N digit numbers divisible by 5 formed from the M digits
Last Updated :
11 Jul, 2025
Given M unique digits and a number N. The task is to find the number of N-digit numbers which can be formed from the given M digits, which are divisible by 5 and none of the digits is repeated.
Note: If it is not possible to form a N digit number from the given digits, print -1.
Examples:
Input : N = 3, M = 6, digits[] = {2, 3, 5, 6, 7, 9}
Output : 20
Input : N = 5, M = 6, digits[] = {0, 3, 5, 6, 7, 9}
Output : 240
For a number to be divisible by 5, the only condition is that the digit at the unit place in the number must be either 0 or 5.
So, to find the count of numbers that are divisible by 5 and can be formed from the given digits, do the following:
- Check if the given digits contain both 0 and 5.
- If the given digits contain both 0 and 5, then the unit place can be filled in 2 ways otherwise the unit place can be filled in 1 way.
- Now, the tens place can now be filled by any of the remaining M-1 digits. So, there are (M-1) ways of filling the tens place.
- Similarly, the hundred's place can now be filled by any of the remaining (M-2) digits and so on.
Therefore, if the given digits have both 0 and 5:
Required number of numbers = 2 * (M-1)* (M-2)...N-times.
Otherwise, if the given digits have either one of 0 and 5 and not both:
Required number of numbers = 1 * (M-1)* (M-2)...N-times.
Below is the implementation of the above approach.
C++
// CPP program to find the count of all
// possible N digit numbers which are
// divisible by 5 formed from M digits
#include <bits/stdc++.h>
using namespace std;
// Function to find the count of all
// possible N digit numbers which are
// divisible by 5 formed from M digits
int numbers(int n, int arr[], int m)
{
int isZero = 0, isFive = 0;
int result = 0;
// If it is not possible to form
// n digit number from the given
// m digits without repetition
if (m < n) {
return -1;
}
for (int i = 0; i < m; i++) {
if (arr[i] == 0)
isZero = 1;
if (arr[i] == 5)
isFive = 1;
}
// If both zero and five exists
if (isZero && isFive) {
result = 2;
// Remaining N-1 iterations
for (int i = 0; i < n - 1; i++) {
result = result * (--m);
}
}
else if (isZero || isFive) {
result = 1;
// Remaining N-1 iterations
for (int i = 0; i < n - 1; i++) {
result = result * (--m);
}
}
else
result = -1;
return result;
}
// Driver code
int main()
{
int n = 3, m = 6;
int arr[] = { 2, 3, 5, 6, 7, 9 };
cout << numbers(n, arr, m);
return 0;
}
// Java program to find the count of all
// possible N digit numbers which are
// divisible by 5 formed from M digits
class GFG {
// Function to find the count of all
// possible N digit numbers which are
// divisible by 5 formed from M digits
static int numbers(int n, int arr[], int m) {
int isZero = 0, isFive = 0;
int result = 0;
// If it is not possible to form
// n digit number from the given
// m digits without repetition
if (m < n) {
return -1;
}
for (int i = 0; i < m; i++) {
if (arr[i] == 0) {
isZero = 1;
}
if (arr[i] == 5) {
isFive = 1;
}
}
// If both zero and five exists
if (isZero == 1 && isFive == 1) {
result = 2;
// Remaining N-1 iterations
for (int i = 0; i < n - 1; i++) {
result = result * (--m);
}
} else if (isZero == 1 || isFive == 1) {
result = 1;
// Remaining N-1 iterations
for (int i = 0; i < n - 1; i++) {
result = result * (--m);
}
} else {
result = -1;
}
return result;
}
// Driver code
public static void main(String[] args) {
int n = 3, m = 6;
int arr[] = {2, 3, 5, 6, 7, 9};
System.out.println(numbers(n, arr, m));
}
}
// This code is contributed by RAJPUT-JI
# Python 3 program to find the count
# of all possible N digit numbers which
# are divisible by 5 formed from M digits
# Function to find the count of all
# possible N digit numbers which are
# divisible by 5 formed from M digits
def numbers(n, arr, m):
isZero = 0
isFive = 0
result = 0
# If it is not possible to form
# n digit number from the given
# m digits without repetition
if (m < n) :
return -1
for i in range(m) :
if (arr[i] == 0):
isZero = 1
if (arr[i] == 5):
isFive = 1
# If both zero and five exists
if (isZero and isFive) :
result = 2
# Remaining N-1 iterations
for i in range( n - 1):
m -= 1
result = result * (m)
elif (isZero or isFive) :
result = 1
# Remaining N-1 iterations
for i in range(n - 1) :
m -= 1
result = result * (m)
else:
result = -1
return result
# Driver code
if __name__ == "__main__":
n = 3
m = 6
arr = [ 2, 3, 5, 6, 7, 9]
print(numbers(n, arr, m))
# This code is contributed by ChitraNayal
// C# program to find the count of all
// possible N digit numbers which are
// divisible by 5 formed from M digits
using System;
public class GFG {
// Function to find the count of all
// possible N digit numbers which are
// divisible by 5 formed from M digits
static int numbers(int n, int []arr, int m) {
int isZero = 0, isFive = 0;
int result = 0;
// If it is not possible to form
// n digit number from the given
// m digits without repetition
if (m < n) {
return -1;
}
for (int i = 0; i < m; i++) {
if (arr[i] == 0) {
isZero = 1;
}
if (arr[i] == 5) {
isFive = 1;
}
}
// If both zero and five exists
if (isZero == 1 && isFive == 1) {
result = 2;
// Remaining N-1 iterations
for (int i = 0; i < n - 1; i++) {
result = result * (--m);
}
} else if (isZero == 1 || isFive == 1) {
result = 1;
// Remaining N-1 iterations
for (int i = 0; i < n - 1; i++) {
result = result * (--m);
}
} else {
result = -1;
}
return result;
}
// Driver code
public static void Main() {
int n = 3, m = 6;
int []arr = {2, 3, 5, 6, 7, 9};
Console.WriteLine(numbers(n, arr, m));
}
}
// This code is contributed by RAJPUT-JI
<?php
// PHP program to find the count of all
// possible N digit numbers which are
// divisible by 5 formed from M digits
// Function to find the count of all
// possible N digit numbers which are
// divisible by 5 formed from M digits
function numbers($n, $arr, $m)
{
$isZero = 0;
$isFive = 0;
$result = 0;
// If it is not possible to form
// n digit number from the given
// m digits without repetition
if ($m < $n)
{
return -1;
}
for ($i = 0; $i < $m; $i++)
{
if ($arr[$i] == 0)
$isZero = 1;
if ($arr[$i] == 5)
$isFive = 1;
}
// If both zero and five exists
if ($isZero && $isFive)
{
$result = 2;
// Remaining N-1 iterations
for ($i = 0; $i < $n - 1; $i++)
{
$result = $result * (--$m);
}
}
else if ($isZero || $isFive)
{
$result = 1;
// Remaining N-1 iterations
for ($i = 0; $i < $n - 1; $i++)
{
$result = $result * (--$m);
}
}
else
$result = -1;
return $result;
}
// Driver code
$n = 3;
$m = 6;
$arr = array( 2, 3, 5, 6, 7, 9 );
echo numbers($n, $arr, $m);
// This code is contributed by jit_t
?>
<script>
// Javascript program to find the count of all
// possible N digit numbers which are
// divisible by 5 formed from M digits
// Function to find the count of all
// possible N digit numbers which are
// divisible by 5 formed from M digits
function numbers(n, arr, m) {
let isZero = 0, isFive = 0;
let result = 0;
// If it is not possible to form
// n digit number from the given
// m digits without repetition
if (m < n) {
return -1;
}
for (let i = 0; i < m; i++) {
if (arr[i] == 0) {
isZero = 1;
}
if (arr[i] == 5) {
isFive = 1;
}
}
// If both zero and five exists
if (isZero == 1 && isFive == 1) {
result = 2;
// Remaining N-1 iterations
for (let i = 0; i < n - 1; i++) {
result = result * (--m);
}
} else if (isZero == 1 || isFive == 1) {
result = 1;
// Remaining N-1 iterations
for (let i = 0; i < n - 1; i++) {
result = result * (--m);
}
} else {
result = -1;
}
return result;
}
let n = 3, m = 6;
let arr = [2, 3, 5, 6, 7, 9];
document.write(numbers(n, arr, m));
</script>
Time Complexity: O(m + n), Auxiliary Space: O(1)
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