# Smallest multiple of 3 which consists of three given non-zero digits

Last Updated : 23 Jun, 2022

Given three non-zero digits 0 < A, B, C < 9. The task is to find the smallest number divisible by 3 all of whose digits are in set {A, B, C}.
Note: It is not necessary to include all three digits. The result can be A, AA, AB, CCA etc.
Examples:

Input: A = 2, B = 4, C = 7
Output: 24
24 is the minimum number divisible by 3 that can be formed by the given digits.
Input: A = 1, B = 2, C = 3
Output:

Approach: Take all three numbers in an array and sort them in increasing order. Now check if any number is divisible by 3, if yes then the number will be the answer.
If not then again check if by taking any of two numbers. Finally take smallest number and our result is this number repeated three times.
Why can’t we get answer of length more than three digits?
Even if any number is not divisible by 3, repeating the smallest number 3 times will make it divisible by 3. Note that a number is divisible by 3 if sum of its digits is divisible by 3.
Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach` `#include ` `using` `namespace` `std;`   `// Function to return the minimum number` `// divisible by 3 formed by the given digits` `int` `printSmallest(``int` `a[3])` `{` `    ``int` `sum, sum1;`   `    ``// Sort the given array in ascending` `    ``sort(a, a + 3);`   `    ``int` `i, j, k, num;`   `    ``// Check if any single digit is divisible by 3` `    ``for` `(``int` `i = 0; i < 3; i++) {` `        ``if` `(a[i] % 3 == 0)` `            ``return` `a[i];` `    ``}`   `    ``// Check if any two digit number formed by` `    ``// the given digits is divisible by 3` `    ``// starting from the minimum` `    ``for` `(i = 0; i < 3; i++) {` `        ``for` `(j = 0; j < 3; j++) {`   `            ``// Generate the two digit number` `            ``num = (a[i] * 10) + a[j];` `            ``if` `(num % 3 == 0)` `                ``return` `num;` `        ``}` `    ``}`   `    ``// If none of the above is true, we can` `    ``// form three digit number by taking a[0]` `    ``// three times.` `    ``return` `a[0]*100 + a[0]*10 + a[0];` `}`   `// Driver code` `int` `main()` `{` `    ``int` `arr[] = { 7, 7, 1 };` `    ``cout << printSmallest(arr);`   `    ``return` `0;` `}`

## Java

 `// Java implementation of the approach` `import` `java.util.Arrays;` `public` `class` `GFG {`   `// Function to return the minimum number` `// divisible by 3 formed by the given digits` `    ``static` `int` `printSmallest(``int` `a[]) {` `        ``int` `sum, sum1;`   `        ``// Sort the given array in ascending` `        ``Arrays.sort(a);`   `        ``int` `i, j, k, num;`   `        ``// Check if any single digit is divisible by 3` `        ``for` `(i = ``0``; i < ``3``; i++) {` `            ``if` `(a[i] % ``3` `== ``0``) {` `                ``return` `a[i];` `            ``}` `        ``}`   `        ``// Check if any two digit number formed by` `        ``// the given digits is divisible by 3` `        ``// starting from the minimum` `        ``for` `(i = ``0``; i < ``3``; i++) {` `            ``for` `(j = ``0``; j < ``3``; j++) {`   `                ``// Generate the two digit number` `                ``num = (a[i] * ``10``) + a[j];` `                ``if` `(num % ``3` `== ``0``) {` `                    ``return` `num;` `                ``}` `            ``}` `        ``}`   `        ``// If none of the above is true, we can` `        ``// form three digit number by taking a[0]` `        ``// three times.` `        ``return` `a[``0``] * ``100` `+ a[``0``] * ``10` `+ a[``0``];` `    ``}`   `// Driver code` `    ``public` `static` `void` `main(String[] args) {`   `        ``int` `arr[] = {``7``, ``7``, ``1``};` `        ``System.out.println(printSmallest(arr));`   `    ``}` `}` `// This code is contributed by 29AjayKumar`

## Python3

 `# Python3 implementation of the approach`   `# Function to return the minimum ` `# number divisible by 3 formed by ` `# the given digits` `def` `printSmallest(a, n):`   `    ``sum0, sum1 ``=` `0``, ``0`   `    ``# Sort the given array in ascending` `    ``a ``=` `sorted``(a)`   `    ``# Check if any single digit is` `    ``# divisible by 3` `    ``for` `i ``in` `range``(n):` `        ``if` `(a[i] ``%` `3` `=``=` `0``):` `            ``return` `a[i]`   `    ``# Check if any two digit number ` `    ``# formed by the given digits is ` `    ``# divisible by 3 starting from ` `    ``# the minimum` `    ``for` `i ``in` `range``(n):` `        ``for` `j ``in` `range``(n):` `            `  `            ``# Generate the two digit number` `            ``num ``=` `(a[i] ``*` `10``) ``+` `a[j]` `            ``if` `(num ``%` `3` `=``=` `0``):` `                ``return` `num`   `    ``# If none of the above is true, we can` `    ``# form three digit number by taking a[0]` `    ``# three times.` `    ``return` `a[``0``] ``*` `100` `+` `a[``0``] ``*` `10` `+` `a[``0``]`   `# Driver code` `arr ``=` `[``7``, ``7``, ``1` `]` `n ``=` `len``(arr)`   `print``(printSmallest(arr, n))`   `# This code is contributed` `# by mohit kumar 29`

## C#

 `    `  `// C# implementation of the approach` `using` `System;`   `public` `class` `GFG {` ` `  `// Function to return the minimum number` `// divisible by 3 formed by the given digits` `    ``static` `int` `printSmallest(``int` `[]a) {` ` `  `        ``// Sort the given array in ascending` `        ``Array.Sort(a);` ` `  `        ``int` `i, j, num;` ` `  `        ``// Check if any single digit is divisible by 3` `        ``for` `(i = 0; i < 3; i++) {` `            ``if` `(a[i] % 3 == 0) {` `                ``return` `a[i];` `            ``}` `        ``}` ` `  `        ``// Check if any two digit number formed by` `        ``// the given digits is divisible by 3` `        ``// starting from the minimum` `        ``for` `(i = 0; i < 3; i++) {` `            ``for` `(j = 0; j < 3; j++) {` ` `  `                ``// Generate the two digit number` `                ``num = (a[i] * 10) + a[j];` `                ``if` `(num % 3 == 0) {` `                    ``return` `num;` `                ``}` `            ``}` `        ``}` ` `  `        ``// If none of the above is true, we can` `        ``// form three digit number by taking a[0]` `        ``// three times.` `        ``return` `a[0] * 100 + a[0] * 10 + a[0];` `    ``}` ` `  `// Driver code` `    ``public` `static` `void` `Main() {` ` `  `        ``int` `[]arr = {7, 7, 1};` `        ``Console.Write(printSmallest(arr));` ` `  `    ``}` `}` `//This code is contributed by Rajput-Ji`

## PHP

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## Javascript

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Output:

`111`

Time Complexity: O(1)

Auxiliary Space: O(1)

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