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# If y > 0, is (y^3 – y) divisible by 4?

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Manager
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If y > 0, is (y^3 – y) divisible by 4? [#permalink]

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26 Aug 2010, 05:31
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If y > 0, is (y^3 – y) divisible by 4?

(1) y^2 + y is divisible by 10.
(2) For a certain integer k, y = 2k + 1.
[Reveal] Spoiler: OA

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Manager
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26 Aug 2010, 05:34
zisis wrote:
If y > 0, is (y3 – y) divisible by 4?

(1) y2 + y is divisible by 10.

(2) For a certain integer k, y = 2k + 1.

my method

$$1. y(y+1)$$ is div by 10

therefore if y is div by 10 (y^3 -y) is div by 4 BUT if y+1 is div by 10 therefore y= 9, 19 etc thus y+1 is not div by 4

2. y is odd therefore is not div by 4

any better methods?

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If y > 0, is (y^3 – y) divisible by 4? [#permalink]

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26 Aug 2010, 07:32
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If y>0, is y^3-y divisible by 4?

$$y^3-y=y(y^2-1)=y(y-1)(y+1)$$

(1) y^2+y is divisible by 10 --> if $$y=10$$ then $$y(y-1)(y+1)=10*9*11$$, so not divisible by 4 (product of 2 odd and not multiple of 4 will not be divisible by 4) BUT if $$y=100$$ then $$y(y-1)(y+1)=100*99*101$$, so divisible by 4 (100 is a multiple of 4, hence product will be divisible by 4). Not sufficient.

(2) For a certain integer k, y=2k+1 --> $$y=odd$$ --> $$y-1=even$$ and $$y+1=even$$ --> $$y(y-1)(y+1)$$ is a multiple of 4 (product of 2 even numbers is a multiple of 4, OR as $$y-1$$ and $$y-1$$ are consecutive even numbers, thus one of them must be multiple of 4). Sufficient.

Pleas use "^" for powers, and formating for formulas.
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26 Aug 2010, 12:18
B for me. I believe this problem can also be solved by using prime boxes.

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26 Aug 2010, 12:30
the key point to this question is to find out if y is an odd#...i pick B

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27 Aug 2010, 10:03
I pick B and I agree with tt11234...The key here is to pick that y is odd!

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03 Mar 2014, 04:57
Bunuel wrote:
If y>0, is y^3-y divisible by 4?

$$y^3-y=y(y^2-1)=y(y-1)(y+1)$$

(1) y^2+y is divisible by 10 --> if $$y=10$$ then $$y(y-1)(y+1)=10*9*11$$, so not divisible by 4 (product of 2 odd and not multiple of 4 will not be divisible by 4) BUT if $$y=100$$ then $$y(y-1)(y+1)=100*99*101$$, so divisible by 4 (100 is a multiple of 4, hence product will be divisible by 4). Sufficient.

(2) For a certain integer k, y=2k+1 --> $$y=odd$$ --> $$y-1=even$$ and $$y+1=even$$ --> $$y(y-1)(y+1)$$ is a multiple of 4 (product of 2 even numbers is a multiple of 4, OR as $$y-1$$ and $$y-1$$ are consecutive even numbers, thus one of them must be multiple of 4). Sufficient.

Pleas use "^" for powers, and formating for formulas.

answer is definitely B , as the product of 2 consecutive even numbers is divisible by 4.
There is a small typo above , it should be (y-1) and (y+1) are 2 consecutive even numbers and not (y-1 ) and ( y-1) are 2 consecutive even numbers ,

Hope this prevents any confusion.
Thank you.
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Re: If y > 0, is (y^3 – y) divisible by 4? [#permalink]

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10 Nov 2015, 09:02
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Bunuel wrote:
If y>0, is y^3-y divisible by 4?

$$y^3-y=y(y^2-1)=y(y-1)(y+1)$$

(1) y^2+y is divisible by 10 --> if $$y=10$$ then $$y(y-1)(y+1)=10*9*11$$, so not divisible by 4 (product of 2 odd and not multiple of 4 will not be divisible by 4) BUT if $$y=100$$ then $$y(y-1)(y+1)=100*99*101$$, so divisible by 4 (100 is a multiple of 4, hence product will be divisible by 4). Not sufficient.

(2) For a certain integer k, y=2k+1 --> $$y=odd$$ --> $$y-1=even$$ and $$y+1=even$$ --> $$y(y-1)(y+1)$$ is a multiple of 4 (product of 2 even numbers is a multiple of 4, OR as $$y-1$$ and $$y-1$$ are consecutive even numbers, thus one of them must be multiple of 4). Sufficient.

Pleas use "^" for powers, and formatting for formulas.

I think one more case is to be considered, i.e when k = 0, y = 1. Even then 0/4 = 0 a integer. B will remain the answer

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If y is an integer greater than 0, is (y^3 - y) divisible by 4? [#permalink]

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15 Jan 2017, 08:20
If $$y$$ is an integer greater than 0, is ($$y^3$$ - $$y$$) divisible by 4?

(1) $$y^2$$ + $$y$$ is divisible by 10.

(2) For a certain integer $$k$$, $$y$$ = 2$$k$$ + 1
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Re: If y > 0, is (y^3 – y) divisible by 4? [#permalink]

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15 Jan 2017, 08:34
HKD1710 wrote:
If $$y$$ is an integer greater than 0, is ($$y^3$$ - $$y$$) divisible by 4?

(1) $$y^2$$ + $$y$$ is divisible by 10.

(2) For a certain integer $$k$$, $$y$$ = 2$$k$$ + 1

Merging topics. Please refer to the discussion above.
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Re: If y > 0, is (y^3 – y) divisible by 4? [#permalink]

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09 Mar 2017, 11:04
Bunuel wrote:
If y>0, is y^3-y divisible by 4?

$$y^3-y=y(y^2-1)=y(y-1)(y+1)$$

(1) y^2+y is divisible by 10 --> if $$y=10$$ then $$y(y-1)(y+1)=10*9*11$$, so not divisible by 4 (product of 2 odd and not multiple of 4 will not be divisible by 4) BUT if $$y=100$$ then $$y(y-1)(y+1)=100*99*101$$, so divisible by 4 (100 is a multiple of 4, hence product will be divisible by 4). Not sufficient.

(2) For a certain integer k, y=2k+1 --> $$y=odd$$ --> $$y-1=even$$ and $$y+1=even$$ --> $$y(y-1)(y+1)$$ is a multiple of 4 (product of 2 even numbers is a multiple of 4, OR as $$y-1$$ and $$y-1$$ are consecutive even numbers, thus one of them must be multiple of 4). Sufficient.

Pleas use "^" for powers, and formating for formulas.

Hello Bunnel or GMATCLUB members,

For Statement 2 we say since y is odd means y-1 is even and y+1 is even, so the given statement is div by 4. But consider if y = 1 since it is given y>0.
Now y-1 is 0 and y+1 = 2. Could this be a valid case ? Then the answer will be C. Request your help?
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If y > 0, is (y^3 – y) divisible by 4? [#permalink]

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09 Mar 2017, 11:07
coolkl wrote:
Bunuel wrote:
If y>0, is y^3-y divisible by 4?

$$y^3-y=y(y^2-1)=y(y-1)(y+1)$$

(1) y^2+y is divisible by 10 --> if $$y=10$$ then $$y(y-1)(y+1)=10*9*11$$, so not divisible by 4 (product of 2 odd and not multiple of 4 will not be divisible by 4) BUT if $$y=100$$ then $$y(y-1)(y+1)=100*99*101$$, so divisible by 4 (100 is a multiple of 4, hence product will be divisible by 4). Not sufficient.

(2) For a certain integer k, y=2k+1 --> $$y=odd$$ --> $$y-1=even$$ and $$y+1=even$$ --> $$y(y-1)(y+1)$$ is a multiple of 4 (product of 2 even numbers is a multiple of 4, OR as $$y-1$$ and $$y-1$$ are consecutive even numbers, thus one of them must be multiple of 4). Sufficient.

Pleas use "^" for powers, and formating for formulas.

Hello Bunnel or GMATCLUB members,

For Statement 2 we say since y is odd means y-1 is even and y+1 is even, so the given statement is div by 4. But consider if y = 1 since it is given y>0.
Now y-1 is 0 and y+1 = 2. Could this be a valid case ? Then the answer will be C. Request your help?

I got my answer, if y =1 then (y-1)y(y+1) is 0 and 0 div by 4 is still 0. This is good question.
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Re: If y > 0, is (y^3 – y) divisible by 4? [#permalink]

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30 May 2017, 06:15
I got to understand that 'y' is odd as given in statement (2) but if y=1, the expression given would become zero. So it shouldn't be considered as divisible by 4.

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Re: If y > 0, is (y^3 – y) divisible by 4? [#permalink]

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30 May 2017, 07:57
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TheMastermind wrote:
I got to understand that 'y' is odd as given in statement (2) but if y=1, the expression given would become zero. So it shouldn't be considered as divisible by 4.

You should brush up fundamentals.

ZERO:

1. 0 is an integer.

2. 0 is an even integer. An even number is an integer that is "evenly divisible" by 2, i.e., divisible by 2 without a remainder and as zero is evenly divisible by 2 then it must be even.

3. 0 is neither positive nor negative integer (the only one of this kind).

4. 0 is divisible by EVERY integer except 0 itself.

Check more here: http://gmatclub.com/forum/number-proper ... 74996.html
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Re: If y > 0, is (y^3 – y) divisible by 4?   [#permalink] 30 May 2017, 07:57
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