# Print N lines of 4 numbers such that every pair among 4 numbers has a GCD K

• Last Updated : 09 Jun, 2022

Given N and K, the task is to print N lines where each line contains 4 numbers such that every among those 4 numbers has a GCD K and the maximum number used in N*4 should be minimized.
Note: In case of multiple outputs, print any one.
Examples:

Input: N = 1, K = 1
Output: 1 2 3 5
Every pair among 1, 2, 3 and 5 gives a GCD K and the largest number among these is 5 which the minimum possible.

Input: 2 2
Output
2 4 6 22
14 18 10 16
In the above input, the maximum number is 22, which is the minimum possible to make 2 lines of 4 numbers.

Approach: The first observation is that if we can solve the given problem for K=1, we can solve the problem with GCD K by simply multiplying the answers with K. We know that any three consecutive odd numbers have a GCD 1 always when paired, so three numbers of every line can be easily obtained. Hence the lines will look like:

```1 3 5 _
7 9 11 _
13 15 17 _
.
.
.```

An even number cannot be inserted always, because inserting 6 in third line will give GCD(6, 9) as 3. So the best number that can be inserted is a number between the first two off numbers of every line. Hence the pattern looks like:

```1 2 3 5
7 8 9 11
13 14 15 17
.
.
.```

To obtain given GCD K, one can easily multiply K to the obtained numbers. Hence for i-th line:

1. the first number will be k * (6*i+1)
2. the second number will be k * (6*i+1)
3. the third number will be k * (6*i+3)
4. the fourth number will be k * (6*i+5)

The maximum number among N*4 numbers will be k * (6*i – 1)
Below is the implementation of the above approach.

## C++

 `// C++ implementation of the``// above approach` `#include ``using` `namespace` `std;` `// Function to print N lines``void` `printLines(``int` `n, ``int` `k)``{``    ``// Iterate N times to print N lines``    ``for` `(``int` `i = 0; i < n; i++) {``        ``cout << k * (6 * i + 1) << ``" "``             ``<< k * (6 * i + 2) << ``" "``             ``<< k * (6 * i + 3) << ``" "``             ``<< k * (6 * i + 5) << endl;``    ``}``}``// Driver Code``int` `main()``{``    ``int` `n = 2, k = 2;``    ``printLines(n, k);``    ``return` `0;``}`

## Java

 `// Java implementation of the``// above approach` `import` `java.util.*;``import` `java.lang.*;``import` `java.io.*;` `class` `GFG``{``// Function to print N lines``static` `void` `printLines(``int` `n, ``int` `k)``{``    ``// Iterate N times to print N lines``    ``for` `(``int` `i = ``0``; i < n; i++) {``        ``System.out.println ( k * (``6` `* i + ``1``) + ``" "``            ``+ k * (``6` `* i + ``2``) + ``" "``            ``+ k * (``6` `* i + ``3``) + ``" "``            ``+ k * (``6` `* i + ``5``) );``    ``}``}``// Driver Code``public` `static` `void` `main(String args[])``{``    ``int` `n = ``2``, k = ``2``;``    ``printLines(n, k);``}``}`

## Python3

 `# Python implementation of the``# above approach.` `# Function to print N lines``def` `printLines(n, k) :` `    ``# Iterate N times to print N lines``    ``for` `i ``in` `range``(n) :``        ``print``( k ``*` `(``6` `*` `i ``+` `1``),``                ``k ``*` `(``6` `*` `i ``+` `2``),``               ``k ``*` `(``6` `*` `i ``+` `3``),``               ``k ``*` `(``6` `*` `i ``+` `5``))``        ` `# Driver code    ``if` `__name__ ``=``=` `"__main__"` `:` `    ``n, k ``=` `2``, ``2``    ``printLines(n, k)` `# This code is contributed by ANKITRAI1`

## PHP

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## C#

 `// C# implementation of the``// above approach``using` `System;` `class` `GFG``{``// Function to print N lines``static` `void` `printLines(``int` `n, ``int` `k)``{``    ``// Iterate N times to print N lines``    ``for` `(``int` `i = 0; i < n; i++)``    ``{``        ``Console.WriteLine ( k * (6 * i + 1) + ``" "` `+``                            ``k * (6 * i + 2) + ``" "` `+``                            ``k * (6 * i + 3) + ``" "` `+``                            ``k * (6 * i + 5) );``    ``}``}` `// Driver Code``public` `static` `void` `Main()``{``    ``int` `n = 2, k = 2;``    ``printLines(n, k);``}``}` `// This code is contributed``// by Akanksha Rai(Abby_akku)`

## Javascript

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Output:

```2 4 6 10
14 16 18 22```

Time Complexity: O(N*4), as we are using a loop to traverse N times and doing 4 operations in each traversal.

Auxiliary Space: O(1), as we are not using any extra space.

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