If x, y, and z are positive integers, what is the value of x?

Question

GMAT Club Tests' Fresh Question:

(1) xyz = z!

(2) x! + y! = 3

Bunuel · Accepted Answer

Official Solution:

If  x ,  y , and  z  are positive integers, what is the value of  x ? 
 
(1)  xyz = z! 
 
If  z=1 , then  xy=1  and thus  x=y=1 ;
 
If  z=3 , then  xy=2  and in this case  x=1  and  y=2  or  x=2  and  y=1 .
 
Not sufficient. 
 
(2)  x! + y! = 3 
 
This is only possible if  x=1  and  y=2  or  x=2  and  y=1 .
 
Not sufficient. 
 
(1)+(2) It's still possible to have two different cases:  x=1 ,  y=2  and  z=2  or  x=2 ,  y=1  and  z=2 . Not sufficient.

Answer: E

DavidTutorexamPAL · Answer

As all we're given are equations, we'll work with them.
This is a Precise approach.

(1) xyz = z! can be rewritten as xyz = z(z -1)(z - 2)….. which simplifies to xy = (z - 1)(z - 2)…. we cannot solve for x so this is insufficient.
Insufficient.

(2) Let's look at a few examples. if x = 1 then x! = 1 so y! = 3 - 1 = 2 and y = 2. If x = 2 then x! = 2 so y! = 1 and y = 1. If x = 3 then x! = 6 and y! = 3 - 6 which is negative and therefore impossible. That is, we have two solutions: x = 1 and y = 2 or x = 2 and y = 1.
Insufficient.

Combined.
(1) lets us solve for z (as xyz = z! --> 2z = z! --> z = 3) but does not help us separate between x = 1, y =2 and y = 2, x = 1.
That is, we still have two possible values for x.
Insufficient.

(E) is our answer.

RB15 · Answer

(1) xyz = z!
=> xy = (z-1)! 
value of x cannot be determined using this information, therefore not sufficient

(2) x! + y! = 3
only possible solution is 1! + 2!
But x can take any value i.e. 1 or 2
Therefore not sufficient

Both statements together does not solve our problem of statement 2, i.e. x can take 2 values

Answer: E

sumitkrocks · Answer

1)

XYZ=z! =1.2.3.4.5.......(z-1)Z

Not sufficient..can not infer much ...

2)X!+y!=3

two possibilities either X=1,Y=2 or X=2,Y=1

Not sufficient

1 & 2..

XYZ=1.2.....(Z-1)Z
i.e 1.2.Z=1.2......(z-1)Z
Z=3

we can't tell X is 1 or 2 ..hence answer is E

MathRevolution · Answer

Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.

Since we have 3 variables (x, y and z) and 0 equations, E is most likely to be the answer. So, we should consider conditions 1) & 2) together first. After comparing the number of variables and the number of equations, we can save time by considering conditions 1) & 2) together first.

Conditions 1) & 2)
Since x! + y! = 3, we have x = 1, y = 2 or x = 2, y = 1.
If x = 1, y = 2 or x = 2, y = 1, we have z = 3 from xyz = z! or 2z = z!.
Then we have x = 1, y = 2, z = 3 or x = 2, y = 1, z = 3.
Thus we don't have a unique solution for the variable x.
Both conditions together are not sufficient.

Therefore, E is the answer.

In cases where 3 or more additional equations are required, such as for original conditions with “3 variables”, or “4 variables and 1 equation”, or “5 variables and 2 equations”, conditions 1) and 2) usually supply only one additional equation. Therefore, there is an 80% chance that E is the answer, a 15% chance that C is the answer, and a 5% chance that the answer is A, B or D. Since E (i.e. conditions 1) & 2) are NOT sufficient, when taken together) is most likely to be the answer, it is generally most efficient to begin by checking the sufficiency of conditions 1) and 2), when taken together. Obviously, there may be occasions on which the answer is A, B, C or D.

Bunuel · Answer

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If x, y, and z are positive integers, what is the value of x?

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