Generate Array of distinct elements with GCD as 1 and no Coprime pair
Last Updated :
23 Jul, 2025
Given an integer N, the task is to generate an array of N distinct integers. Such that the GCD of all the elements of the array is 1, but no pair of elements is co-prime i.e., for any pair (A[i], A[j]) of array GCD(A[i], A[j]) ≠ 1.
Note: If more than 1 answers are possible, print any of them.
Examples:
Input: N = 4
Output: 30 70 42 105
Explanation: GCD(30, 70, 42, 105) = 1.
GCD(30, 70) = 10, GCD(30, 42) = 6, GCD(30, 105) = 15, GCD(70, 42) = 14, GCD(70, 105) = 35, GCD(42, 105) = 21.
Hence, it can be seen that GCD of all elements is 1, whereas GCD of all possible pairs is greater than 1.
Input: N = 6
Output: 10 15 6 12 24 48
Approach: The problem can be solved based on the following observation:
Suppose A1, A2, A3 . . . AN are N prime numbers. Let their product be P, i.e. P = A1 * A2 * A3 . . .*AN.
Then, the sequence P/A1 , P/A2 . . . P/AN = (A2 * A3 . . . AN), (A1 * A3 . . . AN), . . ., (A1 * A2 . . . AN-1)
This sequence have at least one common factor between any pair of elements but the overall GCD is 1
This can be done following the steps given below:
- If N = 2, return -1, as no such sequence is possible for N = 2.
- Store the first N primes using the Sieve of Eratosthenes (say in a vector v).
- Calculate the product of all elements of vector "v" and store it in a variable (say "product").
- Now, iterate from i = 0 to N-1 and at each iteration store the ith element of answer as (product/v[i]).
- Return the resultant vector.
Below is the implementation of the above approach.
C++
// C++ code to implement the approach
#include <bits/stdc++.h>
using namespace std;
#define int unsigned long long
const int M = 100001;
bool primes[M + 1];
// Function to form an array such that
// pairwise gcd of all elements is not 1
// and gcd of whole sequence is 1
void printArray(int N)
{
// Such sequence is not possible for N=2
if (N == 2) {
cout << -1 << endl;
return;
}
// storing primes upto M using sieve
for (int i = 0; i < M; i++)
primes[i] = 1;
primes[0] = 0;
primes[1] = 0;
for (int i = 2; i * i <= M; i++) {
if (primes[i] == 1) {
for (int j = i * i; j <= M;
j += i) {
primes[j] = 0;
}
}
}
// Storing first N primes in vector v
vector<int> v;
for (int i = 0; i < M; i++) {
if (v.size() < N
&& primes[i] == 1) {
v.push_back(i);
}
}
// Calculating product of
// first N prime numbers
int product = 1;
vector<int> answer;
for (auto it : v) {
product *= it;
}
// Calculating answer sequence
for (int i = 0; i < N; i++) {
int num = product / v[i];
answer.push_back(num);
}
// Printing the answer
for (int i = 0; i < N; i++) {
cout << answer[i] << " ";
}
}
// Driver Code
int32_t main()
{
int N = 4;
// Function call
printArray(N);
return 0;
}
// Java code to implement the approach
import java.util.ArrayList;
class GFG {
static int M = 100001;
static int[] primes = new int[M + 1];
// Function to form an array such that
// pairwise gcd of all elements is not 1
// and gcd of whole sequence is 1
static void printArray(int N)
{
// Such sequence is not possible for N=2
if (N == 2) {
System.out.println(-1);
return;
}
// storing primes upto M using sieve
for (int i = 0; i < M; i++)
primes[i] = 1;
primes[0] = 0;
primes[1] = 0;
for (int i = 2; i * i <= M; i++) {
if (primes[i] == 1) {
for (int j = i * i; j <= M; j += i) {
primes[j] = 0;
}
}
}
// Storing first N primes in arraylist v
ArrayList<Integer> v = new ArrayList<Integer>();
for (int i = 0; i < M; i++) {
if (v.size() < N && primes[i] == 1) {
v.add(i);
}
}
// Calculating product of
// first N prime numbers
int product = 1;
ArrayList<Integer> answer
= new ArrayList<Integer>();
;
for (int i = 0; i < v.size(); i++) {
product *= v.get(i);
}
// Calculating answer sequence
for (int i = 0; i < N; i++) {
int num = product / v.get(i);
answer.add(num);
}
// Printing the answer
for (int i = 0; i < N; i++) {
System.out.print(answer.get(i) + " ");
}
}
// Driver Code
public static void main(String[] args)
{
int N = 4;
// Function call
printArray(N);
}
}
// This code is contributed by phasing.
# Python code to implement the approach
M = 100001
primes = [0]*(M + 1)
# Function to form an array such that
# pairwise gcd of all elements is not 1
# and gcd of whole sequence is 1
def printArrayint(N):
# Such sequence is not possible for N=2
if N == 2:
print(-1)
return
# storing primes upto M using sieve
for i in range(M):
primes[i] = 1
primes[0] = 0
primes[1] = 0
i = 2
while i * i <= M:
if primes[i] == 1:
j = i*i
while j <= M:
primes[j] = 0
j += i
i += 1
# Storing first N primes in vector v
v = []
for i in range(M):
if len(v) < N and primes[i] == 1:
v.append(i)
# Calculating product of
# first N prime numbers
product = 1
answer = []
for it in v:
product *= it
# Calculating answer sequence
for i in range(N):
num = product // v[i]
answer.append(num)
# Printing the answer
for i in range(N):
print(answer[i], end=" ")
# Driver Code
if __name__ == "__main__":
N = 4
printArrayint(N)
# This code is contributed by amnindersingh1414.
// C# code to implement the approach
using System;
using System.Collections;
class GFG {
static int M = 100001;
static int[] primes = new int[M + 1];
// Function to form an array such that
// pairwise gcd of all elements is not 1
// and gcd of whole sequence is 1
static void printArray(int N)
{
// Such sequence is not possible for N=2
if (N == 2) {
Console.WriteLine(-1);
return;
}
// storing primes upto M using sieve
for (int i = 0; i < M; i++)
primes[i] = 1;
primes[0] = 0;
primes[1] = 0;
for (int i = 2; i * i <= M; i++) {
if (primes[i] == 1) {
for (int j = i * i; j <= M; j += i) {
primes[j] = 0;
}
}
}
// Storing first N primes in vector v
ArrayList v = new ArrayList();
for (int i = 0; i < M; i++) {
if (v.Count < N && primes[i] == 1) {
v.Add(i);
}
}
// Calculating product of
// first N prime numbers
int product = 1;
ArrayList answer = new ArrayList();
foreach(int it in v) { product *= (int)it; }
// Calculating answer sequence
for (int i = 0; i < N; i++) {
int num = product / (int)v[i];
answer.Add(num);
}
// Printing the answer
for (int i = 0; i < N; i++) {
Console.Write(answer[i] + " ");
}
}
// Driver Code
public static void Main()
{
int N = 4;
// Function call
printArray(N);
}
}
// This code is contributed by Samim Hossain Mondal.
<script>
// JavaScript code for the above approach
let M = 100001;
let primes = new Array(M + 1);
// Function to form an array such that
// pairwise gcd of all elements is not 1
// and gcd of whole sequence is 1
function printArray(N)
{
// Such sequence is not possible for N=2
if (N == 2) {
document.write(-1 + "<br>")
return;
}
// storing primes upto M using sieve
for (let i = 0; i < M; i++)
primes[i] = 1;
primes[0] = 0;
primes[1] = 0;
for (let i = 2; i * i <= M; i++) {
if (primes[i] == 1) {
for (let j = i * i; j <= M;
j += i) {
primes[j] = 0;
}
}
}
// Storing first N primes in vector v
let v = [];
for (let i = 0; i < M; i++) {
if (v.length < N
&& primes[i] == 1) {
v.push(i);
}
}
// Calculating product of
// first N prime numbers
let product = 1;
let answer = [];
for (let it of v) {
product *= it;
}
// Calculating answer sequence
for (let i = 0; i < N; i++) {
let num = Math.floor(product / v[i]);
answer.push(num);
}
// Printing the answer
for (let i = 0; i < N; i++) {
document.write(answer[i] + " ")
}
}
// Driver Code
let N = 4;
// Function call
printArray(N);
// This code is contributed by Potta Lokesh
</script>
Time Complexity : O(M*(log(logM))) where M is a large positive number [here, 100000]
Auxiliary Space : O(M)
Better Approach: The above approach will fail for larger values of N, as the product of more than 15 primes can not be stored. This can be avoided based on the following observation:
Fix the first three distinct numbers with all possible products formed taking two among first three primes (say the numbers formed are X, Y and Z).
Then for the next elements keep on multiplying any one (say X) by one of the primes (say it becomes X') and changing X to X'.
As the GCD of X, Y and Z is 1 the overall GCD will be one but no two elements will be coprime to each other.
For own convenience use the first 3 primes 2, 3, 5 and make the first 3 numbers as 10, 15, 6, and the others by multiplying with 2 i.e. 12, 24, 48, . . .
Below is the implementation of the above approach.
C++
// C++ code for above approach
#include <bits/stdc++.h>
using namespace std;
#define int unsigned long long
// Function to form an array such that
// pairwise gcd of all elements is not 1
// and gcd of whole sequence is 1
void printArray(int N)
{
if (N == 1) {
cout << 2 << endl;
}
// Such sequence is not possible for N = 2
if (N == 2) {
cout << -1 << endl;
return;
}
int ans[N];
// Initializing initial 3 terms
// of the sequence
ans[0] = 10, ans[1] = 15, ans[2] = 6;
// Calculating next terms of the sequence
// by multiplying previous terms by 2
for (int i = 3; i < N; i++) {
ans[i] = 2LL * ans[i - 1];
}
// Printing the final sequence
for (int i = 0; i < N; i++) {
cout << ans[i] << " ";
}
}
// Driver Code
int32_t main()
{
int N = 4;
// Function call
printArray(N);
return 0;
}
// JAVA code for above approach
import java.util.*;
class GFG {
// Function to form an array such that
// pairwise gcd of all elements is not 1
// and gcd of whole sequence is 1
public static void printArray(int N)
{
if (N == 1) {
System.out.println(2);
}
// Such sequence is not possible for N = 2
if (N == 2) {
System.out.println(-1);
return;
}
int ans[] = new int[N];
// Initializing initial 3 terms
// of the sequence
ans[0] = 10;
ans[1] = 15;
ans[2] = 6;
// Calculating next terms of the sequence
// by multiplying previous terms by 2
for (int i = 3; i < N; i++) {
ans[i] = 2 * ans[i - 1];
}
// Printing the final sequence
for (int i = 0; i < N; i++) {
System.out.print(ans[i] + " ");
}
}
// Driver Code
public static void main(String[] args)
{
int N = 4;
// Function call
printArray(N);
}
}
// This code is contributed by Taranpreet
# Python code for above approach
# Function to form an array such that
# pairwise gcd of all elements is not 1
# and gcd of whole sequence is 1
def printArray(N):
if (N == 1):
print(2)
# Such sequence is not possible for N = 2
if (N == 2):
print(-1)
return
ans = [0]*N
# Initializing initial 3 terms
# of the sequence
ans[0] = 10
ans[1] = 15
ans[2] = 6
# Calculating next terms of the sequence
# by multiplying previous terms by 2
for i in range(3, N):
ans[i] = 2 * ans[i - 1]
# Printing the final sequence
for i in range(N):
print(ans[i], end=" ")
# Driver Code
if __name__ == '__main__':
N = 4
printArray(N)
# This code is contributed by amnindersingh1414.
// C# code to implement the above approach
using System;
class GFG
{
// Function to form an array such that
// pairwise gcd of all elements is not 1
// and gcd of whole sequence is 1
public static void printArray(int N)
{
if (N == 1) {
Console.Write(2);
}
// Such sequence is not possible for N = 2
if (N == 2) {
Console.Write(-1);
return;
}
int[] ans = new int[N];
// Initializing initial 3 terms
// of the sequence
ans[0] = 10;
ans[1] = 15;
ans[2] = 6;
// Calculating next terms of the sequence
// by multiplying previous terms by 2
for (int i = 3; i < N; i++) {
ans[i] = 2 * ans[i - 1];
}
// Printing the final sequence
for (int i = 0; i < N; i++) {
Console.Write(ans[i] + " ");
}
}
// Driver code
public static void Main()
{
int N = 4;
// Function call
printArray(N);
}
}
// This code is contributed by sanjoy_62.
<script>
// Javascript code for above approach
// Function to form an array such that
// pairwise gcd of all elements is not 1
// and gcd of whole sequence is 1
function printArray( N)
{
if (N == 1) {
document.write(2);
}
// Such sequence is not possible for N = 2
if (N == 2) {
document.write(-1);
return;
}
let ans = new Array(N);
// Initializing initial 3 terms
// of the sequence
ans[0] = 10;
ans[1] = 15;
ans[2] = 6;
// Calculating next terms of the sequence
// by multiplying previous terms by 2
for (let i = 3; i < N; i++) {
ans[i] = 2 * ans[i - 1];
}
// Printing the final sequence
for (let i = 0; i < N; i++) {
document.write(ans[i] + " ");
}
}
// Driver Code
let N = 4;
// Function call
printArray(N);
// This code is contributed by jana_sayantan.
</script>
Time Complexity: O(N)
Auxiliary Space: O(N)
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