Problem
Problem Link - UVa 11388 - GCD LCM
Given two positive integers $(G, L)$, we have to find a pair of integers $(a,b)$ such that $gcd(a, b)=G$ and $\text{lcm}(a, b)=L$. If there are multiple such pairs, we have to find the pair where $a$ is minimum. Also, both $a$ and $b$ needs to be positive.
Solution
We don't know the value of $(a,b)$ yet. Here is how I approached the problem.
Value of $a$
What we know that $gcd(a,b) = G$. So, $G$ divides $a$ and $b$. Therefore, $a$ is a multiple of $G$. We also need to make sure that $a$ is as small as possible. So, what is the smallest positive number that can be divided by $G$? The answer is $G$ itself.
$$\therefore a = G$$
Existence of Solution
Next, we know that $lcm(a,b)$ is the smallest positive number which is divisible by both $a$ and $b$. Since $a=G$ and $a \: | \: lcm(a,b)$, it follows that $a = G$ should divide $lcm(a,b) = L$. If $L$ is not divisible by $G$, then no solution exists.
$$\therefore \text{if } (G \not| L), \text{no solution exists}$$
Value of $b$
The value of $b$ can be derived in the following way. We know that:
$gcd(a,b) \times lcm(a,b) = a \times b$
$G \times L = G \times b$
$b = \frac{G\times L}{G}$
$\therefore b = L$.
Summary
- $a = G$
- If $G \not| L$, no solution exists
- $b = L$
- $\therefore (a,b) = (G,L)$
Code
Here is the code in C++
#include <bits/stdc++.h>
int main () {
int kase;
scanf ( "%d", &kase ); //Input number of case
while ( kase-- ) {
int G, L;
scanf ( "%d %d", &G, &L ); //Take input
int a, b; //Need to find their value
a = G;
if ( L % G != 0 ) {
printf ( "-1\n" ); //No Solution
continue;
}
b = L;
printf ( "%d %d\n", a, b );
}
return 0;
}
Labels: Analysis, GCD, LCM, Number Theory, UVa