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Rearrangement of a number which is also divisible by it

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  • Difficulty Level : Medium
  • Last Updated : 07 Jan, 2019

Given a number n, we need to rearrange all its digits such that the new arrangement is divisible by n. Also, the new number should not be equal to x. If no such rearrangement is possible, print -1.


Input : n = 1035
Output : 3105
The result 3105 is divisible by
given n and has the same set of digits.

Input : n = 1782
Output : m = 7128

Simple Approach : Find all the permutation of given n and then check whether it is divisible by n or not also check that new permutation should not be equal to n.

Efficient Approach : Let’s suppose that y is our result then y = m * n, also we know that y must be a rearrangement of digits of n so we can say now restrict m (the multiplier) as per given conditions.
1) y has the same number of digits as n has. So, m must be less than 10.
2) y must not be equal to n. So, m will be greater than 1.
So we get the multiplier m in the range [2,9]. So we will find all the possible y and then check that should y has the same digits as n or not.


// CPP program for finding rearrangement of n
// that is divisible  by n
using namespace std;
// perform hashing for given n
void storeDigitCounts(int n, vector<int> &hash)
    // perform hashing
    while (n)
        n /= 10;
// check whether any arrangement exists
int rearrange (int n)
    // Create a hash for given number n
    // The hash is of size 10 and stores
    // count of each digit in n.
    vector<int> hash_n(10, 0);
    storeDigitCounts(n, hash_n);
    // check for all possible multipliers
    for (int mult=2; mult<10; mult++)
        int curr = n*mult;
        vector<int> hash_curr(10, 0);
        storeDigitCounts(curr, hash_curr);
        // check hash table for both. 
        // Please refer below link for help 
        // of equal()
        if (equal(hash_n.begin(), hash_n.end(),
            return curr;
    return -1;
// driver program
int main()
    int n = 10035;
    cout << rearrange(n);
    return 0;


# Python3 program for finding rearrangement 
# of n that is divisible by n 
# Perform hashing for given n 
def storeDigitCounts(n, Hash): 
    # perform hashing 
    while n > 0
        Hash[n % 10] += 1
        n //= 10
# check whether any arrangement exists 
def rearrange(n): 
    # Create a hash for given number n 
    # The hash is of size 10 and stores 
    # count of each digit in n. 
    hash_n = [0] * 10
    storeDigitCounts(n, hash_n) 
    # check for all possible multipliers 
    for mult in range(2, 10): 
        curr = n * mult 
        hash_curr = [0] * 10
        storeDigitCounts(curr, hash_curr) 
        # check hash table for both. 
        if hash_n == hash_curr: 
            return curr 
    return -1
# Driver Code
if __name__ == "__main__"
    n = 10035
# This code is contributed by Rituraj Jain



This article is contributed by Shivam Pradhan (anuj_charm). If you like GeeksforGeeks and would like to contribute, you can also write an article using or mail your article to See your article appearing on the GeeksforGeeks main page and help other Geeks.

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