If x, y and z are different positive integers,

Question

If x, y and z are different positive integers, and y is the greatest common divisor of x, y and z, which of the following MUST be true?

i) y < x
ii) the greatest common divisor of x and z is greater than y
iii) the greatest common divisor of x and y is y

A) ii only
B) i and ii only
C) i and iii only
D) ii and iii only
E) i, ii and iii

* kudos for all correct solutions

niks18 · Answer

as y is the GCF of x , y & z. so this implies x=yk & z=yq for some integer k & q.

i) y < x ---------It is given that integer are different, hence x will be a multiple of y. Hence  Must be TRUE

ii) the greatest common divisor of x and z is greater than y-----------if x=8, y=2 & z=4, then GCF of x & z will be 4 which will be greater than y. But is x=8, y=2 & z=6, then GCF of x & z will be 2 that is equal to y. Hence  NOT ALWAYS TRUE

iii) the greatest common divisor of x and y is y--------as x is a multiple of y. so this will be  ALWAYS TRUE

Option  C

PP777 · Answer

Given: x, y, z are different integers.
 y= GCD(x,y,z)
=> x = my, for some integer m 
 z = nz, for some integer n

1) y<x  True , as x is multiple of y.
2) the greatest common divisor of x and z is greater than y   Maybe or Maybe not 
3) the greatest common divisor of x and y is y  True , as x is multiple of y

Correct Answer: C

SSM700plus · Answer

Given - y is the GCF of x , y & z. 
We can write x=y*p & z=y*q for some positive integer p & q.

i) y < x 
GCF OR GCD of any three number will always be less than or equal to the smallest number. ( I have modified rule according to the question. Don't remember the general statement).
y is GCD. Therefore, y is the smallest of x,y,z.  It is ALWAYS TRUE. 
Also x and y are different integer AND x will be a multiple of y.

ii) the greatest common divisor of x and z is greater than y
Consider x=8, y=2 & z=4, ==> GCF of x & z is 4 which is greater than y=2. It satisfies the condition.
But what if x=6, y=2 & z=8, then GCF of x & z is 2 which is equal to y. It doesn't satisfies the condition.  Hence NOT ALWAYS TRUE

iii) the greatest common divisor of x and y is y
as y <x. x is a multiple of y.  so this is ALWAYS TRUE

Ans C.

Hero8888 · Answer

I just picked numbers for x,y,z -> 5,1,7, and moved through the options step by step. Only 1 and 3 works. Answer (C).

BrentGMATPrepNow · Answer

Be careful with that strategy; it works best when the question asks "Which of the following COULD be true?"
For example, the values x = 4, y = 2 and z = 8 also satisfy the given conditions. 
However, in this instance, the greatest common divisor of x and z is greater than y, in which case all 3 statements are true.

Cheers,
Brent

vishakha23 · Answer

Hi Brent, 
What should be the standard solution then ?

Regards,
Vishakha

BrentGMATPrepNow · Answer

Good question. See my post below.

Cheers,
Brent

BrentGMATPrepNow · Answer

Let's examine each statement separately...

i) y < x 
If y is the greatest common divisor of x, y and z, we can also say that y is a divisor of x
IMPORTANT CONCEPT: The divisors of a number are always less than or equal to that number. For example, the divisors of 10 are {1, 2, 5, 10). 
So, the divisors of x must be less than or equal to x
If y is a divisor of x, then y must be less than or equal to x
Since we're told that x, y and z are  different  positive integers, we can be certain that y is less than x
So,  statement i is TRUE 
Check the answer choices....ELIMINATE A and D

ii) the greatest common divisor of x and z is greater than y 
Let's test some values of x, y and z that satisfy the given information ( x, y and z are different positive integers, and y is the greatest common divisor of x, y and z)
x = 4, y = 2 and z = 6. Notice that y (2) is the greatest common divisor of 4, 2 and 6
Here, the greatest common divisor of x (4) and z (6) is 2. 
So, the greatest common divisor of x and z is NOT greater than y 
So,  statement ii is FALSE 
Check the answer choices....ELIMINATE B and E

By the process of elimination (and without having to even look at statement iii), the correct answer must be C

Cheers,
Brent

QuabityAssurance · Answer

The question might seem intimidating at first but if you could solve it with some assumptions, it becomes a piece of cake.

Assume that the numbers are 6,2,8 as x,y,z needs to be different and y needs to be an HCF/GCD of x,y,z

Also note that it is apparent that the GCD should be the lowest value amongst all

Now if you follow the choice, you can answer easily : 
1. y<x correct
2.GCD of x and z i.e 6 and 8 should be less than y i.e 2, not true as both are same 
3.GCD of 6 and 2 is 2 which is y

Hence choice I and iii are correct!

Vaishbab · Answer

Hi Brent, what if I tried checking this with multiple sets of integers? Initially, I tried it with single digit evens and then tried it with a mix of single digit odds and evens and finally, a set of single digit odds. Didn't take too much time and I was able to pick the right option. Is this a viable strategy for this and similar questions?

If x, y and z are different positive integers,

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If x, y and z are different positive integers,

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615 to 715 in 15 Days (Ria's Strategies)

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