Count integers in a range which are divisible by their euler totient value

Given 2 integers L and R, the task is to find out the number of integers in the range [L, R] such that they are completely divisible by their Euler totient value.

Examples:

Input: L = 2, R = 3
Output: 1
$\phi$ (2) = 2 => 2 % $\phi$ (2) = 0
$\phi$ (3) = 2 => 3 % $\phi$ (3) = 1
Hence 2 satisfies the condition.

Input: L = 12, R = 21
Output: 3
Only 12, 16 and 18 satisfy the condition.

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach: We know that the euler totient function of a number is given as follows:

Rearranging the terms, we get:

If we take a close look at the RHS, we observe that only 2 and 3 are the primes that satisfy n % $\phi(n)$ = 0. This is because for primes p₁ = 2 and p₂ = 3, p₁ – 1 = 1 and p₂ – 1 = 2. Hence, only numbers of the form 2^p3^q where p >= 1 and q >= 0 need to be counted while lying in the range [L, R].

Below is the implementation of the above approach:

C++

// C++ implementation of the above approach. 
#include <bits/stdc++.h> 
  
#define ll long long 
using namespace std; 
  
// Function to return a^n 
ll power(ll a, ll n) 
{ 
    if (n == 0) 
        return 1; 
  
    ll p = power(a, n / 2); 
    p = p * p; 
  
    if (n & 1) 
        p = p * a; 
  
    return p; 
} 
  
// Function to return count of integers 
// that satisfy n % phi(n) = 0 
int countIntegers(ll l, ll r) 
{ 
  
    ll ans = 0, i = 1; 
    ll v = power(2, i); 
  
    while (v <= r) { 
  
        while (v <= r) { 
  
            if (v >= l) 
                ans++; 
            v = v * 3; 
        } 
  
        i++; 
        v = power(2, i); 
    } 
  
    if (l == 1) 
        ans++; 
  
    return ans; 
} 
  
// Driver Code 
int main() 
{ 
    ll l = 12, r = 21; 
    cout << countIntegers(l, r); 
  
    return 0; 
} 

Java

// Java implementation of the above approach. 
class GFG  
{ 
  
// Function to return a^n 
static long power(long a, long n) 
{ 
    if (n == 0) 
        return 1; 
  
    long p = power(a, n / 2); 
    p = p * p; 
  
    if (n%2== 1) 
        p = p * a; 
  
    return p; 
} 
  
// Function to return count of integers 
// that satisfy n % phi(n) = 0 
static int countIntegers(long l, long r) 
{ 
  
    long ans = 0, i = 1; 
    long v = power(2, i); 
  
    while (v <= r)  
    { 
        while (v <= r)  
        { 
  
            if (v >= l) 
                ans++; 
            v = v * 3; 
        } 
  
        i++; 
        v = power(2, i); 
    } 
  
    if (l == 1) 
        ans++; 
  
    return (int) ans; 
} 
  
// Driver Code 
public static void main(String[] args) 
{ 
    long l = 12, r = 21; 
    System.out.println(countIntegers(l, r)); 
} 
} 
  
// This code contributed by Rajput-Ji 

Python3

# Python3 implementation of the approach  
  
# Function to return a^n  
def power(a, n):  
  
    if n == 0:  
        return 1
  
    p = power(a, n // 2)  
    p = p * p  
  
    if n & 1:  
        p = p * a  
  
    return p  
  
# Function to return count of integers  
# that satisfy n % phi(n) = 0  
def countIntegers(l, r):  
  
    ans, i = 0, 1
    v = power(2, i)  
  
    while v <= r:  
  
        while v <= r:  
  
            if v >= l:  
                ans += 1
              
            v = v * 3
  
        i += 1
        v = power(2, i)  
      
    if l == 1:  
        ans += 1
  
    return ans  
  
# Driver Code  
if __name__ == "__main__": 
  
    l, r = 12, 21
    print(countIntegers(l, r))  
      
# This code is contributed  
# by Rituraj Jain 

C#

// C# implementation of the above approach. 
using System; 
  
class GFG  
{ 
  
// Function to return a^n 
static long power(long a, long n) 
{ 
    if (n == 0) 
        return 1; 
  
    long p = power(a, n / 2); 
    p = p * p; 
  
    if (n % 2 == 1) 
        p = p * a; 
  
    return p; 
} 
  
// Function to return count of integers 
// that satisfy n % phi(n) = 0 
static int countIntegers(long l, long r) 
{ 
  
    long ans = 0, i = 1; 
    long v = power(2, i); 
  
    while (v <= r)  
    { 
        while (v <= r)  
        { 
  
            if (v >= l) 
                ans++; 
            v = v * 3; 
        } 
  
        i++; 
        v = power(2, i); 
    } 
  
    if (l == 1) 
        ans++; 
  
    return (int) ans; 
} 
  
// Driver Code 
public static void Main() 
{ 
    long l = 12, r = 21; 
    Console.WriteLine(countIntegers(l, r)); 
} 
} 
  
/* This code contributed by PrinciRaj1992 */

PHP

<?php 
// PHP implementation of the above approach 
  
// Function to return a^n  
function power($a, $n)  
{  
    if ($n == 0)  
        return 1;  
  
    $p = power($a, $n / 2);  
    $p = $p * $p;  
  
    if ($n & 1)  
        $p = $p * $a;  
  
    return $p;  
}  
  
// Function to return count of integers  
// that satisfy n % phi(n) = 0  
function countIntegers($l, $r)  
{  
    $ans = 0 ; 
    $i = 1;  
    $v = power(2, $i);  
  
    while ($v <= $r) 
    {  
        while ($v <= $r)  
        {  
            if ($v >= $l)  
                $ans++;  
            $v = $v * 3;  
        }  
  
        $i++;  
        $v = power(2, $i);  
    }  
  
    if ($l == 1)  
        $ans++;  
  
    return $ans;  
}  
  
// Driver Code  
$l = 12; 
$r = 21;  
  
echo countIntegers($l, $r);  
  
// This code is contributed by Ryuga 
?> 

Output:

My Personal Notes

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

C++

Java

Python3

C#

PHP

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