Given a large number, the task is to check if the number is divisible by 13 or not.
Examples :
Input : 637
Output : 637 is divisible by 13.
Input : 920
Output : 920 is not divisible by 13.
Input : 83959092724
Output : 83959092724 is divisible by 13.
If the given number num is small, we can easily find whether it is divisible by 13 or not by doing num % 13 and checking whether the result is 0 or not. But what about very large numbers? Let’s discuss large numbers.
Below are some interesting facts about the divisibility of 13.
- A number is divisible by 13 if and if the alternating sum (alternatively adding and subtracting) of blocks of three from right to left is divisible by 13. For example, 2911285 is divisible by 13 because the alternating sum of blocks of size 3 is 2 – 911 + 285 = -650 which is divisible by 13.
- A number is divisible by 13 if and only if the number obtained by adding the last digit multiplied by 4 to the rest is also divisible by 13.
For example, consider 2353. Applying above rule, we get 235 + 3*4 = 247. Again we apply the rule and get 24 + 7*4 = 52. Since 52 is divisible by 13, the given number is divisible by 13.
Below is the implementation based on first fact above (Finding the alternating sum of blocks of size 3)
C++
#include <iostream>
using namespace std;
bool checkDivisibility(string num)
{
int length = num.size();
if (length == 1 && num[0] == '0' )
return true ;
if (length % 3 == 1)
{
num += "00" ;
length += 2;
}
else if (length % 3 == 2)
{
num += "0" ;
length += 1;
}
int sum = 0, p = 1;
for ( int i = length - 1; i >= 0; i--)
{
int group = 0;
group += num[i--] - '0' ;
group += (num[i--] - '0' ) * 10;
group += (num[i] - '0' ) * 100;
sum = sum + group * p;
p *= (-1);
}
sum = abs (sum);
return (sum % 13 == 0);
}
int main()
{
string number = "83959092724" ;
if (checkDivisibility(number))
cout << number << " is divisible by 13." ;
else
cout << number << " is not divisible by 13." ;
return 0;
}
|
Java
class GFG
{
static boolean checkDivisibility(String num)
{
int length = num.length();
if (length == 1 &&
num.charAt( 0 ) == '0' )
return true ;
if (length % 3 == 1 )
{
num += "00" ;
length += 2 ;
}
else if (length % 3 == 2 )
{
num += "0" ;
length += 1 ;
}
int sum = 0 , p = 1 ;
for ( int i = length - 1 ; i >= 0 ; i--)
{
int group = 0 ;
group += num.charAt(i--) - '0' ;
group += (num.charAt(i--) - '0' ) * 10 ;
group += (num.charAt(i) - '0' ) * 100 ;
sum = sum + group * p;
p *= (- 1 );
}
sum = Math.abs(sum);
return (sum % 13 == 0 );
}
public static void main(String[] args)
{
String number = "83959092724" ;
if (checkDivisibility(number))
System.out.println(number +
" is divisible by 13." );
else
System.out.println(number +
" is not divisible by 13." );
}
}
|
Python3
def checkDivisibility( num):
length = len (num)
if (length = = 1 and num[ 0 ] = = '0' ):
return True
if (length % 3 = = 1 ):
num = str (num) + "00"
length + = 2
elif (length % 3 = = 2 ):
num = str (num) + "0"
length + = 1
sum = 0
p = 1
for i in range (length - 1 , - 1 , - 1 ) :
group = 0
group + = ord (num[i]) - ord ( '0' )
i - = 1
group + = ( ord (num[i]) - ord ( '0' )) * 10
i - = 1
group + = ( ord (num[i]) - ord ( '0' )) * 100
sum = sum + group * p
p * = ( - 1 )
sum = abs ( sum )
return ( sum % 13 = = 0 )
if __name__ = = "__main__" :
number = "83959092724"
if (checkDivisibility(number)):
print ( number , "is divisible by 13." )
else :
print ( number , "is not divisible by 13." )
|
C#
using System;
class GFG {
static bool checkDivisibility( string num)
{
int length = num.Length;
if (length == 1 && num[0] == '0' )
return true ;
if (length % 3 == 1)
{
num += "00" ;
length += 2;
}
else if (length % 3 == 2)
{
num += "0" ;
length += 1;
}
int sum = 0, p = 1;
for ( int i = length - 1; i >= 0; i--)
{
int group = 0;
group += num[i--] - '0' ;
group += (num[i--] - '0' ) * 10;
group += (num[i] - '0' ) * 100;
sum = sum + group * p;
p *= (-1);
}
sum = Math.Abs(sum);
return (sum % 13 == 0);
}
static void Main()
{
string number = "83959092724" ;
if (checkDivisibility(number))
Console.Write( number +
" is divisible by 13." );
else
Console.Write( number +
" is not divisible by 13." );
}
}
|
PHP
<?php
function checkDivisibility( $num )
{
$length = strlen ( $num );
if ( $length == 1 &&
$num [0] == '0' )
return true;
if ( $length % 3 == 1)
{
$num += "00" ;
$length += 2;
}
else if ( $length % 3 == 2)
{
$num += "0" ;
$length += 1;
}
$sum = 0; $p = 1;
for ( $i = $length - 1; $i >= 0; $i --)
{
$group = 0;
$group += $num [ $i --] - '0' ;
$group += ( $num [ $i --] - '0' ) * 10;
$group += ( $num [ $i ] - '0' ) * 100;
$sum = $sum + $group * $p ;
$p *= (-1);
}
$sum = abs ( $sum );
return ( $sum % 13 == 0);
}
$number = "83959092724" ;
if (checkDivisibility( $number ))
echo ( $number . " is divisible by 13." );
else
echo ( $number . " is not divisible by 13." );
?>
|
Javascript
<script>
function checkDivisibility(num)
{
let length = num.length;
if (length == 1 &&
num[0] == '0' )
return true ;
if (length % 3 == 1)
{
num += "00" ;
length += 2;
}
else if (length % 3 == 2)
{
num += "0" ;
length += 1;
}
let sum = 0; p = 1;
for (let i = length - 1; i >= 0; i--)
{
group = 0;
group += num[i--] - '0' ;
group += (num[i--] - '0' ) * 10;
group += (num[i] - '0' ) * 100;
sum = sum + group * p;
p *= (-1);
}
sum = Math.abs(sum);
return (sum % 13 == 0);
}
let number = "83959092724" ;
if (checkDivisibility(number))
document.write(number + " is divisible by 13." );
else
document.write(number + " is not divisible by 13." );
</script>
|
Output
83959092724 is divisible by 13.
Time Complexity: O(length(number))
Auxiliary Space: O(1)
Method: Checking given number is divisible by 13 or not by using the modulo division operator “%”.
C++
#include <iostream>
using namespace std;
int main() {
long int n = 83959092724L;
if ((n) % 13 == 0)
{
cout<< "Yes" ;
}
else
{
cout<< "No" ;
}
return 0;
}
|
Java
import java.io.*;
import java.util.*;
class GFG
{
public static void main(String[] args)
{
long n = 83959092724L;
if ((n) % 13 == 0 )
{
System.out.println( "Yes" );
}
else
{
System.out.println( "No" );
}
}
}
|
Python3
n = 83959092724
if int (n) % 13 = = 0 :
print ( "Yes" )
else :
print ( "No" )
|
C#
using System;
public class GFG {
static public void Main()
{
long n = 83959092724L;
if ((n) % 13 == 0) {
Console.Write( "Yes" );
}
else {
Console.Write( "No" );
}
}
}
|
PHP
<?php
$n = 83959092724;
if ( $n % 13 == 0)
{
echo "Yes" ;
}
else
{
echo "No" ;
}
?>
|
Javascript
<script>
n = 83959092724;
if (n % 13 == 0)
{
document.write( "Yes" );
}
else
{
document.write( "No" );
}
</script>
|
Time Complexity: O(1)
Auxiliary Space: O(1)
Method 3:
1. Initialize an integer variable alternating_sum to 0 and a variable multiplier to -1.
2. While num is greater than 0, perform the following steps:
a. Add multiplier * (num % 10) to alternating_sum.
b. Multiply multiplier by -1.
c. Divide num by 10.
3. Check if alternating_sum is divisible by 13. If it is, return true. Otherwise, return false.
C++
#include <iostream>
bool is_divisible_by_13( int num)
{
int alternating_sum = 0;
int multiplier = -1;
while (num > 0) {
alternating_sum += multiplier * (num % 10);
multiplier *= -1;
num /= 10;
}
return alternating_sum % 13 == 0;
}
int main()
{
if (is_divisible_by_13(12))
std::cout << "Yes" << std::endl;
else
std::cout << "No" << std::endl;
return 0;
}
|
Python3
def is_divisible_by_13(num):
alternating_sum = 0
multiplier = - 1
while num > 0 :
alternating_sum + = multiplier * (num % 10 )
multiplier * = - 1
num / / = 10
return alternating_sum % 13 = = 0
if is_divisible_by_13( 12 ):
print ( "Yes" )
else :
print ( "No" )
|
Java
public class Main {
public static boolean isDivisibleBy13( int num)
{
int alternatingSum = 0 ;
int multiplier = - 1 ;
while (num > 0 ) {
alternatingSum += multiplier * (num % 10 );
multiplier *= - 1 ;
num /= 10 ;
}
return alternatingSum % 13 == 0 ;
}
public static void main(String[] args)
{
if (isDivisibleBy13( 12 ))
System.out.println( "Yes" );
else
System.out.println( "No" );
}
}
|
Javascript
function isDivisibleBy13(num) {
let alternatingSum = 0;
let multiplier = -1;
while (num > 0) {
alternatingSum += multiplier * (num % 10);
multiplier *= -1;
num = Math.floor(num / 10);
}
return alternatingSum % 13 === 0;
}
if (isDivisibleBy13(12)) {
console.log( "Yes" );
} else {
console.log( "No" );
}
|
C#
using System;
class Program
{
static bool IsDivisibleBy11( int num)
{
int alternating_sum = 0;
int multiplier = -1;
while (num > 0)
{
alternating_sum += multiplier * (num % 10);
multiplier *= -1;
num /= 10;
}
return alternating_sum % 13 == 0;
}
static void Main( string [] args)
{
if (IsDivisibleBy11(12))
Console.WriteLine( "Yes" );
else
Console.WriteLine( "No" );
}
}
|
Time Complexity: O(log10(n))
Auxiliary Space: O(1)
Method: Using lookup table method
- The remainders list contains the precomputed remainders of all numbers from 0 to 12 when divided by 13. The remainders for negative numbers are obtained by subtracting them from 13.
- The is_divisible_by_13() function takes a number as an input.
- It splits the number into blocks of three digits from right to left using the modulo operator and integer division. Each block represents a number in the range [0, 999].
- It computes the remainders of each block using the precomputed values and updates the remainder variable accordingly. The formula used to compute the remainder of a block is (remainder * 1000 + block) % 13. This is based on the fact that 1000 % 13 is 8, which means that 10^n % 13 is 1 for all n >= 3. Therefore, we can compute the remainder of a block by multiplying the current remainder by 1000, adding the block to it, and taking the remainder when divided by 13.
- If the final remainder is 0, then the function returns True, indicating that the number is divisible by 13. Otherwise, it returns False.
Python3
remainders = [ 0 , 1 , 2 , 3 , 4 , 5 , - 1 , - 2 , - 3 , - 4 , - 5 , 1 , 0 ]
def is_divisible_by_13(number):
blocks = []
while number > 0 :
blocks.append(number % 1000 )
number / / = 1000
remainder = 0
for block in blocks:
remainder = remainders[(remainder * 1000 + block) % 13 ]
return remainder = = 0
number = 83959092724
if (is_divisible_by_13(number)):
print ( number , "is divisible by 13." )
else :
print ( number , "is not divisible by 13." )
|
Output
83959092724 is divisible by 13.
Time complexity:
The time complexity of the is_divisible_by_13() function depends on the number of blocks in the input number. The number of blocks is at most ceil(log10(number)/3), which is O(log(number)). The precomputation of the remainders takes constant time. The computation of each block’s remainder takes constant time. Therefore, the overall time complexity is O(log(number)).
Auxiliary space:
The space complexity of the is_divisible_by_13() function is O(1), as it uses only a constant amount of extra space to store the remainders of all numbers from 0 to 12. The number of blocks is at most ceil(log10(number)/3), which is also O(1) in terms of space complexity.
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