The very first solution that comes to our mind is the one that we learned in school. If the sum of digits in a number is a multiple of 3 then the number is a multiple of 3, e.g., for 612, the sum of digits is 9 so it’s a multiple of 3. But this solution is not efficient. You have to get all decimal digits one by one, add them and then check if the sum is a multiple of 3.
There is a pattern in the binary representation of a number that can be used to find if a number is a multiple of 3. If the difference between the count of odd set bits (Bits set at odd positions) and even set bits is a multiple of 3 then is the number.
Example: 23 (00..10111)
1) Get count of all set bits at odd positions (For 23 it’s 3).
2) Get count of all set bits at even positions (For 23 it’s 1).
3) If the difference between the above two counts is a multiple of 3 then the number is also a multiple of 3.
(For 23 it’s 2 so 23, is not a multiple of 3)
Take some more examples like 21, 15, etc…
Algorithm: isMutlipleOf3(n)
1) Make n positive if n is negative.
2) If number is 0 then return 1
3) If number is 1 then return 0
4) Initialize: odd_count = 0, even_count = 0
5) Loop while n != 0
a) If rightmost bit is set then increment odd count.
b) Right-shift n by 1 bit
c) If rightmost bit is set then increment even count.
d) Right-shift n by 1 bit
6) return isMutlipleOf3(odd_count - even_count)
Proof:
Let the binary representation of the number be: abcde.
In decimal form this number will be: 
Every even power of 2 can be represented as 3n + 1, and every odd power of 2 can be represented as 3n – 1. For example:
ans so on...
ans so on...Therefore, the decimal form becomes:
[Tex](3n)(a+b+c+d+e) + (a - b + c - d + e) \\[/Tex](multiple of 3)
[Tex](3n)(a+b+c+d+e) + (a - b + c - d + e) \\[/Tex](multiple of 3)
To have this number divisible by 3, the term 
should be divisible by 3.
Therefore for the number to be divisible by, the difference between the count of odd set bits (a + c + e) and even set bits (b + d) should be divisible by 3.
Program:
C++
// CPP program to check if n is a multiple of 3#include <bits/stdc++.h>using namespace std;/* Function to check if n is a multiple of 3*/int isMultipleOf3(int n){ int odd_count = 0; int even_count = 0; /* Make no positive if +n is multiple of 3 then is -n. We are doing this to avoid stack overflow in recursion*/ if (n < 0) n = -n; if (n == 0) return 1; if (n == 1) return 0; while (n) { /* If odd bit is set then increment odd counter */ if (n & 1) odd_count++; /* If even bit is set then increment even counter */ if (n & 2) even_count++; n = n >> 2; } return isMultipleOf3(abs(odd_count - even_count));}/* Program to test function isMultipleOf3 */int main(){ int num = 24; if (isMultipleOf3(num)) printf("%d is multiple of 3", num); else printf("%d is not a multiple of 3", num); return 0;} |
Java
// Java program to check if// n is a multiple of 3import java.lang.*;import java.util.*;class GFG { // Function to check if n // is a multiple of 3 static int isMultipleOf3(int n) { int odd_count = 0; int even_count = 0; /* Make no positive if +n is multiple of 3 then is -n. We are doing this to avoid stack overflow in recursion*/ if (n < 0) n = -n; if (n == 0) return 1; if (n == 1) return 0; while (n != 0) { /* If odd bit is set then increment odd counter */ if ((n & 1) != 0) odd_count++; /* If even bit is set then increment even counter */ if ((n & 2) != 0) even_count++; n = n >> 2; } return isMultipleOf3(Math.abs(odd_count - even_count)); } /* Program to test function isMultipleOf3 */ public static void main(String[] args) { int num = 24; if (isMultipleOf3(num) != 0) System.out.println(num + " is multiple of 3"); else System.out.println(num + " is not a multiple of 3"); }}// This code is contributed by Sahil_Bansall |
Python3
# Python program to check if n is a multiple of 3# Function to check if n is a multiple of 3def isMultipleOf3(n): odd_count = 0 even_count = 0 # Make no positive if + n is multiple of 3 # then is -n. We are doing this to avoid # stack overflow in recursion if(n < 0): n = -n if(n == 0): return 1 if(n == 1): return 0 while(n): # If odd bit is set then # increment odd counter if(n & 1): odd_count += 1 # If even bit is set then # increment even counter if(n & 2): even_count += 1 n = n >> 2 return isMultipleOf3(abs(odd_count - even_count))# Program to test function isMultipleOf3 num = 24if (isMultipleOf3(num)): print(num, 'is multiple of 3')else: print(num, 'is not a multiple of 3')# This code is contributed by Danish Raza |
C#
// C# program to check if// n is a multiple of 3using System;class GFG { // Function to check if n // is a multiple of 3 static int isMultipleOf3(int n) { int odd_count = 0, even_count = 0; /* Make no positive if +n is multiple of 3 then is -n. We are doing this to avoid stack overflow in recursion*/ if (n < 0) n = -n; if (n == 0) return 1; if (n == 1) return 0; while (n != 0) { /* If odd bit is set then increment odd counter */ if ((n & 1) != 0) odd_count++; /* If even bit is set then increment even counter */ if ((n & 2) != 0) even_count++; n = n >> 2; } return isMultipleOf3(Math.Abs(odd_count - even_count)); } // Driver Code public static void Main() { int num = 24; if (isMultipleOf3(num) != 0) Console.Write(num + " is multiple of 3"); else Console.Write(num + " is not a multiple of 3"); }}// This code is contributed by Sam007 |
PHP
<?php// PHP program to check if n// is a multiple of 3// Function to check if n // is a multiple of 3function isMultipleOf3( $n){ $odd_count = 0; $even_count = 0; // Make no positive if +n // is multiple of 3 then is -n. // We are doing this to avoid // stack overflow in recursion if($n < 0) $n = -$n; if($n == 0) return 1; if($n == 1) return 0; while($n) { // If odd bit is set then // increment odd counter if($n & 1) $odd_count++; // If even bit is set then // increment even counter if($n & 2) $even_count++; $n = $n >> 2; } return isMultipleOf3(abs($odd_count - $even_count));}// Driver Code$num = 24;if (isMultipleOf3($num)) echo $num, "is multiple of 3";else echo $num, "is not a multiple of 3"; // This code is contributed by vt_m.?> |
Javascript
<script>/*Function to check if n is a multiple of 3*/ function isMultipleof3(n) { odd_count = 0; even_count = 0; /* Make no positive if +n is multiple of 3 then is -n. We are doing this to avoid stack overflowin recursion*/ if(n < 0) n = -n; if(n == 0) return 1; if(n == 1) return 0; while (n) { /*If odd bit is set then increment odd counter*/ if(n & 1) odd_count++; /*If even bit is set then increment even counter*/ if(n & 2) even_count++; n = n>>2; } return isMultipleof3(Math.abs(odd_count-even_count)); } /*Program to test function isMultipleof3*/ num = 24; if(isMultipleof3(num)) document.write(num + " is multiple of 3"); else document.write(num + " is not a multiple of 3"); // This code is contributed by simranarora5sos </script> |
Output
24 is multiple of 3
Time Complexity: O(logn)
Auxiliary Space: O(1)
Efficient Method:
Use Dynamic Programming (Top-Down Approach Using Memoization)
C++
// CPP program to check if n is a multiple of 3#include <bits/stdc++.h>using namespace std;int static dp[1001];/* Function to check if n is a multiple of 3*/int isMultipleOf3(int n){ int odd_count = 0; int even_count = 0; // Base Cases if (n < 0) n = -n; if (n == 0) return 1; if (n == 1) return 0; // If a value is already present // in dp, return it if(dp[n] != -1) return dp[n]; while (n) { /* If odd bit is set then increment odd counter */ if (n & 1) odd_count++; /* If even bit is set then increment even counter */ if (n & 2) even_count++; n = n >> 2; } dp[n] = isMultipleOf3(abs(odd_count - even_count)); // return dp return dp[n];}/* Program to test function isMultipleOf3 */int main(){ int num = 24; memset(dp, -1, sizeof(dp)); if (isMultipleOf3(num)) printf("%d is multiple of 3", num); else printf("%d is not a multiple of 3", num); return 0;} |
C
// C program to check if n is a multiple of 3#include <stdio.h>#include<string.h>int static dp[1001];/* Function to check if n is a multiple of 3*/int isMultipleOf3(int n){ int odd_count = 0; int even_count = 0; // Base Cases if (n < 0) n = -n; if (n == 0) return 1; if (n == 1) return 0; // If a value is already present // in dp, return it if(dp[n] != -1) return dp[n]; while (n) { /* If odd bit is set then increment odd counter */ if (n & 1) odd_count++; /* If even bit is set then increment even counter */ if (n & 2) even_count++; n = n >> 2; } int abs = (odd_count - even_count); if(abs<0) { abs= -1*abs; } dp[n] = isMultipleOf3(abs); // return dp return dp[n];}/* Program to test function isMultipleOf3 */int main(){ int num = 24; memset(dp, -1, sizeof(dp)); if (isMultipleOf3(num)) printf("%d is multiple of 3", num); else printf("%d is not a multiple of 3", num); return 0;}// This code is contributed by akashish_. |
Java
// JAVA program to check if n is a multiple of 3import java.util.*;class GFG{ static int []dp ;/* Function to check if n is a multiple of 3*/static int isMultipleOf3(int n){ int odd_count = 0; int even_count = 0; // Base Cases if (n < 0) n = -n; if (n == 0) return 1; if (n == 1) return 0; // If a value is already present // in dp, return it if(dp[n] != -1) return dp[n]; while (n > 0) { /* If odd bit is set then increment odd counter */ if ((n & 1) != 0) odd_count++; /* If even bit is set then increment even counter */ if ((n & 2) != 0) even_count++; n = n >> 2; } dp[n] = isMultipleOf3(Math.abs(odd_count - even_count)); // return dp return dp[n];}/* Program to test function isMultipleOf3 */public static void main(String[] args){ int num = 24; dp = new int[1001]; Arrays.fill(dp, -1); if (isMultipleOf3(num) == 1) System.out.printf("%d is multiple of 3", num); else System.out.printf("%d is not a multiple of 3", num);}}// This codeiscontributed by Rajput-Ji |
Python3
# Python program to check if n is a multiple of 3dp = [-1 for i in range(1001)];''' Function to check if n is a multiple of 3 '''def isMultipleOf3(n): odd_count = 0; even_count = 0; # Base Cases if (n < 0): n = -n; if (n == 0): return 1; if (n == 1): return 0; # If a value is already present # in dp, return it if (dp[n] != -1): return dp[n]; while (n > 0): ''' * If odd bit is set then increment odd counter ''' if ((n & 1) != 0): odd_count+=1; ''' * If even bit is set then increment even counter ''' if ((n & 2) != 0): even_count+=1; n = n >> 2; dp[n] = isMultipleOf3(abs(odd_count - even_count)); # return dp return dp[n];''' Program to test function isMultipleOf3 '''if __name__ == '__main__': num = 24; if (isMultipleOf3(num) == 1): print(num,"is multiple of 3"); else: print(num," is not a multiple of 3");# This code is contributed by Rajput-Ji |
C#
// C# program to check if// n is a multiple of 3using System;class GFG { static int []dp = new int[1001]; /* Function to check if n is a multiple of 3*/static int isMultipleOf3(int n){ int odd_count = 0; int even_count = 0; // Base Cases if (n < 0) n = -n; if (n == 0) return 1; if (n == 1) return 0; // If a value is already present // in dp, return it if(dp[n] != -1) return dp[n]; while (n > 0) { /* If odd bit is set then increment odd counter */ if ((n & 1) != 0) odd_count++; /* If even bit is set then increment even counter */ if ((n & 2) != 0) even_count++; n = n >> 2; } dp[n] = isMultipleOf3(Math.Abs(odd_count - even_count)); // return dp return dp[n];} // Driver Codepublic static void Main(){ int num = 24; for(int i = 0; i < 1001; i++) { dp[i] = -1; } if (isMultipleOf3(num) == 1) Console.Write(num + " is multiple of 3"); else Console.Write(num + " is not a multiple of 3");}}// This code is contributed by Samim Hossain Mondal. |
Javascript
<script>// JavaScript program to check if n is a multiple of 3let dp = [];for(let i = 0; i < 1001; i++) { dp[i] = -1;}/* Function to check if n is a multiple of 3*/function isMultipleOf3(n){ let odd_count = 0; let even_count = 0; // Base Cases if (n < 0) n = -n; if (n == 0) return 1; if (n == 1) return 0; // If a value is already present // in dp, return it if(dp[n] != -1) return dp[n]; while (n) { /* If odd bit is set then increment odd counter */ if (n & 1) odd_count++; /* If even bit is set then increment even counter */ if (n & 2) even_count++; n = n >> 2; } dp[n] = isMultipleOf3(Math.abs(odd_count - even_count)); // return dp return dp[n];}/* Program to test function isMultipleOf3 */let num = 24; if (isMultipleOf3(num)) document.write(num + " is multiple of 3");else document.write(num + " is not a multiple of 3");// This code is contributed by Samim Hossain Mondal.</script> |
Output
24 is multiple of 3
Time Complexity: O(nlogn)
Auxiliary Space: O(n)
Method: Checking given number is multiple of 3 or not using modulo division.
C++
// C++ program to check if given number is multiple of 3 or// not using modulo division#include <iostream>using namespace std;int main(){ // input number int num = 24; cout << num; // checking if the given number if multiple of 3 or not // using modulo division operator if the output of num%3 // is equal to 0 then it is considered as multiple of 3 // otherwise not a multiple of 3 if (num % 3 == 0) { cout << " is multiple of 3"; } else { cout << " is not multiple of 3"; } return 0;}// this code is contributed by gangarajula laxmi |
Java
/*package whatever //do not write package name here */// Java code// To check whether the given number is divisible by 3 or notimport java.io.*;import java.util.*; class GFG{ public static void main(String[] args) { //input int num = 24; System.out.println(num); /// checking if the given number if multiple of 3 or not // using modulo division operator if the output of num%3 // is equal to 0 then it is considered as multiple of 3 // otherwise not a multiple of 3 if ((n) % 3 == 0) { System.out.println(" is multiple of 3"); } else { System.out.println(" is not multiple of 3"); } }}// This code is contributed by satwik4409. |
Python3
# Python3 program to check if given number is multiple of 3 or# not using modulo division# input numbernum = 24print(num,end="")# checking if the given number if multiple of 3 or not# using modulo division operator if the output of num%3# is equal to 0 then it is considered as multiple of 3# otherwise not a multiple of 3if (num % 3 is 0): print(" is multiple of 3")else: print( " is not multiple of 3")# This code is contributed by akashish__ |
C#
// C# program to check if given number is multiple of 3 or// not using modulo divisionusing System;public class GFG { static public void Main() { // input number int num = 24; // checking if the given number if multiple of 3 or // not using modulo division operator if the output // of num%3 is equal to 0 then it is considered as // multiple of 3 otherwise not a multiple of 3 if (num % 3 == 0) { Console.WriteLine(num+" is multiple of 3"); } else { Console.WriteLine(num+" is not multiple of 3"); } }}// This code is contributed by laxmigangarajula03 |
Javascript
<script> // JavaScript code for the above approach // input number let num = 24; document.write(num); // checking if the given number if multiple of 3 or not // using modulo division operator if the output of num%3 // is equal to 0 then it is considered as multiple of 3 // otherwise not a multiple of 3 if (num % 3 == 0) { document.write(" is multiple of 3"); } else { document.write(" is not multiple of 3"); } // This code is contributed by Potta Lokesh </script> |
Output
24 is multiple of 3
Time Complexity: O(1)
Auxiliary Space: O(1)
Method 4:
1. It takes a number ‘num’ as input.
2. It calculates the digital root of the number by repeatedly adding the digits of the number until it becomes a single-digit number.
3. It then checks if the digital root is 3, 6, or 9, and returns True if it is, and False otherwise.
For example, for the number 123, the digital root would be calculated as follows:
1 + 2 + 3 = 6
Since 6 is divisible by 3, the function will return True for the input 123.
C++
#include <iostream>#include <string>using namespace std;bool is_multiple_of_3(int num) { // Calculate digital root while (num > 9) { string num_str = to_string(num); int sum = 0; for (int i = 0; i < num_str.length(); i++) { sum += num_str[i] - '0'; } num = sum; } // Check if digital root is 3, 6, or 9 if (num % 3 == 0) { return true; } else { return false; }}int main() { cout << boolalpha << is_multiple_of_3(123) << endl; cout << boolalpha << is_multiple_of_3(789) << endl; return 0;} |
Java
public class GFG { public static boolean isMultipleOf3(int num) { // Calculate digital root while (num > 9) { int sum = 0; while (num > 0) { sum += num % 10; num /= 10; } num = sum; } // Check if digital root is 3, 6, or 9 if (num % 3 == 0) { return true; } else { return false; } } public static void main(String[] args) { int num1 = 123; int num2 = 789; System.out.println( isMultipleOf3(num1)); // Output: true System.out.println( isMultipleOf3(num2)); // Output: true }} |
Python3
def is_multiple_of_3(num): # Calculate digital root while num > 9: num = sum(map(int, str(num))) # Check if digital root is 3, 6, or 9 if num % 3 == 0: return True else: return Falseprint(is_multiple_of_3(123))print(is_multiple_of_3(789)) |
C#
using System;namespace IsMultipleOf3 { class Program { static bool IsMultipleOf3(int num) { // Calculate digital root while (num > 9) { int sum = 0; foreach (char c in num.ToString()) { sum += c - '0'; } num = sum; } // Check if digital root is 3, 6, or 9 if (num % 3 == 0) { return true; } else { return false; } } static void Main(string[] args) { Console.WriteLine(IsMultipleOf3(123)); Console.WriteLine(IsMultipleOf3(789)); } }} |
Javascript
function is_multiple_of_3(num) { // Calculate digital root while (num > 9) { let num_str = num.toString(); let sum = 0; for (let i = 0; i < num_str.length; i++) { sum += parseInt(num_str[i]); } num = sum; } // Check if digital root is 3, 6, or 9 if (num % 3 === 0) { return true; } else { return false; }}console.log(is_multiple_of_3(123));console.log(is_multiple_of_3(789)); |
Output
True True
Time complexity: O(log n), where n is the input value
Auxiliary Space: O(1)
Related Articles:
Check divisibility in a binary stream
DFA based division
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