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While Loop based Programs

/ Programming Questions

Question 1

WAP to take a number (n) as input and then find the sum of all it’s digits. For example if the input n = 2673, then the output should be 18 i.e. 2 + 6 + 7 + 3

java 

        
            
import java.util.*;
public class Test
{
  public static void main()
  {
    int s = 0, n, d;
    Scanner sc = new Scanner(System.in);
    System.out.println("Enter a number");
    n = sc.nextInt();
    
    // extract digits
    while (n != 0)
    {
        d = n % 10;
        n = n / 10;
        s += d;
    }
    System.out.println("Sum of the digits = " + s);
  } 
}

python 

        
            
n = int(input('enter the number  : '))
s = 0
while n != 0:
  d = n % 10
  n = int(n / 10)
  s += d
print('sum of digits =', s)

Question 2

WAP to input a number and then reverse the number. For example if the original number is 2356, the reverse of the number will be 6532. Then check if the number is a Palindrome i.e. the original and the reversed number are the same. For example, the number 23432 is a Palindrome, since the reverse of the number is also 23432

java 

        
            
import java.util.*;
public class MyClass 
{
  public static void main(String args[]) 
  {
  
  int n, temp, d, rev = 0;
  Scanner sc = new Scanner(System.in);
  System.out.println("Enter a number");
  n = sc.nextInt();
  
  temp = n;
  
  // extract digits
  while (temp != 0)
  {
      d = temp % 10;
      temp /= 10;
      rev = (10 * rev) + d;
  }
  
  System.out.println("Reversed Number : " + rev);
  
  if (n == rev)
    System.out.println("The number is a Palindrome");
  }
}

python 

        
            
n = int(input("Enter a Number: "))
temp = n
rev = 0

# reverse the number
while temp != 0:
  d = temp % 10
  temp = int(temp / 10)
  rev = (rev * 10) + d

print ("Reverse of the number =", rev)

# check if palindrom
if n == rev:
  print("Number is a Palindrome")

Question 3

Use a while loop to keep entering numbers except 0 - the loop will stop once the user enters a 0. Within the loop, keep summing up all the positive even numbers and the negative odd numbers. Once the loop stops, print out both the sums

java 

        
            
import java.util.*;
public class Test
{
  public static void main()
  {
    int pe = 0, no = 0, n = 1;
    Scanner sc = new Scanner(System.in);
    while (n != 0)
    {
        System.out.println("Enter number (0 to exit)");
        n = sc.nextInt();
        
        // positive even
        if ((n > 0) && (n % 2 == 0))
            pe += n;
        // negative odd
        else if ((n < 0) && (n % 2 != 0))
            no += n;
    }
    System.out.println("Sum of +ve even: " + pe);
    System.out.println("Sum of -ve odd: " + no);
  }
}

python 

        
            
n = 1
pe = 0
no = 0

while n != 0:

  n = int(input("Enter a number (0 to exit) : "))

  # positive even
  if n > 0 and n % 2 == 0:
    pe += n
  # negative odd
  elif n < 0 and n % 2 != 0:
    no += n

print("sum of +ve even numbers : ", pe)
print("sum of -ve odd numbers : ", no)

Question 4

WAP to input a number and check if it is a Harshad number i.e. a number that is divisible by the sum of its digits.
For example, if the number is 156, then sum of its digits will be 1 + 5 + 6 = 12. Since 156 is divisible by 12. So, 156 is a Harshad number

java 

        
            
import java.util.*;
public class MyClass {
  public static void main(String args[]) {
  
  Scanner sc = new Scanner(System.in);
  int rem = 0, sum = 0, n, num;  
  System.out.println("Enter a Number");
  num = sc.nextInt();
  
  //Make a copy of num and store it in variable n  
  n = num;  
  
  //Calculates sum of digits  
  while(num > 0)
  {
      rem = num % 10;  
      sum = sum + rem;
      num = num/10;
      
  }
  //Checks whether number is divisible by sum of digits  
  if(n % sum == 0)
    System.out.println(n + " is a harshad number");
  else
    System.out.println(n + " is not a harshad number");  
  }
}

python 

        
            
n = int(input("Enter a Number: "))
s = 0
num = n
while n != 0:
  d = n % 10
  n = int(n / 10)
  s += d
if num % s == 0:
  print("Harshad Number")
else:
  print("Not a Harshad Number")

Question 5

An Automorphic Number is a number which is contained in the last digit(s) of it’s square
For example: 25, since 252 = 625 and 25 is contained in the last digit(s) of 625
WAP to input a number and check whether it is an Automorphic Number or not

java 

        
            
import java.util.*;
import java.lang.*;
public class Test
{
  public static void main()
  {
    int n, d, temp;
    double count = 0, last, s;
    Scanner sc = new Scanner(System.in);
    System.out.println("Enter the number: ");
    n = sc.nextInt();
    
    // square the number
    s = n * n;
    System.out.println("Square = " + s);
    
    // count how many digits
    temp = n;
    while (temp != 0)
    {
        d = temp % 10;
        temp = temp / 10;
        count++;
    }
    
    // extract the last digits
    last = s % (Math.pow(10,count));
    System.out.println("Last Digit(s) = " + last);
    if (last == n)
        System.out.println("Automorphic");
    else
        System.out.println("Not Automorphic");
  }
}

python 

        
            
# enter the number
n = int(input('enter number: '))

# square the number
s = n * n
print("Square =", s)
    
# count how many digits
temp = n
count = 0
while temp != 0:
  d = temp % 10
  temp = int(temp / 10)
  count += 1  

# extract the last digits
print("Digits: ", count)
last = s % (10**count)
print("Last Digit(s) =", last)

if last == n:
  print("Automorphic")
else:
  print("Not Automorphic")

Question 6

An Armstrong number of 3 digits is a 3-digit number , the sum of cubes of each digits is equal to the number itself. For example: 153 = 1x1x1 + 5x5x5 + 3x3x3
WAP to input a 3 digit number and check if it is an Armstrong Number

java 

        
            
import java.util.*;
public class Test
{
  public static void main()
  {
    int s = 0, n, d, a;
    Scanner sc = new Scanner(System.in);
    System.out.println("Enter a number");
    n = sc.nextInt();
    
    // preserve the original number
    a = n;
    
    // extract digits
    while (n != 0)
    {
        d = n % 10;
        n = n / 10;
        s += (d*d*d);
    }
    System.out.println("Sum of the cube of the digits = " + s);
    
    // check if armstrong
    if (a == s)
        System.out.println("Armstrong number");
    else
        System.out.println("Not an Armstrong number");
  }
}

python 

        
            
n = int(input("enter the number : "))

# preserve the original number
a = n

s = 0
# digit extraction
while n != 0:
  d = n % 10
  n = int(n / 10)
  s += (d**3)

print("sum of cube of digits =", s)

if s == a:
  print("armstrong number")
else:
  print("not an armstrong number")

Question 7

WAP to take a 5 digit number from user and display the product of largest and second largest digit of the number
For Example: If the input is 69231, the product will be (9 * 6) = 54

java 

        
            
import java.util.*;
public class Test
{
  public static void main()
  {
    Scanner sc = new Scanner(System.in);
    int n, r, ld, sld, count = 0;
    boolean flag;
    
    System.out.println("Enter the number: ");
    n = sc.nextInt();

    // initialize
    ld = -1;
    sld = -1;
    
    // extract digits
    while (n != 0)
    {
        r = n % 10;
        n = n/10;
        
        // count the digits
        count++;
        
        if ( ld < r )
        {
          if ( sld < ld)
            sld = ld;
          ld = r;
        }
        else if ( ld > r && sld < r)
        {
          sld = r;
        }
    }
    
    // check if 5 digits
    if (count == 5)
    {
        System.out.println("Largest Digit: " + ld);
        System.out.println("Second Largest Digit: " + sld);
        System.out.println("Product = " + (ld * sld));
    }
    else
        System.out.println("Number must be of 5 digits");
  }
}

python 

        
            
# enter the number
n = int(input('enter number: '))

count = 0

ld = -1
sld = -1

# extract digits
while n != 0:

  r = n % 10
  n = int(n/10)

  # count the number of digits
  count += 1

  if ld < r:
    if sld < ld:
        sld = ld
    ld = r
  elif ld > r and sld < r:
    sld = r

# print out if 5 digits
if count == 5:
  print('Largest Digit: ', ld)
  print('Second Largest Digit: ', sld)
  print('Product: ', ld * sld)
else:
  print('Number should be of 5 digits')

Question 8

WAP to enter a 3 digit positive number from user and print the LCM of the digits of the number
For Example: If the input is 664, LCM of digits of the number 6, 6, 4 is 12

java 

        
            
import java.util.*;
public class Test
{
  public static void main()
  {
    Scanner sc = new Scanner(System.in);
    int n, d1, d2, d3, max, lcm;
    System.out.println("Enter the number: ");
    n = sc.nextInt();
    
    // get the 3 digits
    d1 = n % 10;
    n /= 10;
    d2 = n % 10;
    d3 = n / 10;
    System.out.println(d1 + "," + d2 + "," + d3);
    
    // find the largest
    max = d1;
    if ((d1 >= d2) && (d1 >= d3)) 
        max = d1;
    else if ((d2 >= d1) && (d2 >= d3)) 
        max = d2;
    else if ((d3 >= d1) && (d3 >= d2)) 
        max = d3;
        
    // find lcm
    lcm = max;
    while (true)
    {
        if ((lcm % d1 == 0) && (lcm % d2 == 0) && (lcm % d3 == 0))
            break;
        
        lcm += max;
    }
    
    System.out.println("lcm is " + lcm);
  }
}

python 

        
            
# enter the number
n = int(input('enter number: '))

# get the 3 digits
d1 = n % 10
n = int(n / 10)
d2 = n % 10
d3 = int(n / 10)
print(d1, d2, d3)
    
# find the largest
mx = d1
if (d1 >= d2) and (d1 >= d3):
  mx = d1
elif (d2 >= d1) and (d2 >= d3):
  mx = d2
elif (d3 >= d1) and (d3 >= d2):
  mx = d3
        
# find lcm
lcm = mx
while True:
  if (lcm % d1 == 0) and (lcm % d2 == 0) and (lcm % d3 == 0):
    break
        
  lcm += mx

print('LCM is', lcm)

Question 9

WAP to input a number and find if it is Disarium or not. A number is called Disarium if the sum of its digits powered with their respective positions is equal to the number itself
For Example: 135 is a Disarium number as 11 + 32 + 53 = 135

java 

        
            
import java.util.*;
import java.lang.*;
public class MyClass {
  public static void main(String args[]) 
  {
  int n, temp, count_digits = 0, digit;
  double sum = 0, x;
  Scanner sc = new Scanner(System.in);
  System.out.println("Enter a Number");
  n = sc.nextInt();
  
  // count the number of digits
  temp = n;
  while (temp != 0)
  {
      temp = temp / 10;
      count_digits++; 
  }
  System.out.println("Number of digits: " + count_digits);
  
  // calculate sum
  temp = n;
  while (temp != 0)
  {
      // Get the digit
      digit = temp % 10;
      
      // Sum the digits by powering according to
      // the positions
      x = Math.pow(digit, count_digits);
      sum = sum + x;
      count_digits--;
      temp = temp/10;
      
  }
  
  System.out.println("Sum = " + sum);
  if (sum == n)
    System.out.println("Disarium Number");
  
  }
}

python 

        
            
n = int(input("Enter a Number: "))
count_digits = 0
temp = n

# count number of digits
while temp != 0:

  temp = int(temp / 10)
  count_digits += 1

# calculate sum
temp = n
s = 0
while temp != 0:
  # Get the digit
  digit = temp % 10

  # Sum the digits by powering according to
  # the positions

  x = digit**count_digits
  s = s + x
  count_digits -= 1

  temp = int(temp / 10)

print('Sum =', s)

if s == n:
  print("Disarium Number")

Question 10

WAP to input a number and check whether it is a Spy Number
A number is a Spy Number if the sum of it’s digits equals the product of it’s digits . For example, consider the number 1124 - the sum of it’s digits is 1 + 1 + 2 + 4 = 8 and the product of it’s digits is 1 * 1 * 2 * 4 = 8

java 

        
            
import java.util.*;
import java.lang.*;
public class Test
{
  public static void main()
  {
    int n, d, s = 0, p = 1;
    Scanner sc = new Scanner(System.in);
    System.out.println("Enter the number: ");
    n = sc.nextInt();
    
    // extract digits
    while (n != 0)
    {
        d = n % 10;
        n = n / 10;
        
        // calculate sum and product
        s += d;
        p *= d;
    }
    System.out.println("Sum of digits: " + s);
    System.out.println("Product of digits: " + p);
    
    if (s == p)
    {
        System.out.println("Spy Number");
    }
    else
    {
        System.out.println("Not a Spy Number");
    }
  }
}

python 

        
            
# enter the number
n = int(input('enter number: '))
    
# extract digits
s = 0
p = 1
while n != 0:
  d = n % 10
  n = int(n / 10)

  # calculate sum and product
  s += d
  p *= d

print("sum of digits:", s)
print("product of digits:", p)

if s == p:
  print("Spy Number")
else:
  print("Not a Spy Number")

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