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x, y, and z are consecutive integers, where x < y < z. Whic
[#permalink]
12 Aug 2017, 09:59

Expert Reply

3

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Question Stats:

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)

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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)

Show: :: OA

A,D,E,F

Any explanation please, especially for the last one (F) and (D)?

Re: x, y, and z are consecutive integers, where x < y < z. Whic
[#permalink]
08 Jul 2018, 10:30

2

Expert Reply

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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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?

Re: x, y, and z are consecutive integers, where x < y < z. Whic
[#permalink]
13 Jul 2018, 22:47

1

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.

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...

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

Re: x, y, and z are consecutive integers, where x < y < z. Whic
[#permalink]
06 Sep 2019, 10:04

4

Expert Reply

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...

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

_________________

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...

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!

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?

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?

Re: x, y, and z are consecutive integers, where x < y < z. Whic
[#permalink]
16 Apr 2020, 01:39

Expert Reply

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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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]
26 Nov 2020, 14:34

2

A key fact to remember: Product of N sequential numbers is always divisible by N!

So since there are 3 consecutive numbers, the product of them must be divisible by 3!=6. By extension, the product must be divisible by 3 (remember divisibility rules).

Let's try this for each of the answer choices.

A) xyz - True, by definition above

B) (x + 1)yz - False, This is equivalent to y,y,z, which aren't 3 consecutive numbers.

C) (x + 2)yz - False - same reasoning as above

D) (x + 3)yz - True, This is the same as 3 consecutive numbers that start at y.

E) (x + 1)(y + 1)(z + 1) - True, these are 3 consecutive numbers

F) (x + 1)(y + 2)(z + 3) - True, this one is a bit more tricky. This is essentially skipping every other number, which you can factor out into the form \(Xn(n+1)(n+2)\), which is divisible by 6, and by extension, 3.

Re: x, y, and z are consecutive integers, where x < y < z. Whic
[#permalink]
24 May 2021, 07:40

hongj77 wrote:

A key fact to remember: Product of N sequential numbers is always divisible by N!

So since there are 3 consecutive numbers, the product of them must be divisible by 3!=6. By extension, the product must be divisible by 3 (remember divisibility rules).

Let's try this for each of the answer choices.

A) xyz - True, by definition above

B) (x + 1)yz - False, This is equivalent to y,y,z, which aren't 3 consecutive numbers.

C) (x + 2)yz - False - same reasoning as above

D) (x + 3)yz - True, This is the same as 3 consecutive numbers that start at y.

E) (x + 1)(y + 1)(z + 1) - True, these are 3 consecutive numbers

F) (x + 1)(y + 2)(z + 3) - True, this one is a bit more tricky. This is essentially skipping every other number, which you can factor out into the form \(Xn(n+1)(n+2)\), which is divisible by 6, and by extension, 3.

Hi, thanks for the great explanation. for F, how can you factor out consecutive odd numbers? if they are consecutive even numbers, let's say 4.6.8=2 (2.3.4), it can be factored out, but for odd numbers, let's say 3.5.7 , is it possible?

x, y, and z are consecutive integers, where x < y < z. Whic
[#permalink]
24 May 2021, 07:45

1

Hi,

The stem says \(x , y , z\) are consecutive nos. Hence they are either ODD, EVEN, ODD or EVEN, ODD, EVEN

We can't assume consecutive even or odd. Stick to the stem.

So since there are 3 consecutive numbers, the product of them must be divisible by 3!=6. By extension, the product must be divisible by 3 (remember divisibility rules).

Let's try this for each of the answer choices.

A) xyz - True, by definition above

B) (x + 1)yz - False, This is equivalent to y,y,z, which aren't 3 consecutive numbers.

C) (x + 2)yz - False - same reasoning as above

D) (x + 3)yz - True, This is the same as 3 consecutive numbers that start at y.

E) (x + 1)(y + 1)(z + 1) - True, these are 3 consecutive numbers

F) (x + 1)(y + 2)(z + 3) - True, this one is a bit more tricky. This is essentially skipping every other number, which you can factor out into the form \(Xn(n+1)(n+2)\), which is divisible by 6, and by extension, 3.

Hi, thanks for the great explanation. for F, how can you factor out consecutive odd numbers? if they are consecutive even numbers, let's say 4.6.8=2 (2.3.4), it can be factored out, but for odd numbers, let's say 3.5.7 , is it possible?

The stem says \(x , y , z\) are consecutive nos. Hence they are either ODD, EVEN, ODD or EVEN, ODD, EVEN

We can't assume consecutive even or odd. Stick to the stem.

gre29979245 wrote:

hongj77 wrote:

A key fact to remember: Product of N sequential numbers is always divisible by N!

So since there are 3 consecutive numbers, the product of them must be divisible by 3!=6. By extension, the product must be divisible by 3 (remember divisibility rules).

Let's try this for each of the answer choices.

A) xyz - True, by definition above

B) (x + 1)yz - False, This is equivalent to y,y,z, which aren't 3 consecutive numbers.

C) (x + 2)yz - False - same reasoning as above

D) (x + 3)yz - True, This is the same as 3 consecutive numbers that start at y.

E) (x + 1)(y + 1)(z + 1) - True, these are 3 consecutive numbers

F) (x + 1)(y + 2)(z + 3) - True, this one is a bit more tricky. This is essentially skipping every other number, which you can factor out into the form \(Xn(n+1)(n+2)\), which is divisible by 6, and by extension, 3.

Hi, thanks for the great explanation. for F, how can you factor out consecutive odd numbers? if they are consecutive even numbers, let's say 4.6.8=2 (2.3.4), it can be factored out, but for odd numbers, let's say 3.5.7 , is it possible?

x, y, and z are consecutive integers, where x < y < z. Whic
[#permalink]
24 May 2021, 08:14

rx10 wrote:

Hi,

The stem says \(x , y , z\) are consecutive nos. Hence they are either ODD, EVEN, ODD or EVEN, ODD, EVEN

We can't assume consecutive even or odd. Stick to the stem.

So since there are 3 consecutive numbers, the product of them must be divisible by 3!=6. By extension, the product must be divisible by 3 (remember divisibility rules).

Let's try this for each of the answer choices.

A) xyz - True, by definition above

B) (x + 1)yz - False, This is equivalent to y,y,z, which aren't 3 consecutive numbers.

C) (x + 2)yz - False - same reasoning as above

D) (x + 3)yz - True, This is the same as 3 consecutive numbers that start at y.

E) (x + 1)(y + 1)(z + 1) - True, these are 3 consecutive numbers

F) (x + 1)(y + 2)(z + 3) - True, this one is a bit more tricky. This is essentially skipping every other number, which you can factor out into the form \(Xn(n+1)(n+2)\), which is divisible by 6, and by extension, 3.

Hi, thanks for the great explanation. for F, how can you factor out consecutive odd numbers? if they are consecutive even numbers, let's say 4.6.8=2 (2.3.4), it can be factored out, but for odd numbers, let's say 3.5.7 , is it possible?

The stem says \(x , y , z\) are consecutive nos. Hence they are either ODD, EVEN, ODD or EVEN, ODD, EVEN

We can't assume consecutive even or odd. Stick to the stem.

gre29979245 wrote:

hongj77 wrote:

A key fact to remember: Product of N sequential numbers is always divisible by N!

So since there are 3 consecutive numbers, the product of them must be divisible by 3!=6. By extension, the product must be divisible by 3 (remember divisibility rules).

Let's try this for each of the answer choices.

A) xyz - True, by definition above

B) (x + 1)yz - False, This is equivalent to y,y,z, which aren't 3 consecutive numbers.

C) (x + 2)yz - False - same reasoning as above

D) (x + 3)yz - True, This is the same as 3 consecutive numbers that start at y.

E) (x + 1)(y + 1)(z + 1) - True, these are 3 consecutive numbers

F) (x + 1)(y + 2)(z + 3) - True, this one is a bit more tricky. This is essentially skipping every other number, which you can factor out into the form \(Xn(n+1)(n+2)\), which is divisible by 6, and by extension, 3.

Hi, thanks for the great explanation. for F, how can you factor out consecutive odd numbers? if they are consecutive even numbers, let's say 4.6.8=2 (2.3.4), it can be factored out, but for odd numbers, let's say 3.5.7 , is it possible?

right. We have to stick to the stem. but my question is regarding the last point in hongj77's post. so, the last option is basically consecutive even or odd numbers . that is also written in the manhattan solution. But that is not my question here. As he explained we can factor out the term (x + 1)(y + 2)(z + 3) into X * n(n+1)(n+2)(n+2), but I am not able to do it for any odd numbers (e.g. 5.7.9), though for even numbers (e.g. 4.6.8) it is doable. So for a series of odd numbers, how it can be done. That is my question.

gmatclubot

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