Question

Find FOUR positive DISTINCT INTEGERS a b c and d such that

1) their sum is fixed and = 1000,
and ,
2) their LCM is the least amongst LCMs of all such combinations of a, b, c and d.
3) Prove that it is the lowest LCM possible.

Answer 1

Answer = 480, if a, b, c and d are to be distinct. a = 480. b = 240. c = 160. d = 120.

.
Such questions are degree level questions but not in school syllabus.
.

Let us try some combinations of a b c and d.
.

1) 1,9,90,900. LCM 900.
2) 100,200,300,400. lcm 1200.
3) 50,200,250,500. LCM 1000.
4) 20,30,50,900. lcm 900.
5) 100,150,300,450. lcm 900.
6) 50,150,200,600. lcm 600.
7) 40 times {12, 6, 4 , 3}... LCM 480.

.
This way it is not possible to prove that 480 is the least of all LCMs of all combinations of four positive integers.
.

Let 'a' be the largest of the four distinct numbers. Suppose other three numbers are factors of 'a' then their LCM will be ' a'. So then we have to find smallest 'a'.
.

Since b, c and d are factors of a, they can be written as : a/p , a/q, and a/r, for some positive integers p, q and r. p, q and r are not equal to 1 and are distinct.
.

Given a + a/p + a /q + a/r = 1000.
For 'a' to be the smallest , then a/p, a/q and a/r are to be largest possible numbers.
.

That means that p, q and r are to be the least possible numbers. So choose p= 2, q = 3, r = 4.
.

So a + a/2 + a/3 + a/4 = 1000
=> 25 * a /12 = 1000
=> a = 480.
.

NOTE:

If RHS is given as 950 or 1024 or 480 etc., it is not easy to solve even using this method. It will require extra efforts.
.

Answer 2

.
Such questions are degree level questions but not in school syllabus.
.

Let us try some combinations of a b c and d.
.

1) 1,9,90,900. LCM 900.
2) 100,200,300,400. lcm 1200.
3) 50,200,250,500. LCM 1000.
4) 20,30,50,900. lcm 900.
5) 100,150,300,450. lcm 900.
6) 50,150,200,600. lcm 600.
7) 40 times {12, 6, 4 , 3}... LCM 480.

.
This way it is not possible to prove that 480 is the least of all LCMs of all combinations of four positive integers.
.

Let 'a' be the largest of the four distinct numbers. Suppose other three numbers are factors of 'a' then their LCM will be ' a'. So then we have to find smallest 'a'.
.

Since b, c and d are factors of a, they can be written as : a/p , a/q, and a/r, for some positive integers p, q and r. p, q and r are not equal to 1 and are distinct.
.

Given a + a/p + a /q + a/r = 1000.
For 'a' to be the smallest , then a/p, a/q and a/r are to be largest possible numbers.
.

That means that p, q and r are to be the least possible numbers. So choose p= 2, q = 3, r = 4.
.

So a + a/2 + a/3 + a/4 = 1000
=> 25 * a /12 = 1000
=> a = 480.
.

NOTE:

If RHS is given as 950 or 1024 or 480 etc., it is not easy to solve even using this method. It will require extra efforts.
.