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 :
Let be any number such that =100a+10b+c .
Now assume that is divisible by 19. Then
0 (mod 19)
100a+10b+c0 (mod 19)
10(10a+b)+c0 (mod 19)
10+c0 (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 of 1.
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+c0 (mod 19)
by 2 and simplify it:
20+2c0 (mod 19)
+2c0 (mod 19)
We have found out that if 0 (mod 19) then,
+2c0 (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++
#include <bits/stdc++.h>
using namespace std;
bool isDivisible( long long int n)
{
while (n / 100)
{
int d = n % 10;
n /= 10;
n += d * 2;
}
return (n % 19 == 0);
}
int main()
{
long long int n = 101156;
if (isDivisible(n))
cout << "Yes" << endl;
else
cout << "No" << endl;
return 0;
}
|
Java
import java.io.*;
class GFG {
static boolean isDivisible( long n)
{
while (n / 100 > 0 )
{
long d = n % 10 ;
n /= 10 ;
n += d * 2 ;
}
return (n % 19 == 0 );
}
public static void main (String[] args) {
long n = 101156 ;
if (isDivisible(n))
System.out.println( "Yes" );
else
System.out.println( "No" );
}
}
|
Python3
def isDivisible(n):
while n / / 100 :
d = n % 10
n / / = 10
n + = d * 2
return (n % 19 = = 0 )
if __name__ = = "__main__" :
n = 101156
if isDivisible(n):
print ( "Yes" )
else :
print ( "No" )
|
C#
using System;
class GFG
{
static bool isDivisible( long n)
{
while (n / 100 > 0)
{
long d = n % 10;
n /= 10;
n += d * 2;
}
return (n % 19 == 0);
}
public static void Main()
{
long n = 101156;
if (isDivisible(n))
Console.WriteLine( "Yes" );
else
Console.WriteLine( "No" );
}
}
|
Javascript
function isDivisible(n)
{
while (parseInt(n / 100)>0)
{
var d = n % 10;
n = parseInt(n/ 10);
n += d * 2;
}
return (n % 19 == 0);
}
var n = 101156;
if (isDivisible(n))
console.log( "Yes" );
else
console.log( "No" );
|
PHP
<?php
function isDivisible( $n )
{
while (1)
{
$d = $n % 10;
$n = $n / 10;
$n = $n + $d * 2;
if ( $n < 100)
break ;
}
return ( $n % 19 == 0);
}
$n = 38;
if (isDivisible( $n ))
echo "Yes" ;
else
echo "No" ;
?>
|
Time Complexity: O(log10n), time required to check if number is divisible by 19
Auxiliary Space: O(1), as no extra space is required
Approach 2: Using Modular Arithmetic
A third way to check if a number is divisible by 19 is to use modular arithmetic. This approach involves calculating the remainder when the large number is divided by 19, and checking if the remainder is zero. If the remainder is zero, then the number is divisible by 19.
For example, let’s say we want to check if the number 7,654,321 is divisible by 19.
We calculate the remainder when 7,654,321 is divided by 19 using modular arithmetic.
To do this, we first find the remainder when the first digit, 7, is divided by 19. Since 7 is less than 19, its remainder is simply 7.
Next, we multiply the remainder by 10 and add the next digit, 6, to get 76. We then find the remainder when 76 is divided by 19, which is 0.
We repeat this process for the remaining digits, and find that the remainder when 7,654,321 is divided by 19 is also 0.
Therefore, we can conclude that 7,654,321 is divisible by 19.
These are three different ways to check if a large number is divisible by 19.
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.
C++
#include <iostream>
#include <string>
using namespace std;
bool is_divisible_by_19( int num) {
int remainder = 0;
for ( char digit : to_string(num)) {
remainder = (remainder * 10 + (digit - '0' )) % 19;
}
return remainder == 0;
}
int main() {
int n = 101156;
if (is_divisible_by_19(n)) {
cout << "Yes" << endl;
}
else {
cout << "No" << endl;
}
return 0;
}
|
Java
import java.util.*;
public class Gfg {
public static boolean is_divisible_by_19( int num) {
int remainder = 0 ;
for ( char digit : Integer.toString(num).toCharArray()) {
remainder = (remainder * 10 + (digit - '0' )) % 19 ;
}
return remainder == 0 ;
}
public static void main(String[] args) {
int n = 101156 ;
if (is_divisible_by_19(n)) {
System.out.println( "Yes" );
}
else {
System.out.println( "No" );
}
}
}
|
Python
def is_divisible_by_19(num):
remainder = 0
for digit in str (num):
remainder = (remainder * 10 + int (digit)) % 19
return remainder = = 0
if __name__ = = "__main__" :
n = 101156
if (is_divisible_by_19(n)) :
print ( "Yes" )
else :
print ( "No" )
|
C#
using System;
public class Gfg
{
public static bool IsDivisibleBy19( int num)
{
int remainder = 0;
foreach ( char digit in num.ToString())
{
remainder = (remainder * 10 + (digit - '0' )) % 19;
}
return remainder == 0;
}
public static void Main()
{
int n = 101156;
if (IsDivisibleBy19(n))
{
Console.WriteLine( "Yes" );
}
else
{
Console.WriteLine( "No" );
}
}
}
|
Javascript
function is_divisible_by_19(num) {
let remainder = 0;
for (const digit of num.toString()) {
remainder = (remainder * 10 + (digit.charCodeAt() - '0' .charCodeAt())) % 19;
}
return remainder === 0;
}
const n = 101156;
if (is_divisible_by_19(n)) {
console.log( 'Yes' );
} else {
console.log( 'No' );
}
|
Time Complexity:
The time complexity of this approach is O(n), where n is the number of digits in the input number num. This is because we need to iterate over each digit in num and perform a constant amount of arithmetic operations (multiplication, addition, and modulo) for each digit.
Space Complexity:
The space complexity of this approach is O(1), as we only use a constant amount of extra space to store the remainder variable. We do not use any data structures that grow with the size of the input.
Last Updated :
31 Oct, 2023
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