If the least common multiple of distinct positive integers

Hi quangviet512,

Can you pls elaborate on how you arrived at those figures?

Given LCM(a,b,c) =100

Or 100 = ak, 100 = bm, 100=cn where k, m, n are integers.

Factors of 100 = 2^2 x 5^2
# of factors = 3x3 = 9

Atleast one of the three number a,b,c has to be 100, otherwise the LCM cannot be 100. Note that other two can be equal.
Max value of any term = 100-1 = 99
Min value of any term = 100-50 = 50


Sums with two values equal (and third 100) can range from 198, 196, 192, 190, 180, 160, 150, 100, 0.

These sums cover all possible values with atleast one value of the three (a,b,c) as 100.

Therefore, Answer: A

Hey guys, tough one isn't it? Ok the answer must be 18 ieB

Folks, this is really good one.........
Since the a, b and c are distinct ,
Without loss of generality we can suppose that a>b>c
So clearly |a-b| + |b-c| + |a-c| = a-b+b-c+a-c = 2(a-c).
So friends, first of all it is clear that the answer does not depent upon b.

For example
100,4,1 the above sum will be 198.
100,5,1 the above sum will be 198.

So I think this is clear to all of u.

Since LCM of a,b,c is 100 it is clear that a,b and c must be factors of 100.
Clearly 100 has 9 factors (use the formula)
ie S={1,2,4,5,10,20,25,50,100}.

Now I am not worried about the value of b
c=1,b= {2,4,5,10,20,25,50},a=100(Note:all these results will give same sum)
c=1,b= {2,4,5,10,20,},a=50
c=1,b= {2,4,5,10,20},a=25
This is the end as far as c=1 is concerned because
If i take c=1, and a=20 there is no value for b in Swhich is in between 2&20
So taking c=1 we get 3 sums.

Now going to the next values of S i.e
taking c=2 we get 3 sums
c=2, b={4,5,10,20,25,50}, a=100
c=2, b={4,20}, a=50
c=2, b={4,10,20}, a=25

taking c=4 we get 3 sums
c=4, b={5,10,20,25,50}, a=100
c=4, b={5,10,20,25}, a=50
c=4, b={5,10,20}, a=25

taking c=5 we get 3 sums
c=5, b={10,20,25,50}, a=100
c=5, b={20}, a=50
c=5, b={20}, a=25

Similary for c=10 we get 3 more sums.
So till here totally 15 sums are possible.

Now taking
c=20, b=25, a =100
c=20, b=25,a=50
we get only 2 sums.

Finally taking
c=25, b=50, a=100 we get one more sum.

Finally 18 sums are possible.

Hey kevincan please don't say that my answer is wrong huh..............

Great work, Cicerone! I especially appreciate the time you take to provide clear explanations OA=B

That was excellent explanation....thanks a bunch dude....