@jain4444 said:How many ordered quadruples of distinct positive integers (a,b,c,d) are there such that 1/a + 1/b + 1/c + 1/d = 1 ?
Lets say that a > b > c > d, then
1/a + 1/b + 1/c + 1/d
4/d > 1
d
So, possible values of d are 2 and 3, as d can not be 1
When d = 2
1/a + 1/b + 1/c = 1/2
1/a + 1/b + 1/c
3/c > 1/2
c
c = 3
1/a + 1/b = 1/6
(a - 6)(b - 6) = 36
(b, a) = (7, 42), (8, 24), (9, 18), (10, 15)
c = 4
1/a + 1/b = 1/4
(a - 4)(b - 4) = 16
(b, a) = (5, 20), (6, 12)
c = 5
1/a + 1/b = 3/10
(3a - 10)(3b - 10) = 100
(a, b) = (20, 4), (10, 5) ....(not possible as b should be greater than 5)
When d = 3
1/a + 1/b + 1/c = 2/3
1/a + 1/b + 1/c 3/c > 2/3
c
c = 4
1/a + 1/b = 5/12
(5a - 12)(5b - 12) = 144
(a, b) = (12, 3), (6, 4) (not possible as b should be greater than 4)
So, possible solutions are (2, 3, 7, 42), (2, 3, 8, 24), (2, 3, 9, 18), (2, 3, 10, 15), (2, 4, 5, 20), (2, 4, 6, 12) and permutations
So, 6*4! = 144 ordered quadruplets