Choose the correct one OR Answer yourself for a tricky C Program

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

 Answers:
1) 

#include<stdio.h>

#include<conio.h>

void main()

{

int n,temp,rem,sum=0;

printf("enter the two disit number");

scanf("n%d",&n);

for(temp=n;temp>0;temp/=10)

{

rem=temp%10;

sum+=rem;

}

if(n==(sum*3))

printf("the number which is equal to the three times the sum of its disits is %d",n);

else

printf("the number is not three times the sum of its disits");

getch();

}

 

 

 

 

 

 

2)

#include using namespace std; int main()
{
int n a sum=0 dig=0; cout << Enter a number << endl;
cin>>n; if(n<=0||n>99)
{
cout<< Enter 2 digit pecular number ;
return 0;
}
a=n; while(n!=0)
{
dig=n 10; sum=sum+dig; n=n/10;
}
if(a==(3*sum))
{
cout<< Pecular number ;
}
else cout<< Not a pecular number ;
return 0;
}

 

 

 

 

3)

#include<stdio.h>
int main()
{
int iTemp iRemainder iCount;
for(iCount=10;iCount<=99;iCount++){
iRemainder=iCount 10;
iTemp=iCount/10;
if(iCount==3*(iRemainder+iTemp)){
printf(" d" iCount);
}
return 0;
}

 

 

4) 

 

 /*********************************************************************************************************************/
/* The formula to be used is 10x+y 3(x+y) */
/* 10x-3x 3y-y */
/* 7x 2y */
/********************************************************************************************************************/

#include StdAfx.h
#include stdio.h 　
int main(int argc char* argv[])
{
int x y;
for(x 1; x < 9 ; x++)
{
for (y 1; y < 9; y++)
{
if (7*x 2*y){ printf(
number d d n x y);
}
}
}
getchar();
return 0;
}

 

 

 

 

5)

 

bool TrickyCProgram (int num)
{

int tens units digitSum;
bool condition = FALSE;

tens = num/10;

if ( (tens >= 10) || ( tens == 0) )
{
/* if there is ten or more tens or less than one ten
this is not a two digit number */
printf("ERROR:: d is not a 2 digit number" num);
}
else
{
units = num 10;
digitSum = tens + units;

if ( num == 3*digitSum)
{
condition = TRUE;
}

}

return condition;
}

 

 

6)

 

Here is the simple 2 line program

for(int i=10;i<100;i++)
printf("Peculier number d sn" i (i==(((i/10)+(i 10))*3)?"Yes":"No") );

 

 

 

7)

 

Shortest Working Code

Tested for C++ (microsoft visual C++ compiler).

#include <iostream>

int main()

{

using namespace std;

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

{

int a = 3*((i 10) + ((i-(i 10))/10)) ;

if (a==i)

cout<<i;

}

return 0;

}

8)

/**
* Print all 2-digit integers where the value is
* N times the sum of the digits (wnere N < 10).
*
* Start with the formula
* 10 * d0 + d1 == N * (d0 + d1)
*
* where d0 represents the tens digit and d1 represents the units digit we find that
*
* (10 - N) * d0 == (N - 1) * d1 or
* * d0 == (N - 1) * d1 / (10 - N)
*
* So for example if N is 3 then
* (10 - 3) * d0 == (3 - 1) * d1 or
* d0 == 2 * d1 / 7
*
* Then it's just a matter of checking that 27 == 3 * (2 + 7)
* The following code cycles through the unit digits 0 through 9 computes the tens digit
according to the formula above and if the result is valid prints it to stdout.
*/
void digisum(int N)
{
int d0 d1;
int c0 = 10 - N; // it's assumed that N < 10
int c1 = N - 1;

printf( All 2-digit numbers where the value is d times the sum of the digitsn N);

for (d1 = 0; d1 < 10; d1++)
{
d0 = c1 * d1 / c0;
if (d0 < 10)
{
int val = d0 * 10 + d1;
if (val > 0 && val == N * (d0 + d1))
printf( d * ( d + d) == d N d0 d1 val);
}
}
}

To find the number that's 3 times the sum of its digits just call the above as

digisum(3);




9)

#include<stdio.h>

#include<conio.h>

void main()

{

int n temp rem sum=0;

printf("enter the two disit number");

scanf("n d" &n);

for(temp=n;temp>0;temp/=10)

{

rem=temp 10;

sum+=rem;

}

if(n==(sum*3))

printf("the number which is equal to the three times the sum of its disits is d" n);

else

printf("the number is not three times the sum of its disits");

getch();

}

 

 

 

10)

#include<stdio.h>
int main()
{
int x y z;
for(z=10;z<=99;z++)
{
x = z/10;
y = z 10;
if(z==(3*(x+y)))
printf("npec no: dn" z);
}
return 0;
}