If x, y, and z are different positive integers, is x prime?

Question

(1) xyz = 30
(2) z < x < y

Kudos for a correct  solution .

eddyki · Answer

y=/=x=/=z but all >0 and int. (1) -> x could be 15 (z=1 and y=2) or 3 (z=5 and y=2) , its insuff. (2) -> 100<200<300 (no prime) or 1<2<3 (prime!) -> insuff. (1/2) -> we need 3 distinct factors of 30 -> 1/2/15 or 1/5/6 or 2/3/5 or 1/3/10 . every middle number is a prime, and since z is the smallest pos. int. and y the biggest pos. int., x is a prime number. C

chetan2u · Answer

ans C..
1) it tells us that x is a divisor of 30..... x,y,z can be 1,2,3,5,6,10,15
2) insufficient..tells us x is middle value
 combined it tells us that x can be only 2,3,5 or 1,3,10, or 1,2,15 or 1,5,6 or .. so sufficient

gmat6nplus1 · Answer

1) case_1 xyz=3*2*5 answer=yes; case_2 xyz=1*2*15 answer no.

Not sufficient

2) Not sufficient

1+2) x is every time going to be a prime number. Sufficient

Answer C.

DesiGmat · Answer

Here we go:

x, y, and z are different positive integers

St1: xyz = 30

30 = 2*3*5

x = 2, y = 3 and z = 5 (Is X prime? -> true)
x = 1, y = 6 and z = 5 (Is X prime? -> false)

Not sufficient

St2: z < x < y

Clearly not sufficient

Combining both

z = 2, x = 3, y = 5 (Is X prime? -> true)
z = 1, x = 2, y = 15 (Is X prime? -> true)
z = 1, x = 5, y = 6 (Is X prime? -> true)
z = 1, x = 3, y = 10 (Is X prime? -> true)

So C is the answer

santorasantu · Answer

Answer c:

from stem: x,y and z are positive integers

from 1: xyz = 30, prime factors of 30 are 1,2,3 and 5
consider xyz = 2*3*5 = 30 => x is prime, consider xyz = 6*5*1 = 30, x is not prime, 2 answers NSF
from 2: z<x<y, considering only this statement, nohing is known, so NSF

1+2: 
z<x<y, the possible multiples are (note that z, x and y are written accordingly below)
1*2*15 = 30 
1*5*6 = 30
1*3*10 = 30
2*3*5 = 30
are the only possible solutions and for all the solutions x is prime, so sufficient. 
C

Derkus · Answer

1) X*Y*Z = 30 so X*Y*Z must be two factors of 30 * 1
X = 5, Y = 6, Z = 1: Yes X is prime
X = 6, Y = 5, Z = 1: No X is not prime

2) Z<X<Y
Z = 1 < X = 2 < Y = 3: Yes X is prime
Z = 5 < X = 6 < Y = 7: No X is not prime

1+2) All possible pairs of factors of 30 are: (1,30) (2,15) (3,10) (5,6) but you can eliminate (1,30) from our choices because all numbers must be different. Thus ordered triples are:
Z = 1 < X = 2 < Y = 15
Z = 1 < X = 3 < Y = 10
Z = 1 < X = 5 < Y = 6
Z = 2 < X = 3 < Y = 5

X = 2, 3, or 5, all of which are Prime. X is Prime, so C.

Final Answer: C

Bunuel · Answer

VERITAS PREP OFFICIAL SOLUTION:

Solution: C

Let’s take statement (2) first: it gives us no values, so it’s INSUFFICIENT. Move on to statement (1). If three different positive integers have a product of 30, those integers must be one of the following sets:{ 1, 3, 10}; {2, 3, 5}; {1, 2, 15}; {1, 5, 6}. X could be prime or not prime; INSUFFICIENT. Combined, consider the four sets of possible integers. The median value is ALWAYS a prime number, so x must be prime. (C).

pacifist85 · Answer

xyz = 30 | NS because:
2*3*5 = 30, x prime
2*1*15 = 30, x prime
15*1*2 = 30, x not prime

z<x<y | NS because:
2<3<6, x prime
3<8<10, x not prime

and  , S because:
1<1<30, x prime
1<2<15, x prime
1<3<10, x prime
2<3<5, x prime

So, there is no way for x not to be a prime, since, no matter what, y must be the greatest value of the 3 values that multilied together yield 30. Similarly, trying any factor of 30 (it has 8 factors: 1,2,3,5,6,10,15,30) there is no way that the middle one (x) will not be a prime.

shmba · Answer

1+2 combined: there should be 4 sets of combinations as stated in other posts. not just 2,3,5.

bumpbot · Answer

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If x, y, and z are different positive integers, is x prime?

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