In this c program we need to find all numbers that are divisible by 5 means when we take mod of a number by 5 we have to get the reminder zero.

this c program logic is very simple. firstly we declares the variables count sum etc as integers. Now we accept the value for the range from the user.

Then we start a for loop from starting range value to end range value, inside the for loop we take each number and do a mod operation on that number, if the reminder is zero increment count plus 1 and add sum equal to sum + that number. also print that number as integer divisible by five.

#include <stdio.h> #include <conio.h> void main() { int i, N1, N2, count = 0, sum = 0; /* declares count, sum and two variables as integer */ clrscr(); printf ("Enter the value of N1 and N2\n"); /* user gives the value for lower and upper range */ scanf ("%d %d", &N1, &N2); /*Count the number and compute their sum*/ printf ("Integers divisible by 5 are\n"); for (i = N1; i < N2; i++) { if (i%5 == 0) { printf("%3d,", i); /* using mod operator check the number is divisible by 5 */ count++; sum = sum + i; /* add the numbers divisible by 5 to sum variable */ } } printf ("\nNumber of integers divisible by 5 between %d and %d = %d\n", N1,N2,count); printf ("Sum of all integers that are divisible by 5 = %d\n", sum); /* displays the output of program */ } /* End of main()*/

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**OUTPUT of C program to find the number of integers divisible by 5**

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Enter the value of N1 and N2

2

27

Integers divisible by 5 are

5, 10, 15, 20, 25,

Number of integers divisible by 5 between 2 and 27 = 5

Sum of all integers that are divisible by 5 = 75

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Enter the value of N1 and N2

2

27

Integers divisible by 5 are

5, 10, 15, 20, 25,

Number of integers divisible by 5 between 2 and 27 = 5

Sum of all integers that are divisible by 5 = 75