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# For positive integer Z, is the expression (Z + 2)(Z^2 + 4Z +

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For positive integer Z, is the expression (Z + 2)(Z^2 + 4Z + [#permalink]  10 Oct 2009, 21:25
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For positive integer Z, is the expression (Z + 2)(Z^2 + 4Z + 3) divisible by 4?

(1) Z is divisible by 8.
(2) (Z+1)/3 is an odd integer.
[Reveal] Spoiler: OA

Last edited by Bunuel on 23 Jan 2013, 03:11, edited 1 time in total.
Edited the question.
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Re: Z divisible by 4 [#permalink]  10 Oct 2009, 21:57
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yangsta8 wrote:
For positive integer Z, is the expression (Z+2) (Z^2+4Z+3) divisible by 4?

1. Z is divisible by 8.
2. (Z+1)/3 is an odd integer.

Looking for an efficient way to solve this. I did it by picking numbers, however OA looks at evens and odds.

1. If Z is divisible by 8, z = 8k

= (Z+2) (Z^2+4Z+3)
= (8k+2) [(8k)^2 + 4(8k) + 3]
= (8k+2) [64k^2 + 32k + 3]

(8k+2) [64k^2 + 32k + 3] is not divisible by 4 as (8k+2) is only divisible by 2 and [64k^2 + 32k + 3] is an odd. Suff...

2. If (Z+1)/3 = an odd integer

Z+1 = 3(odd integer)
Z = 3(odd integer) - 1
Here Z is an even but it could be twice of an odd integer or an even integer. If z is twice the odd integer, (Z+2) (Z^2+4Z+3) is divisible by 4 otherwise not. NSF..

A...
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For positive integer Z, is the expression (Z + 2)(Z^2 + 4Z + [#permalink]  10 Oct 2009, 22:31
Expert's post
yangsta8 wrote:
For positive integer Z, is the expression (Z + 2)(Z^2 + 4Z + 3) divisible by 4?
1. Z is divisible by 8.
2. (Z+1)/3 is an odd integer.

Looking for an efficient way to solve this. I did it by picking numbers, however OA looks at evens and odds.

For positive integer Z, is the expression (Z + 2)(Z^2 + 4Z + 3) divisible by 4?

$$(z + 2)(z^2 + 4z + 3)=(z+2)((z+2)^2-1)=(z+2)(z+3)(z+1)$$

Question: is $$(z+1)(z+2)(z+3)$$ divisible by 4?

(1) Z is divisible by 8 --> z=8k --> (8k+1)(8k+2)(8k+3) --> 2(8k+1)(4k+1)(8k+3) --> 3 odd integers (8k+1, 4k+1 and 8k+3)*2 --> not divisible by 4. SUFFICIENT

(2) (Z+1)/3 is an odd integer --> (z+1)/3=odd --> z=3*odd-1 --> z=3n-1 (n odd integer) --> 3n(3n+1)(3n+2) --> 3 odd integers (3, n and 3n+2) * 3n+1(even) --> 3n+1=odd+1=even, may or may not divisible by 4. NOT SUFFICIENT

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For positive integer Z, is the expression (Z + 2)(Z^2 + 4Z + [#permalink]  11 Oct 2009, 08:56
Expert's post
This problem can be solved even easier I stated earlier:

Well we know that (Z + 2)(Z^2 + 4Z + 3) can be simplified=(z+1)(z+2)(z+3)

Q: is (z+1)(z+2)(z+3) divisible by 4?

This expression is divisible by 4 only in two cases:
I. z is odd (z+1 - even and z+3 also even, thus their multiplication is divisible by 4)
II. When z+2 is divisible by 4.

so let's check statements:

(1) z=8k --> z is not odd, z+2=8k+2 is not divisible by 4 --> expression is not divisible by 4. Sufficient

(2) z=3n-1 --> z is even, but not sure whether it's divisible by 4 or not. Not sufficient

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Re: For positive integer Z, is the expression (Z + 2)(Z^2 + 4Z + [#permalink]  22 Jan 2013, 22:38
yangsta8 wrote:
For positive integer Z, is the expression (Z + 2)(Z^2 + 4Z + 3) divisible by 4?
1. Z is divisible by 8.
2. (Z+1)/3 is an odd integer.

This is my approach...

GIVEN:
(z+1)(z+2)(z+3) are consecutive integers...
We know that the multiple of 4 appears in this pattern: 1,2,3,(4),5,6,7,(8),9,10,11,(12),...
As you have noticed, after a multiple of 4, it will be followed by 3 non-multiple of 4...

1.
If z is a multiple of 4, then z+1,z+2,z+3 are all non-multiple of 4 based on the pattern above...
SUFFICIENT.
2.
Let z+1 = 3: (3)(4)(5) is divisible by 4
Let z+1 = 9: (9)(10)(11) is not divisible by 4
INSUFFICIENT.

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Re: For positive integer Z, is the expression (Z + 2)(Z^2 + 4Z + [#permalink]  29 Jul 2014, 00:06
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Re: For positive integer Z, is the expression (Z + 2)(Z^2 + 4Z +   [#permalink] 29 Jul 2014, 00:06
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# For positive integer Z, is the expression (Z + 2)(Z^2 + 4Z +

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