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++

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// 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

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// 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#

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// 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

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<?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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