C interview programs: even or odd, prime or not, greatest number

Here are few of the frequently asked programs in C language interviews. Each of the questions have been answered with code and , output and explanation.Question.27(even/odd) is a good starter in an C interview for selecting entry level resources. Question.28(greatest of three) tests whether interviewee candidate is able to write a simple program using control structures. Question.29(prime or not) tests whether the candidate is able to write an algorithm for recognizing a prime number. It gives a fair idea of candidate’s programming ability at entry level. Question.30 tests the analytical and programming ability of the candidate. It is also a fair way to know about C programming knowledge of the candidate depending upon his/her choice of storing the variables. See all C interview questions in this blog in one page.

27. Write a program to check whether a given number is even or odd.

Program:

#include <stdio.h>
int main() {
int a;
printf("Enter a: \n");
scanf("%d", &a);
/* logic */
if (a % 2 == 0) {
 printf("The given number is EVEN\n");
}
else {
 printf("The given number is ODD\n");
}
return 0;
}

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Output:

Enter a: 2
The given number is EVEN

Explanation with examples: 
Example 1: If entered number is an even number
          Let value of 'a' entered is 4
          if(a%2==0) then a is an even number, else odd.
  i.e.    if(4%2==0) then 4 is an even number, else odd.

To check whether 4 is even or odd, we need to calculate (4%2).
/* % (modulus) implies remainder value. */
/* Therefore if the remainder obtained when 4 is divided by 2
is 0, then 4 is even. */
    4%2==0 is true
    Thus 4 is an even number.

Example 2: If entered number is an odd number.
          Let value of 'a' entered is 7
          if(a%2==0) then a is an even number, else odd.
  i.e.    if(7%2==0) then 4 is an even number, else odd.

   To check whether 7 is even or odd, we need to calculate (7%2).
   7%2==0 is false  /* 7%2==1 condition fails and else part is
 executed */
   Thus 7 is an odd number.

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28. Write a program to find the greatest of three numbers.

Program:

#include <stdio.h>
int main(){
int a, b, c;
printf("Enter a,b,c: \n");
scanf("%d %d %d", &a, &b, &c);
if (a > b && a > c) {
printf("a is Greater than b and c");
}
else if (b > a && b > c) {
printf("b is Greater than a and c");
}
else if (c > a && c > b) {
printf("c is Greater than a and b");
}
else {
printf("all are equal or any two values are equal");
}
return 0;
}

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Output:

Enter a,b,c: 3 5 8
 c is Greater than a and b

Explanation with examples: 
consider three numbers a=5,b=4,c=8

if(a>b && a>c) then a is greater than b and c
now check this condition for the three numbers 5,4,8 i.e.
if(5>4 && 5>8) /* 5>4 is true but 5>8 fails */

so the control shifts to else if condition
else if(b>a && b>c) then b is greater than a and c
now checking this condition for 5,4,8 i.e.
else if(4>5 && 4>8) /* both the conditions fail */

now the control shifts to the next else if condition
else if(c>a && c>b) then c is greater than a and b
now checking this condition for 5,4,8 i.e.
else if(8>5 && 8>4)  /* both conditions are satisfied */

Thus c is greater than a and b.

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29. Write a program to check whether a given number is a prime.

A prime number is a natural number that has only one and itself as factors. Examples: 2, 3, 5, 7,… are prime numbers. Program:

#include <stdio.h>
main() {
int n, i, c = 0;
printf("Enter any number n: \n");
scanf("%d", &n);
/*logic*/
for (i = 1; i <= n; i++) {
 if (n % i == 0) {
  c++;
 }
}
if (c == 2) {
 printf("n is a Prime number");
}
else {
 printf("n is not a Prime number");
}
return 0;
}

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Output:

Enter any number n: 7
n is Prime

Explanation with examples:

consider a number n=5
for(i=0;i<=n;i++) /* for loop is executed untill the n value equals i */
i.e. for(i=0;i<=5;i++) /* here the for loop is executed untill i is equal to n */

1st iteration: i=1;i<=5;i++
here i is incremented i.e. i value for next iteration is 2
now if(n%i==0) then c is incremented
i.e.if(5%1==0)then c is incremented, here 5%1=0 thus c is incremented.
now c=1;

2nd iteration: i=2;i<=5;i++
here i is incremented i.e. i value for next iteration is 3
now if(n%i==0) then c is incremented
i.e.if(5%2==0) then c is incremented, but 5%2!=0 and so c is not incremented, c remains 1
c=1;

3rd iteration: i=3;i<=5;i++
here i is incremented i.e. i value for next iteration is 4
now if(n%i==0) then c is incremented
i.e.if(5%3==0) then c ic incremented, but 5%3!=0 and so c is not incremented, c remains 1
c=1;

4th iteration: i=4;i<=5;i++
here i is incremented i.e. i value for next iteration is 5
now if(n%i==0) then c is incremented
i.e. if(5%4==0) then c is incremented, but 5%4!=0 and so c is not incremented, c remains 1
c=1;

5th iteration: i=5;i<=5;i++
here i is incremented i.e. i value for next iteration is 6
now if(n%i==0) then c is incremented
i.e. if(5%5==0) then c is incremented, 5%5=0 and so c is incremented.
i.e. c=2

6th iteration: i=6;i<=5;i++
here i value is 6 and 6<=5 is false thus the condition fails and control leaves the for loop.

now if(c==2) then n is a prime number
we have c=2 from the 5th iteration and thus n=5 is a Prime number.

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30. Write a program to find the greatest among ten numbers.

Program:

#include <stdio.h>
int main() {
 int a[10];
 int i;
 int greatest;
 int position =0;
 printf("Enter ten values:");
 //Store 10 numbers in an array
 for (i = 0; i < 10; i++) {
  scanf("%d", &a[i]);
 }
 //Assume that a[0] is greatest
 greatest = a[0];
 for (i = 0; i < 10; i++) {
  if (a[i] > greatest) {
   greatest = a[i];
   position = i+1;
  }
 }
 printf("\n Greatest of ten numbers is %d at position %d \n", greatest, position);
 return 0;
}

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Output:

Enter ten values: 2 53 65 3 88 8 14 5 77 64
Greatest of ten numbers is 88 at position 5

Explanation with example: Entered values are 2, 53, 65, 3, 88, 8, 14, 5, 77, 64 They are stored in an array of size 10. let a[] be an array holding these values. /* how the greatest among ten numbers is found */ Let us consider a variable ‘greatest’. At the beginning of the loop, variable ‘greatest’ is assinged with the value of first element in the array greatest=a[0]. Here variable ‘greatest’ is assigned 2 as a[0]=2.

Below loop is executed until end of the array 'a[]';.
for(i=0; i<10;>greatest)
         {
            greatest= a[i];
         }
     }

For each value of ‘i’, value of a[i] is compared with value of variable ‘greatest’. If any value greater than the value of ‘greatest’ is encountered, it would be replaced by a[i]. After completion of ‘for’ loop, the value of variable ‘greatest’ holds the greatest number in the array. In this case 88 is the greatest of all the numbers.

About the Author:  This post was written by Madhulika Reddy. You can connect with Madhulika on Orkut.