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# Count number of pairs (A <= N, B <= N) such that gcd (A , B) is B

• Difficulty Level : Hard
• Last Updated : 27 Dec, 2022

Given a number n. We need to find the number of ordered pairs of a and b such gcd(a, b) is b itself
Examples :

```Input : n = 2
Output : 3
(1, 1) (2, 2) and (2, 1)

Input : n = 3
Output : 5
(1, 1) (2, 2) (3, 3) (2, 1) and (3, 1)```

Naive approach: gcd(a, b) = b means b is a factor of a. So the total number of pairs will be equal to sum of divisors for each a = 1 to n. Please refer find all divisors of a natural number for implementation.
Efficient approach: gcd(a, b) = b means that a is a multiple of b. So the total number of pairs will be sum of number of multiples of each b (where b varies from 1 to n) which are less than or equal to n.
For a number i, a number of multiples of i is less than or equal to floor(n/i). So what we need to do is just sum the floor(n/i) for each i = 1 to n and print it. But more optimizations can be done. floor(n/i) can have atmost 2*sqrt(n) values for i >= sqrt(n). floor(n/i) can vary from 1 to sqrt(n) and similarly for i = 1 to sqrt(n) floor(n/i) can have values from 1 to sqrt(n). So total of 2*sqrt(n) distinct values

```let floor(n/i) = k
k <= n/i < k + 1
n/k+1 < i <= n/k
floor(n/k+1) < i <= floor(n/k)
Thus for given k the largest value of i for
which the floor(n/i) = k is floor(n/k)
and all the set of i for which the
floor(n/i) = k are consecutive```

## CPP

 `// C++ implementation of counting pairs` `// such that gcd (a, b) = b` `#include ` `using` `namespace` `std;`   `// returns number of valid pairs` `int` `CountPairs(``int` `n)` `{` `    ``// initialize k` `    ``int` `k = n;`   `    ``// loop till imin <= n` `    ``int` `imin = 1;`   `    ``// Initialize result` `    ``int` `ans = 0;`   `    ``while` `(imin <= n) {`   `        ``// max i with given k floor(n/k)` `        ``int` `imax = n / k;`   `        ``// adding k*(number of i with` `        ``// floor(n/i) = k to ans` `        ``ans += k * (imax - imin + 1);`   `        ``// set imin = imax + 1 and k = n/imin` `        ``imin = imax + 1;` `        ``k = n / imin;` `    ``}`   `    ``return` `ans;` `}`   `// Driver function` `int` `main()` `{` `    ``cout << CountPairs(1) << endl;` `    ``cout << CountPairs(2) << endl;` `    ``cout << CountPairs(3) << endl;` `    ``return` `0;` `}`

## Java

 `// Java implementation of counting pairs` `// such that gcd (a, b) = b` `import` `java.io.*;` `public` `class` `GFG {` `    `  `    ``// returns number of valid pairs` `    ``static` `int` `CountPairs(``int` `n) {` `        `  `        ``// initialize k` `        ``int` `k = n;` `    `  `        ``// loop till imin <= n` `        ``int` `imin = ``1``;` `    `  `        ``// Initialize result` `        ``int` `ans = ``0``;` `    `  `        ``while` `(imin <= n) {` `    `  `            ``// max i with given k floor(n/k)` `            ``int` `imax = n / k;` `        `  `            ``// adding k*(number of i with` `            ``// floor(n/i) = k to ans` `            ``ans += k * (imax - imin + ``1``);` `        `  `            ``// set imin = imax + 1 ` `            ``// and k = n/imin` `            ``imin = imax + ``1``;` `            ``k = n / imin;` `        ``}` `    `  `        ``return` `ans;` `    ``}` `    `  `    ``// Driver code` `    ``public` `static` `void` `main(String[] args) {` `        ``System.out.println(CountPairs(``1``));` `        ``System.out.println(CountPairs(``2``));` `        ``System.out.println(CountPairs(``3``));` `    ``}` `}`   `// This code is contributed by Anant Agarwal.`

## Python3

 `# Python implementation of counting` `# pairs such that gcd (a, b) = b`   `# returns number of valid pairs` `def` `CountPairs(n):` `    `  `    ``# initialize k` `    ``k ``=` `n`   `    ``# loop till imin <= n` `    ``imin ``=` `1`   `    ``# Initialize result` `    ``ans ``=` `0`   `    ``while``(imin <``=` `n):`   `        ``# max i with given k floor(n / k)` `        ``imax ``=` `n ``/` `k`   `        ``# adding k*(number of i with` `        ``# floor(n / i) = k to ans` `        ``ans ``+``=` `k ``*` `(imax ``-` `imin ``+` `1``)`   `        ``# set imin = imax + 1 and` `        ``# k = n / imin` `        ``imin ``=` `imax ``+` `1` `        ``k ``=` `n ``/` `imin`   `    ``return` `ans` `    `  `# Driver code` `print``(CountPairs(``1``))` `print``(CountPairs(``2``))` `print``(CountPairs(``3``))`   `# This code is contributed by Anant Agarwal.`

## C#

 `// C# implementation of counting ` `// pairs such that gcd (a, b) = b` `using` `System;`   `class` `GFG {` `    `  `    ``// returns number of valid pairs` `    ``static` `int` `CountPairs(``int` `n) ` `    ``{` `        `  `        ``// initialize k` `        ``int` `k = n;` `    `  `        ``// loop till imin <= n` `        ``int` `imin = 1;` `    `  `        ``// Initialize result` `        ``int` `ans = 0;` `    `  `        ``while` `(imin <= n) {` `    `  `            ``// max i with given ` `            ``// k floor(n / k)` `            ``int` `imax = n / k;` `        `  `            ``// adding k * (number of i  ` `            ``// with floor(n / i) = k` `            ``// to ans` `            ``ans += k * (imax - imin + 1);` `        `  `            ``// set imin = imax + 1 ` `            ``// and k = n / imin` `            ``imin = imax + 1;` `            ``k = n / imin;` `        ``}` `    `  `        ``return` `ans;` `    ``}` `    `  `    ``// Driver code` `    ``public` `static` `void` `Main(String []args) ` `    ``{` `        ``Console.WriteLine(CountPairs(1));` `        ``Console.WriteLine(CountPairs(2));` `        ``Console.WriteLine(CountPairs(3));` `    ``}` `}`   `// This code is contributed by vt_m.`

## PHP

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## Javascript

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Output

```1
3
5
```

Time complexity: O(n). This is because the while loop takes O(n) time to complete since it is looping over all elements of the array.
Auxiliary space: O(1), as no extra space is used.

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