{
float r;
int x;
clrscr();
printf("Enter the Rs.n");
scanf("%f",&r);
x=r*4/7;
printf("If there r equal no. of 25 paise coins, 50 paise coins and one rupee coins, then the no of each coins is %d n",x);
getch();
return 0;
}

Rohini Dasgupta

 Profile Answers by Rohini Dasgupta 

• Jun 13th, 2011

 

Q.8.
#include<stdio.h>
#include<conio.h>
int main()
{
int x,y,p,q;
clrscr();
printf("Enter the no of eyes and foot in totaln");
scanf("%d %d",&p,&q);
x=(q-p)/2;
y=(p-2*x)/2;
printf("The no of goats in the farm is = %dn",x);
printf("The no of ducks in the farm is = %dn",y);
getch();
return 0;
}

//The equations are
//1. 2*(x+y)=p;
//2. 4*x + 2*y=q;

Rohini Dasgupta

 Profile Answers by Rohini Dasgupta 

• Jun 13th, 2011

 

Q.10.
#include<stdio.h>
#include<conio.h>
int main()
{
int i,a,b;
clrscr();
for(i=10;i<100;i++)
{
b=i%10;
a=i/10;
if(((a-b)==4)&&(i/(a+b)==7))
printf("%dt",i);
}
getch();
return 0;
}

Suyash Patil

• Jul 22nd, 2011

 

Code
#include<stdio.h>


#include<conio.h>


 


 void main()


{clrscr();


  


 int n, x,y,z;


 


 printf("nEnter the number you want to check for");


 


 scanf("%d",&n);


 


 y=n%10;


 x=n/10;


 z=x+y;


 


 if(n==3*z)


{


 printf("nThe number is three times some of its digits....");


}


 else


 


 printf("nnEnter some other number");


 


}

Akul Swamy

 Profile Answers by Akul Swamy 

• Jun 29th, 2012

 

Code
 


#include<stdio.h>


main()


{


  int g,d;


  for(g=30,d=0;g>0,d<30;g--,d++)


    {


       if((((g+d)*2)==60)&&(((g*4)+(d*2))==86)&&(g+d==30))


       { printf("The number of goats and ducks in the farm are %d and %d",g,d);


         break;


       }


    }                         


  system("pause");


}

anusha

• Aug 6th, 2012

 

10th ques answer is
62
73

84

Code
#include<stdio.h>


#include<conio.h>


main()


{


int a,b,i,l;


printf("the numbers are:");


for(i=40;i<100;i++)


{


a=i/10;


b=i%10;


if(a>b)


{


if(a-b==4)


{


l=a+b;


if(i/l==7)


{


printf("


%d",i);


}


}


}


}


getch();


}

Ranjini

• Mar 12th, 2013

 

Given below the code for question 1

Code
/* C program to find a peculiar two digit number which is three times the sum


 * of its digits.


 * Solution :-


 *  let say number is 10x+y


 *  According to the question 10x+y = 3 ( x + y)


 *  10x + y = 3x + 3y


 *  7x = 2y -------> (1)


 *  If any of the two one digit number satisfies the equation (1) , then


 *  the two digit number is (10x+y)


 *


 */


 


#include<stdio.h>


 


int main(void)


{


    int x = 0;


    int y = 0;


 


    for ( x = 1; x < 10; x++)


    {


        for (y = 1; y < 10; y++)


        {


            /*Check the condition 7x == 2y */


            if( (7 * x ) == (2 * y))


            {


                    printf("The two digit number is %d 


", ((10 * x) + y));


            }


        }


    }


 


    return 0;


}


 

Ranjini

• Mar 12th, 2013

 

Answer for Q:2

Code
/* Bacteria are known to multiply very rapidly. If a certain container contains


 * just one bacterium on the first day and there are twice as many on the next


 * day. In this manner the number of bacteria in the container doubles itself


 * everyday. Assuming that the container would be full on the 10th day with


 * 13,312 bacteria, find the number of bacteria that was initially in the


 * container on the first day.


 * Solution :-


 *  It is geometrical progression


 *  eg :- a, a*r, a*r*r, a*r*r*r, ...


 *  The n-th term of a geometric sequence with initial value a and


 *  common ratio r is given by


 *  an = a * (r to the power (n-1))


 *  here r = 2


 *  an = 13312


 *  n = 10


 *  13312 = a * (2 to the power (10 - 1))


 *  13312 = a * (2 to the power 9)


 *  a = 13312 / (2 to the power 9)


 */


 


 


#include<stdio.h>


#include<math.h>


 


int main (void)


{


    int n = 10;


    int an = 13312;     //nth term of gp , here n is 10 .


    int r = 2;          //common ratio


    int power_value;    //this variable is used for storing (r to the power (n-1))


    float a;              //scale factor


 


    power_value = pow(2, 9);


    // a = an/(2 to the power (n - 1))


    a = an / power_value;


    printf(" The number of bacteria initially present in the container on the first day is %f


", a);


    return 0;


}


 

Ranjini

• Mar 15th, 2013

 

Answer to the question 4

Code
/*


 * A number of Cats got together and decided to kill between them 999919 mice.


 * Every cat killed equal number of mice. Write a program to find number of cats


 * soln:-


 *  total number of cats = x


 *  total number of mice = 999919


 *  number of mice killed by cat = y


 *  xy = 999919


 */


 


 


#include<stdio.h>


 


int main(void)


{


    int n_of_cats = 0;


    int total_mice = 999919;


    int n_of_mice_killed_by_each_cat = 0;


    int exit_flag = 0;


 


    for( n_of_mice_killed_by_each_cat = 2; n_of_mice_killed_by_each_cat <= 999919; n_of_mice_killed_by_each_cat++)


    {


        for(n_of_cats = 2; n_of_cats <= 999919; n_of_cats++)


        {


            if((n_of_cats * n_of_mice_killed_by_each_cat) == 999919)


            {


                printf("Each cat killed %d mice and total number of cats present are %d 


", n_of_mice_killed_by_each_cat, n_of_cats);


                exit_flag = 1;


            }


        }


        if(exit_flag == 1)


        {


            break;


        }


    }


 


    return 0;


}


 

Ranjini

• Mar 15th, 2013

 

Answer to the question - 5

Code
/*


 * Consider the following number 45*45=2025; 20+25=45.Write a program to


 * generate number between 32 and 99 that satisfies the above property.


 * Soln:-


 *  x * x = 100y + z


 *  y + z = x


 *  (y + z) * (y + z) = 100y + z


 *


 */


 


#include<stdio.h>


 


int main(void)


{


    int y = 0;


    int z = 0;


    int x = 0;


    int i = 0;


    for(y = 1; y < 98; y++)


    {


        if( y == 20 )


        {


            continue;


        }


        for( z = 1; (z + y) < 98 ; z++)


        {


            x = y + z;


            if ( x < 33 || x > 99)


            {


                continue;


            }


            if(x * x == ((100 * y) + z))


            {


                printf(" The number which follows the propery ( 45 * 45 = 2025 ; 20 + 25 = 45 ) is %d  


  And the property is ( %d * %d = %d ; %d + %d = %d)


", x, x, x, (x * x), (x *x)/100, (x *x )%100, x);


            }


        }


    }


 


    return 0;


}


 

Ranjini

• Mar 15th, 2013

 

Answer to the question 6

Code
/*


 * Rita has a money pouch containing Rs.700. There are equal number of


 * 25 paise coins, 50 paise and one rupee coins. Write a C program to find


 * how many of each are there?


 * Soln:-


 * total amount = 700


 * coins present are 25 paise, 50 paise and 1 rupee


 * Let x be the total count of each coins


 * then


 * x * (25 + 50 + 100) = 700 * 100


 * x * 175 = 700 * 100


 * x = 700 * 100 /175 = 400


 *


 */


 


#include<stdio.h>


 


int main(void)


{


    int total_amt_in_paise = 700 * 100 ;


    int sum_of_three_coins = 25 + 50 + 100 ;


    int count_of_each_coin = 0;


 


    count_of_each_coin = total_amt_in_paise / sum_of_three_coins ;


 


    printf("The total count of each coins(25paise, 50paise, 1rs) is %d 


", count_of_each_coin);


 


    return 0;


}


 

manasa

• Mar 15th, 2013

 

Code
 


#include<stdio.h>


main()


{ int a=27,b,p,sum;


        b=a;


        while(b>0)


        {


                p=b%10;


                sum=sum+p;


                b=b/10;


                }


                if(a==(3*sum))


                


                printf("


 %d is the REQUIRED number:");


                        else


                        printf("


 not THE REQUIRED number:");


                        

ranjini

• Mar 15th, 2013

 

Answer to the Question 9

Code
/*


 *  Write a C program to find a three digit number which is greater than the


 *  aggregate of its third, tenth and the twelfth parts by 58.


 *  Soln:-


 *  Let x be the 3 digit number .


 *  x = (x / 3) + (x /10 ) + (x /12) + 58


 *  x = ( 120 x + 36 x + 30 x )/ 360 + 58


 *  360 x - 186 x = 58 * 360


 *  174 x = 58 * 360


 *  x = (58 * 360 ) / 174


 */


 


 


#include<stdio.h>


 


int main(void)


{


    float x ;


 


    for(x = 100; x < 1000; x++)


    {


        if(x == ((x / 3) + (x / 10) + (x / 12)) + 58)


        {


            printf("%f 


", x);


        }


    }


 


    return 0;


}


 

ranjini

• Mar 15th, 2013

 

Answer to the question 1

Code
/* C program to find a peculiar two digit number which is three times the sum


 * of its digits.


 * Solution :-


 *  let say number is 10x+y


 *  According to the question 10x+y = 3 ( x + y)


 *


 */


 


#include<stdio.h>


 


int main(void)


{


    int x = 0;


    int y = 0;


    int lhs = 0;


    int rhs = 0;


    for ( x = 1; x < 10; x++)


    {


        for (y = 1; y < 10; y++)


        {


            /*Check the condition (10* x )+ y == 3 * ( x + y) */


            lhs = (10 * x) + y;


            rhs = 3 * (x + y);


 


            if( lhs == rhs)


            {


                    printf("The two digit number is %d 


", lhs);


            }


        }


    }


 


    return 0;


}


 

Ranjini

• Mar 15th, 2013

 

Answer to the question 10

Code
/*


 * Write a C program to find a two digit number, the second digit of which is


 * smaller than its first digit by 4, and if the number was divided by the


 * digits sum, the quotient would be 7.


 * Soln:-


 * Let the two digits are x and y and two digit number is (10x+y)


 * y = x - 4


 * (10x+y)/x+y = 7


 */


 


#include<stdio.h>


 


int main(void)


{


    float x = 0;


    float y = 0;


 


    for(x = 1; x < 10; x++)


    {


        for(y = 0; y < 10; y++)


        {


            if(((((10 * x) + y ) / (x + y)) == 7) && (y == (x - 4)))


            {


                printf("The two digit number which satisfy the condition ( (10x+y)/x+y = 7 ) and ( y = x - 4) is %f


", (10 * x) + y);


            }


        }


    }


   return 0;


}


 

muskan

• Jul 29th, 2013

 

Code
#include<stdio.h>


#include<conio.h>


 


int main()


{


    int n;


    printf("enter the number");


    scanf("%d",&n);


    int a = n/10;


    int b= n%10;


    int c = a+b;


    int x= n/c;


    if(x==3)


    printf("the number which is 3 times of sum of its digit is:%d",n);


    getch();


    return 0;


}


 

muskan

• Jul 29th, 2013

 

Q:2;

Code
#include<stdio.h>


#include<conio.h>


 


int main()


{


    int n;


    n=13312;


  int i;


    for(i=0;i<10;i++)


    n=n/2;


    


    printf("no. of bacteria in container : %d",n);


 


    


    getch();


    return 0;


}


 

muskan

• Aug 1st, 2013

 

Q3:Calculate the factorial of a number recursively. From that calculate the value of COS(X) = 1 - X2/2! + X4 /4! - X6/ 6! +. . .+Xn/n!.

Code
#include<stdio.h>


#include<conio.h>


 


long factorial1(int n);


 


int main()


{


    int n;


  int i;


  long fact=1;


 


printf("enter the number:");


scanf("%d",&n);


int x;    


printf("enter X:");


scanf("%d",&x);


 


 float sum=0;


 


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


{


 int m= i%2;


 if(m==0)


 {


    m=i/2;


    if(m%2==0)


     {


       sum = sum+(float)(pow(x,i)/factorial1(i));


       }   


       else


       {


       sum = sum - (float)(pow(x,i)/factorial1(i));                


       }


 


 


}


 


}    


printf("cos(x):%f",sum);


    getch();


    return 0;


}


 


 


 


long factorial1(int n)


{


     


     if(n==0)


     return 1;


     else


   return (factorial1(n-1) * n);


          


     


     


 }


 

sourav punoriyar

 Profile Answers by sourav punoriyar 

• Sep 5th, 2013

 

Code
//answer to 2025 question


 


#include<stdio.h>


 


main()


 


{


 


printf("only for two digit numbers...


");


 


int x,y,i;


 


int b=0;


 


 


for(i=10;i<100;i++)


 


{


 


b=0;


 


x=i*i;


 


 


while(x)


 


{


 


y=x;


 


b=b+(y%100);


 


 


x=x/100;


 


 


}


 


if(b==i)


 


printf("number=%d, num*num=%d 


",i,i*i);


 


}


 


}


 

sourav punoriyar

 Profile Answers by sourav punoriyar 

• Sep 5th, 2013

 

Code
//code for goat and duck


#include<stdio.h>


 


//with command line input...


 


main(int argc,char **argv)


 


{


 


 


int eyes,foot,gf,ge,de,df;


 


int r1,r2,cnt1=0,cnt2=0,r3,r4;


 


if(argc!=3)


 


{


 


printf("wrong entry...


");


 


}


 


else


{


 


eyes=atoi(argv[1]);


 


foot=atoi(argv[2]);


 


ge=eyes/2;


 


printf("two possiblities....


");


 


printf("**************************************************************


");


 


printf("first possblity if goats assumed to be more than ducks 


");


 


printf("no of goats: %d 


",(foot/4));


 


printf("foot= %d 


",foot/4);


 


r1=ge-(foot/4);


 


printf("remaing eyes= %d 


",r1);


 


r2=foot-((foot/4)*4);


 


while(r2>=2)


 


{


 


if(cnt1<r1)


cnt1++;


 


r2=r2/2;


 


}


 


printf("remaining foots= %d 


",r2);


 


printf("no of ducks: %d 


",cnt1);


 


 


printf("*****************************************


");


 


printf("second possblity if ducks assumed to be more than goats 


");


 


r3=foot/2;


 


r4=0;


 


while(r4<r3)


 


{


 


cnt2++;


 


 


r4++;


 


if(cnt2>(eyes/2))


 


break;


 


}


 


printf("no of ducks: %d 


",cnt2);


 


printf("foot= %d 


",foot/2);


 


r1=ge-(foot/2);


 


printf("remaing eyes= %d 


",r1);


 


r2=foot-((foot/2)*2);


 


while(r2>=2)


 


{


 


if(cnt1<r1)


 


cnt1++;


 


r2=r2/2;


 


}


 


printf("remaining foots= %d 


",r2);


 


printf("no of goats: %d 


",cnt1);


 


 


 


}


 


 


 


 


}


 


 

paritosh

• Aug 24th, 2014

 

#include
using namespace std;
int main()
{
int end=13312;
for(int i=0;i<10;i++)
{
end=end/2;
}
cout<<"The number is :"< return 0;
}

~
~

Rohan

• Jun 18th, 2015

 

It says find the magic number, so we need to check all possible combinations.
Maximum number is 3*(9+9) = 54. So we can check for all numbers from 0 to 54.

Code
    #define MAX_POSSIBLE_NUM 3*(9+9)


    int num_to_check;


    float a=0,b=0;


 


    for(a=0;a<=(MAX_POSSIBLE_NUM/10);a++){


        for(b=0;b<=9;b++){


            num_to_check = (int)((a/10 + b/100)*100);


            if(num_to_check == 3*(a+b))


               printf("Number to check = %d


", num_to_check);


        }


    }


 

Rohan

• Jun 18th, 2015

 

I dont know if its optimized but works

Code
{/*There are some goats and ducks in a farm. There are 60 eyes and 86 foot in total. 


   Write a program to find number of goats and ducks in the farm */


 


   int num_goats=0, num_ducks=0, animals=0;


   int eyes = 60, feet = 86;


 


   animals = eyes/2;


 


   for(num_goats=0;num_goats<animals;num_goats++){


       for(num_ducks=0;num_ducks<animals;num_ducks++){


           if((num_goats*4 + num_ducks*2 == feet ) && (num_goats*2 + num_ducks*2 == eyes) && (num_goats + num_ducks == animals))


               printf("Goats :: %d & Ducks :: %d


", num_goats,num_ducks);


 


       }


   }


}   


 

Aneesh

• Jul 29th, 2016

 

27

Code
#include<stdio.h>


#include<conio.h>


 void main()


  {


    int i,k,a,sum=0;


    clrscr();


     for(i=10;i<=99;i++)


      {


          k=i;


          a=k%10;


          k=k/10;


          sum=(a+k)*3;


          if(sum==i)


           printf("%d",i);


       }


     getch();


  }