kevincan wrote:

If the least common multiple of distinct positive integers a, b and c is 100, and Q is the set of all possible values of |a-b| + |b-c| + |a-c|, then the number of elements in Q is

(A) less than 18 (B) 18 (C) 20 (D) 21 (E) higher than 21

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..............

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Averages Accelerated:Guide to solve Averages Quickly(with 10 practice problems)