# Check if N is divisible by a number which is composed of the digits from the set {A, B}

Given three integers **N**, **A** and **B**, the task is to find whether N is divisible by any number that contains only **A** and **B** as it’s digits.

**Examples:**

Input:N = 106, a = 3, b = 5

Output:Yes

106 is divisible by 53

Input:N = 107, a = 3, b = 5

Output:No

**Approach 1 (Recursive):** An efficient solution is to make all the numbers that contains **a** and **b** as their digits using recursive function starting with the numbers **a** and **b**. If function call is **fun(x)** then recursively call for **fun(x * 10 + a)** and **fun(x * 10 + b)**. If **n** is divisible by any of the number then print **Yes** else print **No**.

Below is the implementation of the above approach:

## C++

`// CPP program to find if number N is divisible by a ` `// number that contains only a and b as it's digits ` `#include <bits/stdc++.h> ` `using` `namespace` `std; ` ` ` `// Function to check whether n is divisible ` `// by a number whose digits are either a or b ` `bool` `isDivisibleRec(` `int` `x, ` `int` `a, ` `int` `b, ` `int` `n) ` `{ ` ` ` `// base condition ` ` ` `if` `(x > n) ` ` ` `return` `false` `; ` ` ` ` ` `if` `(n % x == 0) ` ` ` `return` `true` `; ` ` ` ` ` `// recursive call ` ` ` `return` `(isDivisibleRec(x * 10 + a, a, b, n) ` ` ` `|| isDivisibleRec(x * 10 + b, a, b, n)); ` `} ` ` ` `bool` `isDivisible(` `int` `a, ` `int` `b, ` `int` `n) ` `{ ` ` ` `// Check for all numbers beginning with 'a' or 'b' ` ` ` `return` `isDivisibleRec(a, a, b, n) || ` ` ` `isDivisibleRec(b, a, b, n); ` `} ` ` ` `// Driver program ` `int` `main() ` `{ ` ` ` `int` `a = 3, b = 5, n = 53; ` ` ` ` ` `if` `(isDivisible(a, b, n)) ` ` ` `cout << ` `"Yes"` `; ` ` ` `else` ` ` `cout << ` `"No"` `; ` ` ` ` ` `return` `0; ` `} ` |

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## Java

`// Java program to find if number N is divisible by a ` `// number that contains only a and b as it's digits ` ` ` `import` `java.util.*; ` `class` `solution ` `{ ` ` ` `// Function to check whether n is divisible ` `// by a number whose digits are either a or b ` `static` `boolean` `isDivisibleRec(` `int` `x, ` `int` `a, ` `int` `b, ` `int` `n) ` `{ ` ` ` `// base condition ` ` ` `if` `(x > n) ` ` ` `return` `false` `; ` ` ` ` ` `if` `(n % x == ` `0` `) ` ` ` `return` `true` `; ` ` ` ` ` `// recursive call ` ` ` `return` `(isDivisibleRec(x * ` `10` `+ a, a, b, n) ` ` ` `|| isDivisibleRec(x * ` `10` `+ b, a, b, n)); ` `} ` ` ` `static` `boolean` `isDivisible(` `int` `a, ` `int` `b, ` `int` `n) ` `{ ` `// Check for all numbers beginning with 'a' or 'b' ` `return` `isDivisibleRec(a, a, b, n) ` ` ` `||isDivisibleRec(b, a, b, n); ` `} ` ` ` `// Driver program ` `public` `static` `void` `main(String args[]) ` `{ ` ` ` `int` `a = ` `3` `, b = ` `5` `, n = ` `53` `; ` ` ` ` ` `if` `(isDivisible(a, b, n)) ` ` ` `System.out.print(` `"Yes"` `); ` ` ` `else` ` ` `System.out.print(` `"No"` `); ` ` ` `} ` ` ` `} ` `//contributed by Arnab Kundu ` |

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## Python 3

# Python 3 program to find if number N

# is divisible by a number that contains

# only a and b as it’s digits

# Function to check whether n is divisible

# by a number whose digits are either a or b

def isDivisibleRec(x, a, b, n):

# base condition

if (x > n):

return False

if (n % x == 0):

return True

# recursive call

return (isDivisibleRec(x * 10 + a, a, b, n) or

isDivisibleRec(x * 10 + b, a, b, n))

def isDivisible(a, b, n):

# Check for all numbers beginning

# with ‘a’ or ‘b’

return (isDivisibleRec(a, a, b, n) or

isDivisibleRec(b, a, b, n))

# Driver Code

a = 3; b = 5; n = 53;

if (isDivisible(a, b, n)):

print(“Yes”)

else:

print(“No”)

# This code is contributed

# by Akanksha Rai

## C#

`// C# program to find if number N is ` `// divisible by a number that contains ` `// only a and b as it's digits ` `using` `System; ` ` ` `class` `GFG ` `{ ` ` ` `// Function to check whether n is divisible ` `// by a number whose digits are either a or b ` `static` `bool` `isDivisibleRec(` `int` `x, ` `int` `a, ` ` ` `int` `b, ` `int` `n) ` `{ ` ` ` `// base condition ` ` ` `if` `(x > n) ` ` ` `return` `false` `; ` ` ` ` ` `if` `(n % x == 0) ` ` ` `return` `true` `; ` ` ` ` ` `// recursive call ` ` ` `return` `(isDivisibleRec(x * 10 + a, a, b, n) || ` ` ` `isDivisibleRec(x * 10 + b, a, b, n)); ` `} ` ` ` `static` `bool` `isDivisible(` `int` `a, ` `int` `b, ` `int` `n) ` `{ ` ` ` `// Check for all numbers beginning ` `// with 'a' or 'b' ` `return` `isDivisibleRec(a, a, b, n) || ` ` ` `isDivisibleRec(b, a, b, n); ` `} ` ` ` `// Driver Code ` `static` `public` `void` `Main () ` `{ ` ` ` `int` `a = 3, b = 5, n = 53; ` ` ` ` ` `if` `(isDivisible(a, b, n)) ` ` ` `Console.WriteLine(` `"Yes"` `); ` ` ` `else` ` ` `Console.WriteLine(` `"No"` `); ` `} ` `} ` ` ` `// This code is contributed by Sachin ` |

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## PHP

`<?php ` `// PHP program to find if number N is ` `// divisible by a number that contains ` `// only a and b as it's digits ` ` ` `// Function to check whether n is divisible ` `// by a number whose digits are either a or b ` `function` `isDivisibleRec(` `$x` `, ` `$a` `, ` `$b` `, ` `$n` `) ` `{ ` ` ` `// base condition ` ` ` `if` `(` `$x` `> ` `$n` `) ` ` ` `return` `false; ` ` ` ` ` `if` `(` `$n` `% ` `$x` `== 0) ` ` ` `return` `true; ` ` ` ` ` `// recursive call ` ` ` `return` `(isDivisibleRec(` `$x` `* 10 + ` `$a` `, ` `$a` `, ` `$b` `, ` `$n` `) || ` ` ` `isDivisibleRec(` `$x` `* 10 + ` `$b` `, ` `$a` `, ` `$b` `, ` `$n` `)); ` `} ` ` ` `function` `isDivisible(` `$a` `, ` `$b` `, ` `$n` `) ` `{ ` ` ` `// Check for all numbers beginning ` `// with 'a' or 'b' ` `return` `isDivisibleRec(` `$a` `, ` `$a` `, ` `$b` `, ` `$n` `) || ` ` ` `isDivisibleRec(` `$b` `, ` `$a` `, ` `$b` `, ` `$n` `); ` `} ` ` ` `// Driver Code ` `$a` `= 3; ` `$b` `= 5; ` `$n` `= 53; ` ` ` `if` `(isDivisible(` `$a` `, ` `$b` `, ` `$n` `)) ` ` ` `echo` `"Yes"` `; ` `else` ` ` `echo` `"No"` `; ` ` ` `// This code is contributed ` `// by Akanksha Rai ` |

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**Output:**

Yes

**Approach 2 (Queue Based):** The idea is to generate all numbers (smaller than n) containing digits a and b. For every number check if it divides n or not. How to generate all numbers smaller than n? We use queue for this. Initially we push ‘a’ and ‘b’ to the queue. Then we run a loop while front of queue is smaller than n. We pop an item one by one and for ever popped item x, we generate next numbers x*10 + a and x*10 + b and enqueue them. Time complexity of this approach is O(n)

Please refer below post for implementation of this approach.

Count of Binary Digit numbers smaller than N

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