# GCD and LCM Codechef Solution

## Problem

Two integers A and B are the inputs. Write a program to find GCD and LCM of A and B.

## Input

The first line contains an integer T, total number of testcases. Then follow T lines, each line contains an integer A and B.

## Output

Display the GCD and LCM of A and B separated by space respectively. The answer for each test case must be displayed in a new line.

## Constraints

• 1 <= T <= 1000
• 1 <= A, B <= 100000

## Example

Input:

```3
120 140
10213 312
10 30
```

Output:

```20 840
1 3186456
10 30```

## GCD and LCM – CodeChef Solution in Python

```T = int(input())
for i in range(T):
num = [int(x) for x in input().split(' ')]
a = min(num)
b = max(num)
multiply = a*b
while(b):
a,b = b,a%b
hcf = a
lcm = multiply/hcf
print(hcf,int(lcm))```

## GCD and LCM – CodeChef Solution in CPP

```#include <iostream>
#include <algorithm>
using namespace std;
int main() {
// your code goes here
int t=0;
cin>>t;
while(t--)
{
long long int a=0,b=0;
cin>>a>>b;
cout<<__gcd(a,b)<<" "<<((a*b)/__gcd(a,b))<<endl;
}
return 0;
}```

## GCD and LCM – CodeChef Solution in JAVA

```import java.util.*;
public class Main {
public static void main(String[] args) {
Scanner input =new Scanner(System.in);
int t=input.nextInt();
long a,b,Gcd,Lcm;
while(t-->0) {
a=input.nextInt();
b=input.nextInt();
Gcd=getGcd(a,b);
Lcm=(a*b)/Gcd;
System.out.println(Gcd+" "+Lcm);
}
}
public static long getGcd(long a, long b) {
while(a != b) {
if(a>b) a=a-b;
else b=b-a;
}
return a;
}
}```

Disclaimer: The above Problem (GCD and LCM) is generated by CodeChef but the solution is provided by Chase2learn.This tutorial is only for Educational and Learning purpose.