Sum of Numbers Divisible by 4 in C language

Write a Sum of Numbers Divisible by 4 in C program to calculate the sum of all numbers from 0 to 100  that are divisible by 4. Program should directly print the Sum of Numbers Divisible by 4 in C.

Sum of Numbers Divisible by 4 Program will simply start with 0 to 100 and check for numbers Divisible by 4.

Sum of Numbers Divisible by 4  C Code

#include<stdio.h>

#include<conio.h>

void main()
{
int i,sum=0;
clrscr();
for(i=0;i<=100;i++)
{
if(i%4==0)                                                              
sum=sum+i;
}

printf(“\naddition of numbers from 0 to 100 that are divisible by 4 is\n%d”,sum);
getch();
}

 Sum of Numbers Divisible by 4 Output Snapshot

Find triangle is equilateral,isosceles or right angled Code in C Language

Write a C program to Find triangle is equilateral,isosceles or right angled by accepting  the length of three sides of a triangle from input and then test and print the type of triangle.

Program will test the type of triangle is equilateral, isosceles, right angled or none of these and print the output to screen.Program will use the basic mathematics formulas of trigonometric to test type of triangle.

Find triangle is equilateral,isosceles or right angled code

#include “stdio.h”
void main()
{
int i, j; 
int temp; 
int sides[3]; 

int type; 
int isRightAngled; 
clrscr();

printf(“Please enter the three sides of triange\n”);

for(i = 0; i < 3; i++)
{
scanf(“%d”, &sides[i]);
}

for(i = 0; i < 3; i++)
{
for(j = i + 1; j < 3; j++)
{

if(sides[i] > sides[j])
{

temp = sides[i];
sides[i] = sides[j];
sides[j] = temp;
}
}
}

type = 3;
isRightAngled = 0;

if(sides[2] > sides[0] + sides[1]) /*check if triangle is valid */
type = 0;
else if(sides[0] == sides[2]) 
type = 1; 
else
{
if(sides[0] == sides[1]) 
type = 2; 

if(sides[2] * sides[2] == sides[0] * sides[0] + sides[1] * sides[1])
isRightAngled = 1;
}

switch(type)
{
case 0:
printf(“The triangle is Invalid\n”);
break;
case 1:
printf(“The triangle is Equilateral Triangle\n”);
break;
case 2:
if(isRightAngled == 1)
printf(“The triangle is Isosceles and Right Angled Triangle\n”);
else
printf(“The triangle is Isosceles Triangle\n”);
break;
case 3:
if(isRightAngled == 1)
printf(“The triangle is Scalene and Right Angled Triangle\n”);
else
printf(“The triangle is Scalene Triangle\n”);
break;

}
getch();
}

Display numbers as English word in C Language

Write a Display numbers as English word program in C to read an integer and display each of the digit of the integer in English.User has to provide a number which will be printed as English word on the screen.

You can take an example of 2345 will be printed as two three four five.

Display numbers as English word in C Code

#include <stdio.h>

void main ()
{
int number, digit1, digit2, digit3, digit4, digit5;
clrscr();
printf(“Enter a number (max 5 digits): “);
scanf(“%i”, &number);

if (number < 100000)
{
digit1 = number % 100000 / 10000;

switch (digit1)
{
case 0 : printf(“Zero “); break;
case 1 : printf(“One “); break;
case 2 : printf(“Two “); break;
case 3 : printf(“Three “); break;
case 4 : printf(“Four “); break;
case 5 : printf(“Five “); break;
case 6 : printf(“Six “); break;
case 7 : printf(“Seven “); break;
case 8 : printf(“Eight “); break;
case 9 : printf(“Nine “); break;
default : printf(“Unknown Interger “); break;
}

digit2 = number % 10000 / 1000;

switch (digit2)
{
case 0 : printf(“Zero “); break;
case 1 : printf(“One “); break;
case 2 : printf(“Two “); break;
case 3 : printf(“Three “); break;
case 4 : printf(“Four “); break;
case 5 : printf(“Five “); break;
case 6 : printf(“Six “); break;
case 7 : printf(“Seven “); break;
case 8 : printf(“Eight “); break;
case 9 : printf(“Nine “); break;
default : printf(“Unknown Interger “); break;
}

digit3 = number % 1000 / 100;

switch (digit3)
{
case 0 : printf(“Zero “); break;
case 1 : printf(“One “); break;
case 2 : printf(“Two “); break;
case 3 : printf(“Three “); break;
case 4 : printf(“Four “); break;
case 5 : printf(“Five “); break;
case 6 : printf(“Six “); break;
case 7 : printf(“Seven “); break;
case 8 : printf(“Eight “); break;
case 9 : printf(“Nine “); break;
default : printf(“Unknown Interger “); break;
}

digit4 = number % 100 / 10;

switch (digit4)
{
case 0 : printf(“Zero “); break;
case 1 : printf(“One “); break;
case 2 : printf(“Two “); break;
case 3 : printf(“Three “); break;
case 4 : printf(“Four “); break;
case 5 : printf(“Five “); break;
case 6 : printf(“Six “); break;
case 7 : printf(“Seven “); break;
case 8 : printf(“Eight “); break;
case 9 : printf(“Nine “); break;
default : printf(“Unknown Interger “); break;
}

digit5 = number % 10;

switch (digit5)
{
case 0 : printf(“Zero “); break;
case 1 : printf(“One “); break;
case 2 : printf(“Two “); break;
case 3 : printf(“Three “); break;
case 4 : printf(“Four “); break;
case 5 : printf(“Five “); break;
case 6 : printf(“Six “); break;
case 7 : printf(“Seven “); break;
case 8 : printf(“Eight “); break;
case 9 : printf(“Nine “); break;
default : printf(“Unknown Interger “); break;
}
}
else
{
printf(“Entered Number is too long.”);
}
getch();
}