If the least common multiple of distinct positive integers : PS Archive
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# If the least common multiple of distinct positive integers

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If the least common multiple of distinct positive integers [#permalink]

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18 Sep 2006, 05:06
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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

Last edited by kevincan on 18 Sep 2006, 06:46, edited 2 times in total.
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18 Sep 2006, 05:42
Probably and the (E) is higher than 21

My choice is (A)... pure guess here
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18 Sep 2006, 06:36
Q is the set of all possible values of |a-b| + |b-c| + |a-c|
We can figure out set of Q=(16,18,38,46,48,98,198)
And so A is the clear answer
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18 Sep 2006, 07:06
Hi quangviet512,

Can you pls elaborate on how you arrived at those figures?
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18 Sep 2006, 08:10
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.

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Re: PS: Number of Elements in Q [#permalink]

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18 Sep 2006, 10:32
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.

_________________

Last edited by cicerone on 24 Sep 2008, 23:59, edited 1 time in total.
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18 Sep 2006, 14:48
Great work, Cicerone! I especially appreciate the time you take to provide clear explanations OA=B
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18 Sep 2006, 21:58
That was excellent explanation....thanks a bunch dude....
18 Sep 2006, 21:58
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