Check if a number is divisible by 23 or not
Given a number, the task is to quickly check if the number is divisible by 23 or not.
Examples:
Input : x = 46 Output : Yes Input : 47 Output : No
A solution to the problem is to extract the last digit and add 7 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 23, then the given number is divisible by 23.
Approach:
- Extract the last digit of the number/truncated number every time
- Add 7*(last digit of the previous number) to the truncated number
- Repeat the above three steps as long as necessary.
Illustration:
17043-->1704+7*3 = 1725-->172+7*5 = 207 which is 9*23, so 17043 is also divisible by 23.
Mathematical Proof :
Let be any number such that =100a+10b+c .
Now assume that is divisible by 23. Then
0 (mod 23)
100a+10b+c0 (mod 23)
10(10a+b)+c0 (mod 23)
10+c0 (mod 23)
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 23.
It can be observed that the smallest n which satisfies this property is 7 as 701 mod 23.
Now we can multiply the original equation 10+c0 (mod 23)
by 7 and simplify it:
70+7c0 (mod 23)
+7c0 (mod 23)
We have found out that if 0 (mod 23) then,
+7c0 (mod 23).
In other words, to check if a 3-digit number is divisible by 23,
we can just remove the last digit, multiply it by 7,
and then subtract it from the rest of the two digits.
C++
// CPP program to validate above logic #include <iostream> using namespace std; // Function to check if the number is // divisible by 23 or not bool isDivisible( long long int n) { // While there are at least 3 digits while (n / 100) { int d = n % 10; // Extracting the last digit n /= 10; // Truncating the number // Adding seven times the last // digit to the remaining number n += d * 7; } return (n % 23 == 0); } int main() { long long int n = 1191216; if (isDivisible(n)) cout << "Yes" << endl; else cout << "No" << endl; return 0; } |
Java
// Java program to validate above logic class GFG { // Function to check if the // number is divisible by // 23 or not static boolean isDivisible( long n) { // While there are at // least 3 digits while (n / 100 != 0 ) { // Extracting the last digit long d = n % 10 ; n /= 10 ; // Truncating the number // Adding seven times the last // digit to the remaining number n += d * 7 ; } return (n % 23 == 0 ); } // Driver Code public static void main(String[] args) { long n = 1191216 ; if (isDivisible(n)) System.out.println( "Yes" ); else System.out.println( "No" ); } } // This code is contributed by mits |
Python 3
# Python 3 program to validate above logic # Function to check if the number is # divisible by 23 or not def isDivisible(n) : # While there are at least 3 digits while n / / 100 : # Extracting the last d = n % 10 # Truncating the number n / / = 10 # Adding seven times the last # digit to the remaining number n + = d * 7 return (n % 23 = = 0 ) # Driver Code if __name__ = = "__main__" : n = 1191216 # function calling if (isDivisible(n)) : print ( "Yes" ) else : print ( "No" ) # This code is contributed by ANKITRAI1 |
C#
// C# program to validate // above logic class GFG { // Function to check if the // number is divisible by // 23 or not static bool isDivisible( long n) { // While there are at // least 3 digits while (n / 100 != 0) { // Extracting the last digit long d = n % 10; n /= 10; // Truncating the number // Adding seven times the last // digit to the remaining number n += d * 7; } return (n % 23 == 0); } // Driver Code public static void Main() { long n = 1191216; if (isDivisible(n)) System.Console.WriteLine( "Yes" ); else System.Console.WriteLine( "No" ); } } // This code is contributed by mits |
PHP
<?php // PHP program to validate above logic // Function to check if the number // is divisible by 23 or not function isDivisible( $n ) { // While there are at // least 3 digits while ( intval ( $n / 100)) { $n = intval ( $n ); $d = $n % 10; // Extracting the last digit $n /= 10; // Truncating the number // Adding seven times the last // digit to the remaining number $n += $d * 7; } return ( $n % 23 == 0); } $n = 1191216; if (isDivisible( $n )) echo "Yes" . "\n" ; else echo "No" . "\n" ; // This code is contributed // by ChitraNayal ?> |
Javascript
<script> // JavaScript program to validate above logic // Function to check if the // number is divisible by // 23 or not function isDivisible(n) { // While there are at // least 3 digits while (Math.floor(n / 100) != 0) { // Extracting the last digit let d = n % 10; n = Math.floor(n/10); // Truncating the number // Adding seven times the last // digit to the remaining number n += d * 7; } return (n % 23 == 0); } // Driver Code let n = 1191216; if (isDivisible(n)) document.write( "Yes" ); else document.write( "No" ); // This code is contributed by rag2127 </script> |
Output:
Yes
Time Complexity: O(log10N)
Auxiliary Space: O(1)
Note that the above program may not make a lot of sense as could simply do n % 23 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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