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# x, y, and z are consecutive integers, where x < y < z. Whic

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Founder
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x, y, and z are consecutive integers, where x < y < z. Whic [#permalink]  12 Aug 2017, 09:59
Expert's post
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Question Stats:

56% (01:23) correct 43% (00:59) wrong based on 93 sessions

x, y, and z are consecutive integers, where x < y < z. Which of the following must be divisible by 3 ?

Indicate all that apply.

A) xyz

B) (x + 1)yz

C) (x + 2)yz

D) (x + 3)yz

E) (x + 1)(y + 1)(z + 1)

F) (x + 1)(y + 2)(z + 3)
[Reveal] Spoiler: OA

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Kudos [?]: 9 [0], given: 100

Re: x, y, and z are consecutive integers, where x < y < z. Whic [#permalink]  08 Jul 2018, 06:36
Carcass wrote:

x, y, and z are consecutive integers, where x < y < z. Which of the following must be divisible by 3 ?

Indicate all that apply.

❑ xyz

❑ (x + 1)yz

❑ (x + 2)yz

❑ (x + 3)yz

❑ (x + 1)(y + 1)(z + 1)

❑ (x + 1)(y + 2)(z + 3)

[Reveal] Spoiler: OA
A,D,E,F

Any explanation please, especially for the last one (F) and (D)?
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Re: x, y, and z are consecutive integers, where x < y < z. Whic [#permalink]  08 Jul 2018, 10:30
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Expert's post
The stem is pretty straight: 3 consecutive numbers and integers.

Pick 1,2,3

D) $$(x + 3)yz$$ $$= 4*6=24$$ divisible by 3

F) $$(x + 1)(y + 2)(z + 3) = 2*4*6= 48$$ divisible by 3

Each number above has a number divisible by 3 inside. So they must be divisible by 3

Hope this helps

PS: more often than not is useful picking number strategy instead to think theoretically, especially when you are not at that level.

Regards
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Re: x, y, and z are consecutive integers, where x < y < z. Whic [#permalink]  13 Jul 2018, 17:32
Why not B & C?
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Re: x, y, and z are consecutive integers, where x < y < z. Whic [#permalink]  13 Jul 2018, 22:47
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Akash03jain wrote:
Why not B & C?

Thanks Carcass. Back to your question, the reason simply because 3,4,5 in B are 4 * 4 * 5 does not have 3 prime factor, so it's not divisible. Same with C.
Manager
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Re: x, y, and z are consecutive integers, where x < y < z. Whic [#permalink]  06 Sep 2019, 08:00
Why are we not considering the possibility that the integers are negative?

what if we pick x=-3, y=-2, z=-1? Then only A is satisfied...
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Re: x, y, and z are consecutive integers, where x < y < z. Whic [#permalink]  06 Sep 2019, 10:04
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Expert's post
bellavarghese wrote:
Why are we not considering the possibility that the integers are negative?

what if we pick x=-3, y=-2, z=-1? Then only A is satisfied...

If x=-3, y=-2, z=-1, then A, D, E and F are divisible by 3.

KEY CONCEPT: O is divisible by 3 (but I've never seen an OFFICIAL GRE question that tests this)

Cheers,
Brent
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Manager
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Kudos [?]: 28 [0], given: 62

Re: x, y, and z are consecutive integers, where x < y < z. Whic [#permalink]  06 Sep 2019, 10:54
GreenlightTestPrep wrote:
bellavarghese wrote:
Why are we not considering the possibility that the integers are negative?

what if we pick x=-3, y=-2, z=-1? Then only A is satisfied...

If x=-3, y=-2, z=-1, then A, D, E and F are divisible by 3.

KEY CONCEPT: O is divisible by 3 (but I've never seen an OFFICIAL GRE question that tests this)

Cheers,
Brent

oh. That's new info for me. Thanks!
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Re: x, y, and z are consecutive integers, where x < y < z. Whic [#permalink]  16 Apr 2020, 00:24
If we select 1,2 and 3 for x,y and z respectively, B and C can eval to true

B) (x + 1)yz = y * y * z ( since z is 3, any multiple of 3 is divisible by 3 )

C) (x + 2)yz = z * y * z ( following the same logic )

Can anyone explain why this is not possible?
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Re: x, y, and z are consecutive integers, where x < y < z. Whic [#permalink]  16 Apr 2020, 01:39
Expert's post
Now, if either y or z is a multiple of 3, then the expressions in choices B and C will also be divisible by 3, but you do not know for certain which of x, y, and z is the multiple of

B and C are not MUST be true
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Re: x, y, and z are consecutive integers, where x < y < z. Whic   [#permalink] 16 Apr 2020, 01:39
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