It is currently 14 Jul 2020, 11:37

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# x, y, and z are integers such that (x + 2)(y— 3)(z + 4) = 0.

Author Message
TAGS:
Founder
Joined: 18 Apr 2015
Posts: 12104
Followers: 256

Kudos [?]: 3025 [0], given: 11294

x, y, and z are integers such that (x + 2)(y— 3)(z + 4) = 0. [#permalink]  20 May 2020, 09:02
Expert's post
00:00

Question Stats:

30% (00:36) correct 69% (01:25) wrong based on 23 sessions
x, y, and z are integers such that ($$x + 2)(y - 3)(z + 4) = 0$$. Which of the following could be the value of xyz?

Indicate $$all$$ such values.

❑ -9

❑ -5

❑ 0

❑ 6

❑ 7

❑ 12

❑ 27
[Reveal] Spoiler: OA

_________________

Need Practice? 20 Free GRE Quant Tests available for free with 20 Kudos
GRE Prep Club Members of the Month: Each member of the month will get three months free access of GRE Prep Club tests.

Founder
Joined: 18 Apr 2015
Posts: 12104
Followers: 256

Kudos [?]: 3025 [0], given: 11294

Re: x, y, and z are integers such that (x + 2)(y— 3)(z + 4) = 0. [#permalink]  20 May 2020, 09:02
Expert's post
Post A Detailed Correct Solution For The Above Questions And Get A Kudos.
Question From Our New Project: GRE Quant Challenge Questions Daily - NEW EDITION!

_________________

Need Practice? 20 Free GRE Quant Tests available for free with 20 Kudos
GRE Prep Club Members of the Month: Each member of the month will get three months free access of GRE Prep Club tests.

Moderator
Joined: 16 Oct 2019
Posts: 63
Followers: 0

Kudos [?]: 51 [1] , given: 22

Re: x, y, and z are integers such that (x + 2)(y— 3)(z + 4) = 0. [#permalink]  20 May 2020, 14:01
1
KUDOS
(x + 2)(y— 3)(z + 4) = 0
to make it zero; there are three possibilities: x = -2 or y = 3 or z = -4

if x = -2, only products 6, 0 and 12 can provide y and z as integers
if y = 3, only products -9, 0 and 27 can provide x and z as inegers
if z = -4, only product 12 and 0 can provide x and y as integers

Ans: 6, 0, 12, -9, 27
Manager
Joined: 29 Apr 2020
Posts: 67
Followers: 0

Kudos [?]: 5 [0], given: 13

Re: x, y, and z are integers such that (x + 2)(y— 3)(z + 4) = 0. [#permalink]  21 May 2020, 06:21
vndnjn wrote:
(x + 2)(y— 3)(z + 4) = 0
to make it zero; there are three possibilities: x = -2 or y = 3 or z = -4

if x = -2, only products 6, 0 and 12 can provide y and z as integers
if y = 3, only products -9, 0 and 27 can provide x and z as inegers
if z = -4, only product 12 and 0 can provide x and y as integers

Ans: 6, 0, 12, -9, 27

Founder
Joined: 18 Apr 2015
Posts: 12104
Followers: 256

Kudos [?]: 3025 [0], given: 11294

Re: x, y, and z are integers such that (x + 2)(y— 3)(z + 4) = 0. [#permalink]  21 May 2020, 10:55
Expert's post
The explanation above is already very efficient and simple.

We can use another alternative approach

Xyz we need the value which is multiplication

Now

x=-2
Y=3
Z=-4

So Xyz must be a multiple of 2,3, and 4. However, 4 it itself a multiple of 2 4=2^2

So the question boils down that the values possible for Xyz are a multiple of 2,3 or both which is 6

Among the answer choices A, C,D,F, and G are multiple of these factors

Hope now is more clear
_________________

Need Practice? 20 Free GRE Quant Tests available for free with 20 Kudos
GRE Prep Club Members of the Month: Each member of the month will get three months free access of GRE Prep Club tests.

Intern
Joined: 18 May 2020
Posts: 35
Followers: 0

Kudos [?]: 26 [1] , given: 2

Re: x, y, and z are integers such that (x + 2)(y— 3)(z + 4) = 0. [#permalink]  25 May 2020, 15:37
1
KUDOS
This was a tricky one for me to see the solution, or even understand the solutions that were given!

The trick partially lies in recognizing that when determining the values of x, y, and z separately (before you even multiply them all together), you have to remember two things: 1) they're all integers and 2) at least ONE of their values has to generate a value of "0" as per the polynomial given. They don't ALL have to generate a value of 0, just one of them.

Stated differently: For the factors of x, y, and z that - when multiplied together - generate the number x*y*z, you can basically make them any integer as long as ONE of them solves the polynomial.

That means at least one of the following has to be true: x = -2, y = 3, or z = -4.

So when you're looking at the given answers "a" through "g" as a potential value of x*y*z, you know that the factors are all integers. That means that for a given answer value must have 2, 3, or 4 as a factor, otherwise the factors won't have the opportunity to multiply together to generate that answer value. (Remember: the other two-out-of-the-three values can become any integer, and you don't have to solve for them. And because the other two out of the three values can become any integer, you don't have to think about negatives and positives here because those hypothetical values have your back.)

Going through each of the answer options...
(a) -9 ---> 3 is a factor. YES, THIS IS A POSSIBLE ANSWER.
(b) -5 ---> neither 2 nor 3 nor 4 is a factor. NOT AN ANSWER.
(c) 0 ---> Take a step back on this one; you know that x or y or z could be set to 0 and you get 0 as an answer. YES, THIS IS A POSSIBLE ANSWER.
(d) 6 ---> 2 and 3 are factors. YES, THIS IS A POSSIBLE ANSWER.
(e) 7 ---> neither 2 nor 3 nor 4 is a factor. NOT AN ANSWER.
(f) 12 ---> 2 and 3 and 4 are all factors. YES, THIS IS A POSSIBLE ANSWER.
(g) 27 ---> 3 is a factor. YES THIS IS A POSSIBLE ANSWER.

A,C,D,F,G
Re: x, y, and z are integers such that (x + 2)(y— 3)(z + 4) = 0.   [#permalink] 25 May 2020, 15:37
Display posts from previous: Sort by