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.

  • 2
    $\begingroup$ 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
  • 2
    $\begingroup$ An inverse problem here. $\endgroup$
    – peterh
    Jul 10 '18 at 18:29

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 $(a,b,c,d)=(247,249,251,253)$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.