Minimize $a+b+c+d$ given $\text{LCM}(a,b,c,d)=1000$

Question

It is the inverse of this problem.

Minimize $a+b+c+d$ given $a, b, c, d$ are distinct positive integers, and $\operatorname{lcm}(a,b,c,d) =1000$.

Of course, it would be possible to enumerate all quadruplets with $\operatorname{lcm}(a,b,c,d)=1000$, however it might be interesting to see whether there is a more intelligent approach. Or even if the minimum is found by an exhaustive search, at least the bound can possibly be proved more directly. (As in the answer to the post which inspired this: Minimize $LCM(a,b,c,d)$ given $a+b+c+d=1000$.)

Since $1000=2^3\cdot5^3$, we will need at least one of the numbers $a$, $b$, $c$, $d$ to contain factor $2^3$. And, similarly, at least one of them has to contain $5^3$.

Answer 1

At least one of $a,b,c,d$ must be divisible by $5^3$ and at least one must be divisible by $2^3$. Any extra factors can be eliminated, reducing $a+b+c+d$. Since $1+2^3 \cdot 5^3 > 2^3 + 5^3$, the optimal solution is $1,\; 1,\; 2^3,\; 5^3$.

Answer 2

Prime factorisation of $1000$ is $2^3\cdot5^3$ so how about $$a=1,\quad b=1,\quad c=2^3,\quad d=5^3$$ which gives $a+b+c+d=135$?

Answer 3

Since $1000=2^35^3$, we know that among the numbers $a$, $b$, $c$ and $d$ at least one must have the factor of $2^3$ and another must have the factor $5^3$ (by the definition of LCM). They are either the same number or different ones.

If they are the same number, WLOG¹ let this be $a$. So $a+b+c+d\geq 2^35^3+1+1+1=1003$, which is achievable when $a=1000$ and $b=c=d=1$.

If they are different numbers, WLOG let $8\mid a$ and $125\mid b$. Then $a+b+c+d \geq 135$ which is achievable when $a=8$, $b=125$, and $c=d=1$.

Overall, the smallest possible value is $135$.

¹Without loss of generality.