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 using namespace std; // Function to check if the number is// divisible by 23 or notbool 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 logicclass GFG{     // Function to check if the // number is divisible by // 23 or notstatic 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 Codepublic 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 Codeif __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 logicclass GFG{     // Function to check if the // number is divisible by // 23 or notstatic 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 Codepublic 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

 

Javascript

 

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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