Let $a,b,c$ be three positive integers such that $$\text{lcm}(a,b) \cdot \text{lcm}(b,c) \cdot \text{lcm}(c,a) = a \cdot b \cdot c \cdot \gcd(a,b,c). $$ Given that none of $a,b,c$ is an integer multiple of any other of $a,b,c$, find the minimum possible value of $a+b+c$.

I converted it into a series of minimum/maximum equations but I can't get any farther than that. Can someone provide a hint and the solution?

  • 1
    $\begingroup$ Hint: For any two integers $a$, $b$, we have $gcd(a,b)\cdot lcm(a,b) = a\cdot b$. $\endgroup$
    – rogerl
    Jul 14 '15 at 1:33


I assume that you already get to the $$\forall i\in\{1,2,3\},\quad \max (r_a(i), r_b(i)) + \max (r_b(i), r_c(i)) + \max (r_a(i), r_c(i)) = r_a(i) + r_b(i) + r_c(i) + \min (r_a(i) , r_b(i) , r_c(i))$$ part.

By symmetry, just consider the case $r_a(i) \ge r_b(i) \ge r_c(i)$. We can conclude that

$$ r_a(i) = 2 r_c(i).$$

So the smallest triplet possible for $(r_a(i) , r_b(i) , r_c(i))$ is $(2,1,1)$. Now you can pick the 3 smallest primes to form $a,b$ and $c$.


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