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Minimize $$LCM(a,b,c,d)$$ given $\; a, b, c, d\; $ are distinct positive integers, and $$a+b+c+d=1000$$

Any hint?


Tried the AM-GM inequality $a+b+c+d>4\sqrt[4]{abcd}=4\sqrt[4]{GL}$ but couldn't use this.

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    $\begingroup$ What have you tried? Can you, say, do it for numbers smaller than $1000$? What about $100$? What about $10$? $\endgroup$
    – lulu
    Jul 10 '18 at 18:01
  • $\begingroup$ Tried the AM-GM inequality $a+b+c+d\gt 4\sqrt[4]{abcd}=4\sqrt[4]{GL}$ but coudn't use this $\endgroup$
    – Wolfdale
    Jul 10 '18 at 18:03
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    $\begingroup$ An inverse problem here. $\endgroup$
    – peterh
    Jul 10 '18 at 18:29
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Suppose without loss of generality that $a<b<c<d$. Write $M$ for the minimum possible value of $\text{lcm}(a,b,c,d)$. Since $$120+160+240+480=1000\text{ and }\text{lcm}(120,160,240,480)=480\,,$$ we get $$M\leq 480\,.$$ Suppose that $M$ is attained when $(a,b,c,d)=(A,B,C,D)$.

Set $k:=\text{lcm}(A,B,C)$ and $t:=\gcd(k,D)$. Clearly, $t\mid k$. We have $$M=\text{lcm}(A,B,C,D)=\text{lcm}\big(\text{lcm}(A,B,C),D\big)=\frac{kD}{t}\,.$$ If $k\geq 4t$, then $M\geq 4D$. From $$1000=A+B+C+D\leq (D-3)+(D-2)+(D-1)+D=4D-6\,,$$ we have $D\geq\frac{1006}{4}$ or $D\geq 252$; i.e.,$$M\geq 4D\geq 1008\,,$$ which is a contradiction. Hence, $k\in \{t,2t,3t\}$.

If $k=3t$, then $A$, $B$, and $C$ are divisors of $3D$ that are less than $D$. Hence, $3D\geq 4C$, $3D\geq 5B$, and $3D\geq 6A$. Hence, $A+B+C+D=1000$ implies that $$1000=A+B+C+D\leq \frac{D}{2}+\frac{3D}{5}+\frac{3D}{4}+D=\frac{57}{20}D\,,$$ whence $D\geq \frac{20000}{57}$, or $D\geq 351$. However, this means $$M=3D\geq 1053\,,$$ which is absurd.

If $k=2t$, then $A$, $B$, and $C$ are divisors of $2D$ that are less than $D$. Hence, $2D\geq 3C$, $2D\geq 4B$, and $2D\geq 5A$. Ergo, $$1000=A+B+C+D\leq \frac{2D}{5}+\frac{D}{2}+\frac{2D}{3}+D=\frac{77}{30}D\,,$$ and so $D\geq \frac{30000}{77}$, or $D\geq 390$. Nonetheless, $$M=2D\geq 780$$ leads to another contradiction.

Now, we have concluded that $k=t$. Hence, $A$, $B$, and $C$ are proper divisors of $D$. This shows that $D\geq 2C$, $D\geq 3B$, and $D\geq 4A$. Therefore, $$1000=A+B+C+D\leq \frac{D}{4}+\frac{D}{3}+\frac{D}{2}+D=\frac{25}{12}D\,,$$ or $D\geq 480$. That means $$M=D\geq 480\,.$$

Consequently, $M=480$. The only possible $(a,b,c,d)\in\mathbb{Z}_{>0}^4$ satisfying the conditions $a<b<c<d$, $a+b+c+d=1000$, and $\text{lcm}(a,b,c,d)=M$ is $$(a,b,c,d)=(120,160,240,480)\,.$$


P.S. It can also be shown that the maximum value of $\text{lcm}(a,b,c,d)$ is $3905625009$. This happens iff $(a,b,c,d)=(247,249,251,253)$.

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