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Find all triplets $(a,b,c)$ less than or equal to 50 such that $a + b +c$ be divisible by $a$ and $b$ and $c$.(i.e $a|a+b+c~~,~~b|a+b+c~~,~~c|a+b+c$) for example $(10,20,30)$ is a good triplet. ($10|60 , 20|60 , 30|60$).

Note: $a,b,cleq 50$ and $a,b,cin N$.

In other way the question says to find all $(a,b,c)$ such that $lcm(a,b,c) | a+b+c$

After writing different situations, I found that if $gcd(a,b,c) = d$ then all triplets are in form of $(d,2d,3d)$ or $(d,d,d)$ or $(d,d,2d)$ are answers. (of course the permutation of these like $(2d,3d,d)$ is also an answer). It gives me $221$ different triplets. I checked this with a simple Java program and the answer was correct but I cannot say why other forms are not valid. I can write other forms and check them one by one but I want a more intelligent solution than writing all other forms. Can anyone help?

My java code: (All of the outputs are in form of $(d,d,d)$ or $(d,2d,3d)$ or $(d,d,2d)$ and their permutations.)

import java.util.ArrayList;
import java.util.Collections;

public class Main 
public static void main(String[] args) 
 int count = 0;
 for (int i = 1; i <= 50; i++) 
 for (int j = 1; j <= 50; j++) 
 for (int k = 1; k <= 50; k++) 
 int s = i + j + k;
 if (s % i == 0 && s % j == 0 && s % k == 0 && i != j && j != k && i != k) 
 ArrayList<Integer> array = new ArrayList<Integer>();
 array.clear();
 int g = gcd(gcd(i, j), k);
 array.add(i / g);
 array.add(j / g);
 array.add(k / g);
 Collections.sort(array);
 int condition = 4; //To find out whether it is (d,d,d) or (d,d,2d) or (d,2d,3d)
 if (array.get(0) == 1 && array.get(1) == 1 && array.get(2) == 1) 
 condition = 1;
 
 if (array.get(0) == 1 && array.get(1) == 1 && array.get(2) == 2) 
 condition = 2;
 
 if (array.get(0) == 1 && array.get(1) == 2 && array.get(2) == 3) 
 condition = 3;
 
 System.out.printf("%d %d %d ::: Condition: %dn", i, j, k, condition);
 count++;
 
 
 
 
 System.out.println(count);


public static int gcd(int a, int b) 
 if (b == 0) 
 return a;
 else
 return gcd(b, a % b);

If $aleq bleq c$ then $cmid a+b+c$ implies $cmid a+b$ and so $a+b=cz$ for some $zinBbbN$. Then
$$cz=a+bleq2bleq2c,$$
and so $zleq2$. If $z=2$ then the inequalities are all equalities and so $a=b=c$. Then the triplet $(a,b,c)$ is of the form $(d,d,d)$.

If $z=1$ then $c=a+b$, and then $bmid a+b+c$ implies that $bmid 2a$. As $bgeq a$ it follows that either $b=a$ or $b=2a$. If $b=a$ then $c=2a$ and the triplet $(a,b,c)$ is of the form $(d,d,2d)$. If $b=2a$ then $c=3a$ and the triplet $(a,b,c)$ is of the form $(d,2d,3d)$.

This allows us to count the total number of triplets quite easily;

1. The number of triplets of the form $(d,d,d)$ is precisely $50$; one for each positive integer $d$ with $dleq50$.


2. The number of triplets of the form $(d,d,2d)$ is precisely $25$; one for each positive integer $d$ with $2dleq50$. Every such triplets has precisely three distinct permutations of its coordinates, yielding a total of $3times25=75$ triplets.


3. The number of triplets of the form $(d,2d,3d)$ is precisely $16$; one for each positive integer $d$ with $3dleq50$. Every such triplets has precisely six distinct permutations of its coordinates, yielding a total of $6times 16=96$ triplets.

This yields a total of $50+75+96=221$ triplets.