Given a number N, the task is to print the first N numbers such that every number and the reverse of the number is divisible by its sum of digits.
Example:
Input: N = 4
Output: 1 2 3 4
Explanation:
The reverse of every single digit number is the same number. And, every number is divisible by itself.
Input: N = 12
Output: 1 2 3 4 5 6 7 8 9 10 12 18
Approach: The idea is to iterate through every number from 1 and compute the sum of the digits. For every such number, check if the number and the reverse of the number are divisible by the sum or not. Therefore, the following steps are followed to compute the answer:
- Initialize the counter to one and iterate through all the numbers one by one.
- For every number, find the reverse of the number.
- While finding the reverse of the number, compute the sum of the digits of the number.
- Now, check if the number and the reverse of the number are divisible by the sum of its digits.
- If it is, then increment the counter. Repeat the above steps until this counter is equal to N.
Below is the implementation of the above approach:
// C++ program to print the first N numbers // such that every number and the reverse // of the number is divisible by its // sum of digits #include <bits/stdc++.h> using namespace std;
// Function to calculate the sum of digits int digit_sum( int n)
{ int sum = 0, m;
// Loop to iterate through every
// digit of the number
while (n > 0) {
m = n % 10;
sum = sum + m;
n = n / 10;
}
// Returning the sum of digits
return (sum);
} // Function to calculate the reverse // of a number int reverse( int n)
{ int r = 0;
// Loop to calculate the reverse
// of the number
while (n != 0) {
r = r * 10;
r = r + n % 10;
n = n / 10;
}
// Return the reverse of the
// number
return (r);
} // Function to print the first N numbers // such that every number and the reverse // of the number is divisible by its // sum of digits void operation( int n)
{ int i = 1, a, count = 0, r;
// Loop to continuously check and
// generate number until there
// are n outputs
while (count < n) {
// Variable to hold the sum of
// the digit of the number
a = digit_sum(i);
// Computing the reverse of the
// number
r = reverse(i);
// Checking if the condition satisfies.
// Increment the count and print the
// number if it satisfies.
if (i % a == 0 && r % a == 0) {
cout << i << " " ;
count++;
i++;
}
else
i++;
}
} // Driver code int main()
{ int n = 10;
operation(n);
} |
// Java program to print the first N numbers // such that every number and the reverse // of the number is divisible by its // sum of digits import java.util.*;
class GFG{
// Function to calculate the sum of digits static int digit_sum( int n)
{ int sum = 0 , m;
// Loop to iterate through
// every digit of the number
while (n > 0 )
{
m = n % 10 ;
sum = sum + m;
n = n / 10 ;
}
// Returning the sum of digits
return (sum);
} // Function to calculate the // reverse of a number static int reverse( int n)
{ int r = 0 ;
// Loop to calculate the
// reverse of the number
while (n != 0 )
{
r = r * 10 ;
r = r + n % 10 ;
n = n / 10 ;
}
// Return the reverse
// of the number
return (r);
} // Function to print the first N numbers // such that every number and the reverse // of the number is divisible by its // sum of digits static void operation( int n)
{ int i = 1 , a, count = 0 , r;
// Loop to continuously check and
// generate number until there
// are n outputs
while (count < n)
{
// Variable to hold the sum
// of the digit of the number
a = digit_sum(i);
// Computing the reverse of the
// number
r = reverse(i);
// Checking if the condition satisfies.
// Increment the count and print the
// number if it satisfies.
if (i % a == 0 && r % a == 0 )
{
System.out.print(i + " " );
count++;
i++;
}
else
i++;
}
} // Driver code public static void main(String args[])
{ int n = 10 ;
operation(n);
} } // This code is contributed by shivanisinghss2110 |
# Python3 program to print the first N numbers # such that every number and the reverse # of the number is divisible by its # sum of digits # Function to calculate the sum of digits def digit_sum(n):
sum = 0
# Loop to iterate through every
# digit of the number
while (n > 0 ):
m = n % 10 ;
sum = sum + m;
n = n / / 10
# Returning the sum of digits
return ( sum )
# Function to calculate the reverse # of a number def reverse(n):
r = 0
# Loop to calculate the reverse
# of the number
while (n ! = 0 ):
r = r * 10
r = r + n % 10
n = n / / 10
# Return the reverse of the
# number
return (r)
# Function to print the first N numbers # such that every number and the reverse # of the number is divisible by its # sum of digits def operation(n):
i = 1
count = 0
# Loop to continuously check and
# generate number until there
# are n outputs
while (count < n):
# Variable to hold the sum of
# the digit of the number
a = digit_sum(i)
# Computing the reverse of the
# number
r = reverse(i)
# Checking if the condition satisfies.
# Increment the count and print the
# number if it satisfies.
if (i % a = = 0 and r % a = = 0 ):
print (i, end = " " )
count + = 1
i + = 1
else :
i + = 1
# Driver code if __name__ = = '__main__' :
n = 10
operation(n)
# This code is contributed by Samarth |
// C# program to print the first N numbers // such that every number and the reverse // of the number is divisible by its // sum of digits using System;
class GFG{
// Function to calculate the sum of digits static int digit_sum( int n)
{ int sum = 0, m;
// Loop to iterate through
// every digit of the number
while (n > 0)
{
m = n % 10;
sum = sum + m;
n = n / 10;
}
// Returning the sum of digits
return (sum);
} // Function to calculate the // reverse of a number static int reverse( int n)
{ int r = 0;
// Loop to calculate the
// reverse of the number
while (n != 0)
{
r = r * 10;
r = r + n % 10;
n = n / 10;
}
// Return the reverse
// of the number
return (r);
} // Function to print the first N numbers // such that every number and the reverse // of the number is divisible by its // sum of digits static void operation( int n)
{ int i = 1, a, count = 0, r;
// Loop to continuously check and
// generate number until there
// are n outputs
while (count < n)
{
// Variable to hold the sum
// of the digit of the number
a = digit_sum(i);
// Computing the reverse of the
// number
r = reverse(i);
// Checking if the condition satisfies.
// Increment the count and print the
// number if it satisfies.
if (i % a == 0 && r % a == 0)
{
Console.Write(i + " " );
count++;
i++;
}
else
i++;
}
} // Driver code public static void Main()
{ int n = 10;
operation(n);
} } // This code is contributed by Code_Mech |
<script> // Javascript program to print the first N numbers
// such that every number and the reverse
// of the number is divisible by its
// sum of digits
// Function to calculate the sum of digits
function digit_sum(n)
{
let sum = 0, m;
// Loop to iterate through every
// digit of the number
while (n > 0) {
m = n % 10;
sum = sum + m;
n = parseInt(n / 10, 10);
}
// Returning the sum of digits
return (sum);
}
// Function to calculate the reverse
// of a number
function reverse(n)
{
let r = 0;
// Loop to calculate the reverse
// of the number
while (n != 0) {
r = r * 10;
r = r + n % 10;
n = parseInt(n / 10, 10);
}
// Return the reverse of the
// number
return (r);
}
// Function to print the first N numbers
// such that every number and the reverse
// of the number is divisible by its
// sum of digits
function operation(n)
{
let i = 1, a, count = 0, r;
// Loop to continuously check and
// generate number until there
// are n outputs
while (count < n) {
// Variable to hold the sum of
// the digit of the number
a = digit_sum(i);
// Computing the reverse of the
// number
r = reverse(i);
// Checking if the condition satisfies.
// Increment the count and print the
// number if it satisfies.
if (i % a == 0 && r % a == 0) {
document.write(i + " " );
count++;
i++;
}
else
i++;
}
}
let n = 10;
operation(n);
</script> |
1 2 3 4 5 6 7 8 9 10
Time Complexity: O(n * log10n)
Auxiliary Space: O(1)