Probability that an arbitrary positive divisor of 10^X is an integral multiple of 10^Y

Given two numbers X and Y, the task is to find the probability that an arbitrary positive divisor of 10^X is an integral multiple of 10^Y.
Note: Y should be <= X.
Examples:

Input: X = 2, Y = 1 
Output: 4/9 
Explanation: 
Positive divisors of 10² are 1, 2, 4, 5, 10, 20, 25, 50, 100. A total of 9. 
Multiples of 10¹ upto 10² are 10, 20, 50, 100. A total of 4. 
P(divisor of 10² is a multiple of 10¹) = 4/9.
Input: X = 99, Y = 88 
Output: 9/625 
Explanation: 
A = 10⁹⁹, B = 10⁸⁸
P(divisor of 10⁹⁹ is a multiple of 10⁸⁸) = 9/625 

Pre-requisites: Total number of divisors of a number
Approach:
In order to solve the problem, we need to observe the steps below: 

• All divisors of 10^X will be of the form:
  

  (2 ^P * 5 ^Q), where 0 <= P, Q <= X

• Find the number of Prime factors of 10^X
  

  10^X = 2^X * 5^X

• Hence, total number of divisors of 10^X will be: ( X + 1 ) * ( X + 1 ).
• Now, consider all multiples of 10^Y which can be potential divisors of 10^X. They are also of the form:
  

  ( 2 ^A * 5 ^B ), where Y <= A, B <= X. 

• So, count of potential divisors of 10^X which are also multiples of 10^Y is ( X – Y + 1 ) * ( X – Y + 1 ).
• Hence, required probability is (( X – Y + 1 ) * ( X – Y + 1 )) / (( X + 1 ) * ( X + 1 )). Computing the value of the expression for given values of X and Y gives us the required probability.

Below is the implementation of the above approach:

C++

// C++ program to find the probability
// of an arbitrary positive divisor of
// Xth power of 10 to be a multiple of
// Yth power of 10

#include <bits/stdc++.h>
using namespace std;
#define int long long int

// Function to calculate and print
// the required probability
void prob(int x, int y)
{
    // Count of potential divisors
    // of X-th power of 10 which are
    // also multiples of Y-th power
    // of 10
    int num = abs(x - y + 1)
              * abs(x - y + 1);

    // Count of divisors of X-th
    // power of 10
    int den = (x + 1) * (x + 1);

    // Calculate GCD
    int gcd = __gcd(num, den);

    // Print the reduced
    // fraction probability
    cout << num / gcd << "/"
         << den / gcd << endl;
}

// Driver Code
signed main()
{
    int X = 2, Y = 1;
    prob(X, Y);
    return 0;
}

Java

// Java program to find the probability
// of an arbitrary positive divisor of
// Xth power of 10 to be a multiple of
// Yth power of 10
import java.util.*;

class GFG{

// Function to calculate and print
// the required probability
static void prob(int x, int y)
{

    // Count of potential divisors
    // of X-th power of 10 which are
    // also multiples of Y-th power
    // of 10
    int num = Math.abs(x - y + 1) *
              Math.abs(x - y + 1);

    // Count of divisors of X-th
    // power of 10
    int den = (x + 1) * (x + 1);

    // Calculate GCD
    int gcd = __gcd(num, den);

    // Print the reduced
    // fraction probability
    System.out.print(num / gcd + "/" +
                     den / gcd + "\n");
}

static int __gcd(int a, int b)
{
    return b == 0 ? a : __gcd(b, a % b);
}

// Driver code
public static void main(String[] args)
{
    int X = 2, Y = 1;

    prob(X, Y);
}
}

// This code is contributed by amal kumar choubey

Python3

# Python3 program to find the probability
# of an arbitrary positive divisor of
# Xth power of 10 to be a multiple of
# Yth power of 10
from math import *

# Function to calculate and print
# the required probability
def prob(x, y):

    # Count of potential divisors
    # of X-th power of 10 which are
    # also multiples of Y-th power
    # of 10
    num = abs(x - y + 1) * abs(x - y + 1)

    # Count of divisors of X-th
    # power of 10
    den = (x + 1) * (x + 1)

    # Calculate GCD
    gcd1 = gcd(num, den)

    # Print the reduced
    # fraction probability
    print(num // gcd1, end = "")
    print("/", end = "")
    print(den // gcd1)

# Driver Code
if __name__ == '__main__':

    X = 2
    Y = 1

    prob(X, Y)

# This code is contributed by Surendra_Gangwar

C#

// C# program to find the probability
// of an arbitrary positive divisor of
// Xth power of 10 to be a multiple of
// Yth power of 10
using System;
class GFG{

// Function to calculate and print
// the required probability
static void prob(int x, int y)
{

    // Count of potential divisors
    // of X-th power of 10 which are
    // also multiples of Y-th power
    // of 10
    int num = Math.Abs(x - y + 1) *
              Math.Abs(x - y + 1);

    // Count of divisors of X-th
    // power of 10
    int den = (x + 1) * (x + 1);

    // Calculate GCD
    int gcd = __gcd(num, den);

    // Print the reduced
    // fraction probability
    Console.Write(num / gcd + "/" +
                  den / gcd + "\n");
}

static int __gcd(int a, int b)
{
    return b == 0 ? a : __gcd(b, a % b);
}

// Driver code
public static void Main(string[] args)
{
    int X = 2, Y = 1;

    prob(X, Y);
}
}

// This code is contributed by AnkitRai01

Javascript

<script>

// JavaScript program to find the probability
// of an arbitrary positive divisor of
// Xth power of 10 to be a multiple of
// Yth power of 10

// Function to calculate and print
// the required probability
function prob(x, y)
{

    // Count of potential divisors
    // of X-th power of 10 which are
    // also multiples of Y-th power
    // of 10
    var num = Math.abs(x - y + 1) *
              Math.abs(x - y + 1);

    // Count of divisors of X-th
    // power of 10
    var den = (x + 1) * (x + 1);

    // Calculate GCD
    var gcd = __gcd(num, den);

    // Print the reduced
    // fraction probability
    document.write(num / gcd + "/" +
                   den / gcd + "\n");
}

function __gcd(a, b)
{
    return b == 0 ? a : __gcd(b, a % b);
}

// Driver Code
var X = 2, Y = 1;

prob(X, Y);

// This code is contributed by Khushboogoyal499

</script>

Output:

4/9

Time Complexity: O(log(N))