Check if any large number is divisible by 19 or not
Given a number, the task is to quickly check if the number is divisible by 19 or not.
Examples:
Input : x = 38 Output : Yes Input : x = 47 Output : No
A solution to the problem is to extract the last digit and add 2 times of last digit to remaining number and repeat this process until a two digit number is obtained. If the obtained two digit number is divisible by 19, then the given number is divisible by 19.
Approach:
- Extract the last digit of the number/truncated number every time
- Add 2*(last digit of the previous number) to the truncated number
- Repeat the above three steps as long as necessary.
Illustration:
101156-->10115+2*6 = 10127-->1012+2*7=1026-->102+2*6=114 and 114=6*19, So 101156 is divisible by 19.
Mathematical Proof :
Letbe any number such that
=100a+10b+c .
Now assume thatis divisible by 19. Then
0 (mod 19)
100a+10b+c0 (mod 19)
10(10a+b)+c0 (mod 19)
10+c
0 (mod 19)
Now that we have separated the last digit from the number, we have to find a way to use it.
Make the coefficient of1.
In other words, we have to find an integer such that n such that 10n1 mod 19.
It can be observed that the smallest n which satisfies this property is 2 as 201 mod 19.
Now we can multiply the original equation 10+c
0 (mod 19)
by 2 and simplify it:
20+2c
0 (mod 19)
+2c
0 (mod 19)
We have found out that if0 (mod 19) then,
+2c
0 (mod 19).
In other words, to check if a 3-digit number is divisible by 19,
we can just remove the last digit, multiply it by 2,
and then add to the rest of the two digits.
C++
// CPP Program to validate the above logic #include <bits/stdc++.h> using namespace std; // Function to check if the number // is divisible by 19 or not bool isDivisible( long long int n) { while (n / 100) // { // Extracting the last digit int d = n % 10; // Truncating the number n /= 10; // Adding twice the last digit // to the remaining number n += d * 2; } // return true if number is divisible by 19 return (n % 19 == 0); } // Driver code int main() { long long int n = 101156; if (isDivisible(n)) cout << "Yes" << endl; else cout << "No" << endl; return 0; } |
Java
// Java Program to validate the above logic import java.io.*; class GFG { // Function to check if the // number is divisible by 19 or not static boolean isDivisible( long n) { while (n / 100 > 0 ) { // Extracting the last digit long d = n % 10 ; // Truncating the number n /= 10 ; // Subtracting the five times the // last digit from the remaining number n += d * 2 ; } // Return n is divisible by 19 return (n % 19 == 0 ); } // Driver code public static void main (String[] args) { long n = 101156 ; if (isDivisible(n)) System.out.println( "Yes" ); else System.out.println( "No" ); } } // This code is contributed by Raj. |
# Python 3 Program to check
# if the number is divisible
# by 19 or not
# Function to check if the number
# is divisible by 19 or not
def isDivisible(n) :
while (n // 100) :
# Extracting the last digit
d = n % 10
# Truncating the number
n //= 10
# Adding twice the last digit
# to the remaining number
n += d * 2
# return true if number
# is divisible by 19
return (n % 19 == 0)
# Driver Code
if __name__ == "__main__" :
n = 101156
if (isDivisible(n)) :
print("Yes" )
else :
print("No")
# This code is contributed
# by ANKITRAI1
C#
// C# Program to validate the // above logic using System; class GFG { // Function to check if the // number is divisible by 19 or not static bool isDivisible( long n) { while (n / 100 > 0) { // Extracting the last digit long d = n % 10; // Truncating the number n /= 10; // Subtracting the five times // the last digit from the // remaining number n += d * 2; } // Return n is divisible by 19 return (n % 19 == 0); } // Driver code public static void Main() { long n = 101156; if (isDivisible(n)) Console.WriteLine( "Yes" ); else Console.WriteLine( "No" ); } } // This code is contributed by ajit |
PHP
<?php // PHP Program to validate // the above logic // Function to check if the number // is divisible by 19 or not function isDivisible( $n ) { while (1) { // Extracting the last digit $d = $n % 10; // Truncating the number $n = $n / 10; // Adding twice the last digit // to the remaining number $n = $n + $d * 2; if ( $n < 100) break ; } // return true if number is // divisible by 19 return ( $n % 19 == 0); } // Driver code $n = 38; if (isDivisible( $n )) echo "Yes" ; else echo "No" ; // This code is contributed by ash264 ?> |
Javascript
<script> // javascript Program to validate the above logic // Function to check if the // number is divisible by 19 or not function isDivisible(n) { while (parseInt(n / 100)>0) { // Extracting the last digit var d = n % 10; // Truncating the number n = parseInt(n/ 10); // Subtracting the five times the // last digit from the remaining number n += d * 2; } // Return n is divisible by 19 return (n % 19 == 0); } // Driver code var n = 101156; if (isDivisible(n)) document.write( "Yes" ); else document.write( "No" ); // This code is contributed by 29AjayKumar </script> |
Yes
Time Complexity: O(log10n), time required to check if number is divisible by 19
Auxiliary Space: O(1), as no extra space is required
Note that the above program may not make a lot of sense as could simply do n % 19 to check for divisibility. The idea of this program is to validate the concept. Also, this might be an efficient approach if input number is large and given as string.
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