GMAT Club World Cup 2022 (DAY 8): If the least common multiple of

LCM is 30, it means number will be 2,5 & 3 in any order. (30= 2*3*5)

It means we have 6 different combination of x,y & z among 2,3 & 5. This implies answer has to be a multiple of 6.

Only one option is a multiple of 6, option B. Hence B is the correct answer .

Note: if we had multiple option with 6 as a factor, we have to solve further to check each multiple.

x,y,z LCM is 30
2*3*5 ; factors are 8
possible pairs can be ( 2,3,5 ) ; ( 6,3,5) ; ( 10,3,5) ; ( 15,6,5) .. so on
best possible pairs possible would be 20
option D

If the least common multiple of positive integers, x, y and z is 30, then how many different values can |x−y|+|x−z|+|y−z||x−y|+|x−z|+|y−z| take ?

x,y,z can be (1,2,15); (1,3,10); (2,3,5)
there are 6 possible ways of assigning values to x,y,z from each set
therefore the given expression can take 18 different values
Answer:B

I think I did it the long way around by taking individual values, which obviously won't be the right way in a competitive exam such as GMAT.

So, what are the possible values of x, y and z? They're the factors of 30, i.e., 1, 2, 3, 5, 6, 10, 15 and 30

Assuming there's no repetition, I solved the whole thing.
Then I assumed that two numbers are repeated, giving me (1,1,30), (2,2,30), (3,3,30), (5,5,30), (6,6,30), (10,10,30), (15,15,30), (2,2,15), (3,3,10), (6,6,10), (6,6,15)
And then lastly, all numbers are the same, i.e, (30, 30, 30)

This gave me a total of 19 values possible for the stated equation.

Possible number combinations are -

1,2,30
1,2,15
1,3,30
1,3,10
1,5,30
1,5,6
1,6,30
1,6,15
1,6,10
1,10,30
1,10,15
2,3,30
2,5,30
2,6,30
2,10,30
2,15,30
2,3,15
2,3,10
2,3,5
3,5,30
3,6,30
3,10,30
3,15,30
3,5,10
3,5,6
5,6,10
5,10,15
6,10,15

Asked: If the least common multiple of positive integers, x, y and z is 30, then how many different values can \(|x - y| + |x - z| + |y - z|\) take ?

30 = 2*3*5
Number of factors of 30 = {1,2,3,5,6,10,15,30} = (1+2)(1+3)(1+5) : 8 factors
LCM(x,y,z) = 30

Without loss of generality, let us assume that x>=y>=z

|x - y| + |x - z| + |y - z| = (x-y) + (x-z) + (y-z) = 2x -2z = 2(x-z)
x-z = {29,28,27,25,24,20,15,14,13,12,10,9,8,7,5,4,3,2,1,0} : 20 different values
|x - y| + |x - z| + |y - z| = 2(x-z) = {58,56,54,50,48,40,30,28,26,24,20,18,16,14,10,8,6,4,2,0} : 20 different values

For illustration: -
If x=30; y=30, z=30; |x - y| + |x - z| + |y - z| = 0
If x=30; y=30, z=15; |x - y| + |x - z| + |y - z| = 30
If x=30; y=30, z=10; |x - y| + |x - z| + |y - z| = 40
If x=30; y=30, z=6; |x - y| + |x - z| + |y - z| = 48
If x=30; y=30, z=5; |x - y| + |x - z| + |y - z| = 50
If x=30; y=30, z=3; |x - y| + |x - z| + |y - z| = 54
If x=30; y=30, z=2; |x - y| + |x - z| + |y - z| = 56
If x=30; y=30, z=1; |x - y| + |x - z| + |y - z| = 58

If x=15; y=15, z=10; |x - y| + |x - z| + |y - z| = 10
If x=15; y=15, z=6; |x - y| + |x - z| + |y - z| = 18
If x=15; y=15, z=5; |x - y| + |x - z| + |y - z| = 20
If x=15; y=15, z=3; |x - y| + |x - z| + |y - z| = 24
If x=15; y=15, z=2; |x - y| + |x - z| + |y - z| = 26
If x=15; y=15, z=1; |x - y| + |x - z| + |y - z| = 28

If x=10; y=10, z=6; |x - y| + |x - z| + |y - z| = 8
If x=10; y=10, z=5; |x - y| + |x - z| + |y - z| = 10
If x=10; y=10, z=3; |x - y| + |x - z| + |y - z| = 14
If x=10; y=10, z=2; |x - y| + |x - z| + |y - z| = 16
If x=10; y=10, z=1; |x - y| + |x - z| + |y - z| = 18

If x=6; y=6, z=5; |x - y| + |x - z| + |y - z| = 2
If x=6; y=6, z=3; |x - y| + |x - z| + |y - z| = 6
If x=6; y=6, z=2; |x - y| + |x - z| + |y - z| = 8
If x=6; y=6, z=1; |x - y| + |x - z| + |y - z| = 10

If x=5; y=5, z=3; |x - y| + |x - z| + |y - z| = 4
If x=5; y=5, z=2; |x - y| + |x - z| + |y - z| = 6
If x=5; y=5, z=1; |x - y| + |x - z| + |y - z| = 8

If x=3; y=3, z=2; |x - y| + |x - z| + |y - z| = 2
If x=3; y=3, z=1; |x - y| + |x - z| + |y - z| = 4

If x=2; y=2, z=1; |x - y| + |x - z| + |y - z| = 2

|x - y| + |x - z| + |y - z| + {0,2,4,6,8,10,14,16,18,20,24,26,28,30,40,48,50,54,56,58} : 20 different values.

IMO D

LCM(x,y,z)=30=2*3*5

We can form LCM triplets by taking at least power 1 of each of the factors:
While finding the sum of difference of absolute value we have the following:
1) The sum will always be even number. Since we are adding 3 even or 2 odd and 1 even.
2) The sum of triplets bear symmetry. For example, 1,1,30 sums up to the same value as 30,30,1.
3) Arrangement of numbers in x, y, z does not bear any meaning since we are only concerned with the sum of absolute values cyclic differences of x, y and z.

So, keeping these points in mind, below are unique triplets whose LCM is 30 and that lead to unique values of the sum:
x -- y -- z ------- Sum
--------------------------
1 -- 1 -- 30 ----- 58
2 -- 2 -- 30 ----- 56
3 -- 3 -- 30 ----- 54
5 -- 5 -- 30 ----- 50
6 -- 6 -- 30 ----- 48
10 - 10 -30 ----- 40
15 - 15 - 30 ---- 30
1 -- 2 -- 15 ---- 28
2 -- 2 -- 15 ---- 26
3 -- 6 -- 15 ---- 24
5 -- 6 -- 15 ---- 20
6 -- 6 -- 15 ---- 18
2 -- 3 -- 10 ---- 16
3 -- 3 -- 10 ---- 14
10--10--15 ---- 10
6 -- 6 -- 10 ---- 8
2 -- 3 -- 5 ---- 6
5 -- 5 -- 6 ---- 2
30--30--30 ---- 0

[quote="Bunuel"]If the least common multiple of positive integers, x, y and z is 30, then how many different values can \(|x - y| + |x - z| + |y - z|\) take ?

A. 17
B. 18
C. 19
D. 20
E. 21

LCM of x,y,z =30
Thus x,y,z can take any value from the set ={1,2,3,5,6,10,15,30}
Now the possible values of x,y,z for which the sum of |x-y|+|x-z|+|y-z| would be unique are
(x,y,z)
(1,1,30) - Sum 58
(2,2,30) - Sum 56
(3,3,30) - Sum 54
(5,5,30) - Sum 50
(6,6,30) - Sum 48
(10,10,30) - Sum 40
(15,15,30) - Sum 30
(1,2,15) - Sum 28
(2,2,15) - Sum 26
(6,6,15) - Sum 18
(3,6,15) -Sum 24
(5,6,15) - Sum 20
(10,10,15) - Sum 10
(3,3,10) - Sum 14
(6,6,10) - Sum 8
(2,6,10) - Sum 16
(5,5,6) - Sum 2
(2,3,5) - Sum 6
(30,30,30) - Sum 0
Thus total number of unique sum =19
Answer C

Possible values of x, y and z are:
30, 30, 30 - 1 value (sum 0)
30, 30, 15/10/6/5/3/2/1 - 7 values (sum 30/40/48/50/54/56/58)
15, 10, 10/6/5/3/2/1 - 6 values (sum 10/18/20/24/26/28)
10, 6, 6/3/2 - 3 values (sum 8/14/16)
6, 5, 5 - 1 value (sum 2)
2, 3, 5 - 1 value (sum 6)

Total 19 different values as apparent from above, hence, answer is C.

We know that the LCM of x, y and z is 30

Therefore each of x or y or z can take the combination of one of the possible values -

\(2^0\) or \(2^1\)
\(3^0\) or \(3^1\)
\(5^0\) or \(5^1\)

Numbers that are divisible by (contains) 2 - 2,6,10,30
Numbers that are divisible by (contains) 3 - 6,6,15,30
Numbers that are divisible by (contains) 5 - 5,10,15,30

Now from the question stem, we know that order of x, y and z does not matter.

For ex: x = 2; y = 3 and z = 5, yields the same result of |x−y|+|x−z|+|y−z| as does x = 3; y = 2 and z = 5

The sum of the distances will be zero when x = y = z = 30 , this is one possibility.

The number of ways of selecting numbers such that none of x , y and z repeat = 4 * 2 = 8
The number of ways of selecting numbers such that two of x , y and z repeat = 5 * 2 = 10

Total number of ways = 10 + 8 + 1 (30-30-30) = 19

IMO C

Answer is C

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If the least common multiple of positive integers, x, y and z is 30, then how many different values can |x−y|+|x−z|+|y−z|take ?

Possible pairs of x,y,z:

[2,3,5] / [1,6,5] / [1,3,10] / [1,2,15] / [1,1,30]

No. of possible values of |x−y|+|x−z|+|y−z||x−y|+|x−z|+|y−z| = 20

IMO Option D

Imo E

Lcm (x, y, z) = 30 = 2*3*5
Better to right down different pairs:
(x, y, z)
(30, 1,1)
(30, 30 ,1)
(30, 30 ,30)
(30, 3, 1)
(30, 5, 1)
(30, 2, 1)
(30, 6, 1)
(30, 10, 1)
(15, 2, 1)
(15, 15, 2)
(15, 2, 2)
(15, 6, 1)
(15, 10, 1)
(15, 15, 6)
(10, 3, 1)
(10, 10, 3)
(10, 3, 3)
(10, 6, 1)
(6, 5, 1)
(6, 5, 5)
(6, 6, 5)

Total set of values = 21

LCM(x,y,z) = 30
30 = 2*3*5
We get 1 case here

Let x=2, y=3, z=5
\(|x - y| + |x - z| + |y - z|\) =6

We can also take
x= \(2^2\) y = 3 z= 5
x= 2, y = \(3^2\), z = 5
so on...

for x*y*z = 60, we get 3 cases
x*y*z = 90, we get 3 cases again

So total cases would be 3n+1 form ---> Ans C

Correct answer : choice D, I am guessing

If the least common multiple of positive integers, x, y and z is 30, then how many different values can |x−y|+|x−z|+|y−z| take ?

30 has 8 factors i.e. 1,2,3,5,6,10,15,30
Now we have to take any three from the above list but also ensure that the |x−y|+|x−z|+|y−z| value is unique

Now removing redundant factors as listed below
15 has 2 factors here 3, 5
10 has 2 factors here 2, 5
6 has 2 factors here 2, 3

Removing these we get 6C3 = 20

IMHO Option D

Answer: D

If the least common multiple of positive integers, x, y and z is 30, then how many different values can |x−y|+|x−z|+|y−z| take ?

Order of x,y,z doesn't matter since we need to add absolute values of difference between two numbers. i.e. x,y,z can be replaced with one another and we will get the same value.

If x, y, z are arranged in descending order, then required value
= 2(x-z)

Since LCM is 30,
possible values of x, y, z can be 1, 2, 3, 5, 6, 10, 15, 30.

For any distinct values of x and z, difference between them could be
(1,2,3,4,5,7,8,9,10,12,13,14,15,20,24,25,27,28,29) = 19 values.
When x and z are repeated i.e. same, then difference = 0 (1value).

For the above x and z, y can be selected in a manner that we would get LCM as 30.
But the required value of the expression = |x−y|+|x−z|+|y−z| = 2(x-z) ...post arrangement
will remain the same.

Thus the different values which expression can take = 19+1 =20 ways.