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

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For positive integer k, is the expression (k + 2)(k^2 + 4k +  [#permalink]

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Updated on: 22 Apr 2014, 02:03
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For positive integer k, is the expression (k + 2)(k^2 + 4k + 3) divisible by 4?

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

Originally posted by kishankolli on 22 Sep 2009, 02:51.
Last edited by Bunuel on 22 Apr 2014, 02:03, edited 1 time in total.
Renamed the topic, edited the question and added the OA.
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Re: For positive integer k, is the expression (k + 2)(k2 + 4k +  [#permalink]

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22 Apr 2014, 02:10
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pretzel wrote:

For second statement, if K + 1/3 or (K+1)/3, K is even. And so, in (K + 2) (K^2 + 4K + 3) = (K+2)(K+3)(K+1) = even x odd x odd

Am I correct?

No, the correct answer is A.

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

$$(k + 2)(k^2 + 4k + 3)=(k+1)(k+2)(k+3)$$, so the expression is the product of three consecutive integers.

(1) k is divisible by 8 --> $$k=8n=even$$ --> $$(k+1)(k+2)(k+3)=odd*even*odd$$. Now, $$k+2=8n+2$$, though even, is not a multiple of 4 (it's 2 greater than a multiple of 8), therefore the expression is not divisible by 4. Sufficient.

(2) (k + 1)/3 is an odd integer --> $$k+1=3*odd=odd$$ --> $$k=even$$ --> $$(k+1)(k+2)(k+3)=odd*even*odd$$. Now, $$k+2=even$$ may or may not be divisible by 8, therefore the expression may or may not be divisible by 8. For example, consider $$k=2$$ and $$k=6$$. Not sufficient.

Hope it's clear.
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Re: integer k ... DS  [#permalink]

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22 Sep 2009, 07:31
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Start off by factoring the original equation in the stem where (k^2+4k+3)=(k+3)(k+1)
So, (k+2(k^2+4k+3)=(k+1)(k+2)(k+3) where (k+1)(k+2)(k+3) are essentially consecutive integers.

As consecutive integers, (k+1)(k+2)(k+3) can be (odd)(even)(odd) in which case k has to be even. It can also be (even)(odd)(even) in which case k has to be odd. If k is odd,(k+1)(k+2)(k+3) will definitely be divisible by 4 as the product terms carry at least two 2s. This is because there are two even numbers in the product terms and they will each carry at least one 2 in their prime factorization and 4=(2)(2).

If k is even however,(k+1)(k+2)(k+3) may or may not be divisible by 4. It depends if the middle term carries two 2s, (i.e. is a multiple of 4) or not.(k+1) and (k+3) will definitely not carry a 2 in this case as they are odd numbers. For example, if k=2,(k+1)(k+2)(k+3) can be (3)(4)(5) which is divisible by 4 as the middle term, 4, is a multiple of 4. The same logic applies for k=6. If k=8 however,(k+1)(k+2)(k+3) = (9)(10)(11) which is NOT divisible by 4 as the middle term is 10=5x2 and there is only one 2 in its prime factorization.

1) All multiples of 8 are even. Hence, k is even. Possible values of (k+1)(k+2)(k+3) are:
If k=8, (k+1)(k+2)(k+3) = (9)(10)(11)
If k =16, (k+1)(k+2)(k+3) =(17)(18)(19)
If k=24,(k+1)(k+2)(k+3) = (25)(26)(27)
If k=32, (k+1)(k+2)(k+3) =(33)(34)(25)

Looking at the middle, or the only even term, of each series of consecutive integers, 10, 18 or 26, we see that in their prime factorization, they only contain one 2. Hence, the entire product of 3 consecutive integers will NOT be divisible by 4 if k is a multiple of 8.

Logically, this is also because multiples of 4 which are 8 or greater include - 8,12,16,20, 24
Multiples of 8 include 8, 16, 24.
The equation given in the stem asks for the product of 3 consecutive integers. The difference of multiples of 4 and 8 are either 0 or a multiple of 8 apart. Never 3 apart. Hence, the middle number of (k+1)(k+2)(k+3) when k is a multiple of 8 will never contain a multiple of 4.
Statement A is sufficient.

2) I assume the statement means (k+1)/3 is odd.
So, k+1=3(odd number)
k+1=odd
k=odd-1
k = even

For reasons explained in the beginning, if k is even,(k+1)(k+2)(k+3) may or may not be divisible by 4.

So, Statement B is insufficient.

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Re: integer k ... DS  [#permalink]

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22 Sep 2009, 16:03
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I agree that the correct answer is A.

Is (k + 2)(k2 + 4k + 3) divisible by 4?

$$(k + 2)(k^2 + 4k + 3)$$
$$= (k + 1)(k + 2)(k + 3)$$

Now to simplify matters, one of the terms NEEDS to be divisible by 4 in order for the whole expression to be divisible by 4. Why?

In order for an integer to be divisible by 4, its product must contain two factors of 2. This would appear to be satisfied by two even numbers. In the instance that we had two even numbers (only possible in this case if k+1 and k+3 are even, then we would already have a number divisible by 4 since those numbers are two consecutive even numbers.

Statement 1: k is divisible by 8:

If k is divisible by 8, then k = 8n, where n is a positive integer. This immediately makes k+1 and k+3 odd numbers, which are not divisible by 4. Now we test the (k+2) term:

$$k+2 = 8n + 2 = 4(2n) + 2$$

As a result, if k is divisible by 8, the expression when divided by 4 will have a remainder of 2. Therefore we can can conclude it IS NOT divisible by 4, and Statement 1 is sufficient.

Statement 2: (k + 1)/3 is an odd integer:

We can represent an odd integer as 2n + 1, where n is an integer.
Therefore,
$$(k+1)/3 = 2n + 1$$
$$k = 6n + 2$$

Subbing this into the original equation:

$$(k+1)(k+2)(k+3)$$
$$= (6n + 3)(6n + 4)(6n + 5)$$

Since 6n + 3 and 6n + 5 both represent odd numbers, if 6n + 4 is divisible by 4, the expression is divisible by 4. Since we cannot determine this (if n = 1, it is not divisible by 4, if n = 2, it is divisible by 4, etc.), Statement B is insufficient.

Therefore, the correct answer is A.
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Re: integer k ... DS  [#permalink]

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24 Sep 2009, 03:17
Yes..Its A... Its a farily simple answer...But I guess due t the timeing I got it messed it up..

My Analysis :

1>Since its divisible by 8, its divisible by 4...SO TRUE

2.K+1 to be divisible by 3 has to be an odd number, so K is even !!

The numbers that leads to K+1 to be divisible by 3 to be odd are 2,8 & 14.

All went well till I tried 8, by inserting it into the expression..I just tried inserting 2...and since I got a positive answer I did not try for 8.

So, only A is right..

Tx guys.
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For positive integer k, is the expression (k +  [#permalink]

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21 Feb 2010, 13:40
1
For positive integer k, is the expression $$(k + 2)(k^2 + 4k + 3)$$ divisible by 4?

(1) k is divisible by 8.

(2) $$\frac{k + 1}{3}$$ is an odd integer.

SOURCE: Manhattan Tests

Spoiler: :: OA
A

Please explain. A little doubtful with this OA
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Re: For positive integer k, is the expressio  [#permalink]

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21 Feb 2010, 13:55
1
1) if k is divisible by 8 (and by 4 too), then k+2 is even but isn't divisible by 4 or 8.
(k^2+4k+3) - an odd integer. ---> (even but not divisible by 4 or 8)*odd ---> the expression isn't divisible by 4.

2)k+1/3 is odd --> k+1 is odd --> k is even --> k+2 is even. if k+2:
- is divisible by 4, the expression is also divisible by 4
- isn't divisible by 4, the expression isn't divisible by 4
insufficient
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Re: For positive integer k, is the expressio  [#permalink]

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22 Feb 2010, 11:57
jeeteshsingh wrote:
For positive integer k, is the expression $$(k + 2)(k^2 + 4k + 3)$$ divisible by 4?

(1) k is divisible by 8.

(2) $$\frac{k + 1}{3}$$ is an odd integer.

SOURCE: Manhattan Tests

Spoiler: :: OA
A

Please explain. A little doubtful with this OA

Another way of looking at it ..
the expression $$(k + 2)(k^2 + 4k + 3)$$ can be written as (K+2)(K+1)(K+3)

which means they are consecutive numbers.

St1. K is divisible by 8 lets say K = 8

then the expression becomes 9*10*11 which is not divisible by 4, ( which holds for any value of K which is a multiple of 8)

Hence Sufficient

St 2 $$\frac{k + 1}{3}$$ is an odd integer.

Which K + 1 is an odd integer lets take K=2

If K =2 then the expression is divisible by 4

but not lets assume K = 8 then then again the expression is not divisible by 4

Hence Insufficient.
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Re: For positive integer k, is the expressio  [#permalink]

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12 Mar 2010, 07:30
IMO A

expression = (k+2)(k+1)(k+3)
(1): (k+1)(k+3) is odd integer, (k+2) is only divisible by 2 but not by 4 so (1) is sufficient!
(2): k=2 then expression=4*odd integer is divisible by 4
k=5 then expression=6*odd integer is not divisible by 4
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Re: For positive integer k, is the expressio  [#permalink]

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12 Mar 2010, 10:59
clearly A.

B) when k=8 8+1/3 = divisible; when k=14 14+1/3 not divisible. Insuff
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Re: For positive integer k, is the expressio  [#permalink]

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13 Jul 2010, 03:56
Another way to look at it:
i)k is divisible by 8, so k is also divisible by 4. From the property of consecutive multiple , we know that the next term that is divisible by 4 after k is (k+4) (that is every 4th number from k). That means (K+2)(K+1)(K+3) is not divisible by 4.

ii) k+2 is even, but we don't know if it is a multiple of 4 or not. Plug in 1 and 3 for k+1 yields different answer
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Re: For positive integer k, is the expression (k + 2)(k2 + 4k +  [#permalink]

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21 Apr 2014, 22:56

For second statement, if K + 1/3 or (K+1)/3, K is even. And so, in (K + 2) (K^2 + 4K + 3) = (K+2)(K+3)(K+1) = even x odd x odd

Am I correct?
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Re: For positive integer k, is the expression (k + 2)(k2 + 4k +  [#permalink]

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29 Apr 2014, 15:52
Bunuel wrote:
pretzel wrote:

For second statement, if K + 1/3 or (K+1)/3, K is even. And so, in (K + 2) (K^2 + 4K + 3) = (K+2)(K+3)(K+1) = even x odd x odd

Am I correct?

No, the correct answer is A.

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

$$(k + 2)(k^2 + 4k + 3)=(k+1)(k+2)(k+3)$$, so the expression is the product of three consecutive integers.

(1) k is divisible by 8 --> $$k=8n=even$$ --> $$(k+1)(k+2)(k+3)=odd*even*odd$$. Now, $$k+2=8n+2$$, though even, is not a multiple of 8 (it's 2 greater than a multiple of 8), therefore the expression is not divisible by 8. Sufficient.

(2) (k + 1)/3 is an odd integer --> $$k+1=3*odd=odd$$ --> $$k=even$$ --> $$(k+1)(k+2)(k+3)=odd*even*odd$$. Now, $$k+2=even$$ may or may not be divisible by 8, therefore the expression may or may not be divisible by 8. For example, consider $$k=2$$ and $$k=6$$. Not sufficient.

Hope it's clear.

I think that we are asked if the product of the three consecutive integers is divisible by 4 not by 8

Let me know
Cheers!
J
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Re: For positive integer k, is the expression (k + 2)(k2 + 4k +  [#permalink]

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30 Apr 2014, 06:38
jlgdr wrote:
Bunuel wrote:
pretzel wrote:

For second statement, if K + 1/3 or (K+1)/3, K is even. And so, in (K + 2) (K^2 + 4K + 3) = (K+2)(K+3)(K+1) = even x odd x odd

Am I correct?

No, the correct answer is A.

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

$$(k + 2)(k^2 + 4k + 3)=(k+1)(k+2)(k+3)$$, so the expression is the product of three consecutive integers.

(1) k is divisible by 8 --> $$k=8n=even$$ --> $$(k+1)(k+2)(k+3)=odd*even*odd$$. Now, $$k+2=8n+2$$, though even, is not a multiple of 8 (it's 2 greater than a multiple of 8), therefore the expression is not divisible by 8. Sufficient.

(2) (k + 1)/3 is an odd integer --> $$k+1=3*odd=odd$$ --> $$k=even$$ --> $$(k+1)(k+2)(k+3)=odd*even*odd$$. Now, $$k+2=even$$ may or may not be divisible by 8, therefore the expression may or may not be divisible by 8. For example, consider $$k=2$$ and $$k=6$$. Not sufficient.

Hope it's clear.

I think that we are asked if the product of the three consecutive integers is divisible by 4 not by 8

Let me know
Cheers!
J

It should have been 4 instead of 8...
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Re: For positive integer k, is the expression (k + 2)(k^2 + 4k +  [#permalink]

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09 May 2014, 08:31
Hi Bunuel,

In statement 2, if we use k=6 then it does not meet the criteria (k+1)/3= Odd integer. However, if we use k=2 or k=8 or k=20 etc which meet the criteria, then we are able to say that statement 2 is sufficient. Is this line of thinking correct?
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Re: For positive integer k, is the expression (k + 2)(k^2 + 4k +  [#permalink]

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29 Jul 2014, 10:20
(1): k is divisible by 8: so k is even
in the expression (k + 2) (k^2 + 4k + 3), k^2 + 4k + 3 is odd if k =8
how? k^2 = 8^2 = even, 4k = even, 3 = odd
so even + even + odd = odd and odd is never divisible by 4
it is sufficient if we test only k+2 is divisible by 4
since k is a multiple of 8, k is also a multiple of 4 (factor foundation rule)
so a multiple of 4 + 2 is not divisible by 4. Sufficient
(2) (k+1) / 3 = odd
k + 1 = 3* odd = odd => k = odd - 1 = even
if k = 2 the expression is divisible by 4
if k = 4 the expression is not divisble by 4. Not sufficient

Ans: A
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Re: For positive integer k, is the expression (k + 2)(k^2 + 4k +  [#permalink]

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01 Feb 2015, 04:14
kishankolli wrote:
For positive integer k, is the expression (k + 2)(k^2 + 4k + 3) divisible by 4?

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

simplifying the expression --> (k^3 + 6k^2 + 11k + 6)

1) k = 8p (say)
=> [(8p)^3 + 6*(8p)^2 + 88p + 6] mod 4 = 2
Not div. Hence sufficient.

2) (k+1)/3 = odd ==> k = even = 2p (say)
=> [8p^3 + 24p^2 + 22p + 6] mod 4
=> 2(p+1) mod 4
if p is even then no ; if p is odd then yes
Insufficient

Hence, A.
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Re: For positive integer k, is the expression (k + 2)(k^2 + 4k +  [#permalink]

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25 Nov 2016, 08:34
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Re: For positive integer k, is the expression (k + 2)(k^2 + 4k +  [#permalink]

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20 Dec 2016, 07:58
Simplify to: (K+1)(K+2)(K+3). This expression will not be divisible by 4 if k is a multiple of 4: factors of 4 repeat once every 4 consecutive integers, thus in between no factors of 4.
Stat1 gives you k=8q. So k=4*2q, k has 4 as a factor. Suff.
Stat2 does not tell you anything about factors of 4 in the expression. Insuff.
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Re: For positive integer k, is the expression (k + 2)(k^2 + 4k +  [#permalink]

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22 Aug 2017, 07:49
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kishankolli wrote:
For positive integer k, is the expression (k + 2)(k^2 + 4k + 3) divisible by 4?

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

Factorise the expression and we get --> (k+1)(k+2)(k+3)
If k is a multiple of 4 then (k+1)(k+2)(k+3) won't be divisible by 4, else for every non-multiple of 4 value of k (k+1)(k+2)(k+3) will be divisible by 4.

Statement A --> k is divisible by 8; that means it is divisible by 4 also. Thus (k+1)(k+2)(k+3) will never be divisible by 4. SUFFICIENT

Statement B --> (k+1)/3 = 2n+1 => k+1=6n+3 =>k=6n+2
Now, for n=1; k=8 ----> (k+1)(k+2)(k+3) won't be divisible by 4.
Now, for n=2; k=14 ---> (k+1)(k+2)(k+3) will be divisible by 4. Thus INSUFFICIENT

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Re: For positive integer k, is the expression (k + 2)(k^2 + 4k + &nbs [#permalink] 22 Aug 2017, 07:49

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