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Joined: 16 Jan 2011
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If k is a positive integer, which of the following must be  [#permalink]

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Question Stats: 45% (02:33) correct 55% (02:20) wrong based on 286 sessions

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If k is a positive integer, which of the following must be divisible by 24?

(A) (k – 4)(k)(k + 3)(k + 7)
(B) (k – 4)(k – 2)(k + 3)(k + 5)
(C) (k – 2)(k + 3)(k + 5)(k + 6)
(D) (k + 1)(k + 3)(k + 5)(k + 7)
(E) (k – 3)(k + 1)(k + 4)(k + 6)
Math Expert V
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If k is a positive integer, which of the following must be divisible by 24?

(A) (k – 4)(k)(k + 3)(k + 7)
(B) (k – 4)(k – 2)(k + 3)(k + 5)
(C) (k – 2)(k + 3)(k + 5)(k + 6)
(D) (k + 1)(k + 3)(k + 5)(k + 7)
(E) (k – 3)(k + 1)(k + 4)(k + 6)

24=8*3.

Note that the product of two consecutive even integers is always divisible by 8 (since one of them is divisible by 4 and another by 2). Only option B offers two consecutive even numbers for any integer value of k: k-4 and k-2, if k=even or k+3 and k+5 if k=odd.

Also from the following 3 consecutive integers: (k-4), (k-3), (k-2) one must be divisible by 3, if it's not k-4 or k-2 then it must be k-3 (if it's k-4 or k-2 option B is divisible by 3 right away). But if it's k-3 then (k-3)+6=k+3 must also be divisible by 3.

So, option B: (k – 4)(k – 2)(k + 3)(k + 5) is divisible by 8 and 3 in any case.

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Re: If k is a positive integer, which of the following must be  [#permalink]

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Galiya wrote:
If k is a positive integer, which of the following must be divisible by 24?

(A) (k – 4)(k)(k + 3)(k + 7)
(B) (k – 4)(k – 2)(k + 3)(k + 5)
(C) (k – 2)(k + 3)(k + 5)(k + 6)
(D) (k + 1)(k + 3)(k + 5)(k + 7)
(E) (k – 3)(k + 1)(k + 4)(k + 6)

Ideally, you should use the logical approach suggested by Bunuel above. If it doesn't come to mind during the test, you can also test the options for values of k.
Try putting k = 5, 6, 7 and 8 (0 is divisible by all numbers so k = 2, 3 and 4 doesn't make sense. Also, divisibility of a negative number does not have physical meaning)

The options will keep falling out one after the other and you will be left will only the answer.
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Re: If k is a positive integer, which of the following must be  [#permalink]

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Bunuel, Karishma
thank you so much for help!
Manager  Joined: 12 Oct 2011
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GMAT 1: 700 Q48 V37 GMAT 2: 720 Q48 V40 Re: If k is a positive integer, which of the following must be  [#permalink]

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I have a question. 0 is a multiple of every number, and thus 0 would also be divisible by 24, correct? So if k=3, E would also be divisble by 24, wouldn't it?
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BN1989 wrote:
I have a question. 0 is a multiple of every number, and thus 0 would also be divisible by 24, correct? So if k=3, E would also be divisble by 24, wouldn't it?

Zero is a multiple of every integer, except zero itself - true.

Next, the question asks "which of the following MUST be divisible by 24" not COULD be divisible. Now, every option COULD be divisible by 24 (for example A if k=4, C if k=2, D if k=-1, and E for k=3. Notice that these are not the only values of k for which these options are divisible by 24). But the only option which is divisible by 24 for ANY integer value of k is B.

Hope it's clear.
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Re: If k is a positive integer, which of the following must be  [#permalink]

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Bumping for review and further discussion*. Get a kudos point for an alternative solution!

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Re: If k is a positive integer, which of the following must be  [#permalink]

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Galiya wrote:
If k is a positive integer, which of the following must be divisible by 24?

(A) (k – 4)(k)(k + 3)(k + 7)
(B) (k – 4)(k – 2)(k + 3)(k + 5)
(C) (k – 2)(k + 3)(k + 5)(k + 6)
(D) (k + 1)(k + 3)(k + 5)(k + 7)
(E) (k – 3)(k + 1)(k + 4)(k + 6)

Yes agree, one needs to add 6 to both 'k-4' and 'k-2' and we will end up with

(k+2)(k+3)(k+4)(k+5) four consecutive integers we have 2 even, 2 odd and of course they are all divisible by 3 (Actually by 4!)

Hope it helps
Cheers
J
Intern  Joined: 05 May 2016
Posts: 35
If k is a positive integer, which of the following must be  [#permalink]

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jlgdr wrote:
If k is a positive integer, which of the following must be divisible by 24?

(A) (k – 4)(k)(k + 3)(k + 7)
(B) (k – 4)(k – 2)(k + 3)(k + 5)
(C) (k – 2)(k + 3)(k + 5)(k + 6)
(D) (k + 1)(k + 3)(k + 5)(k + 7)
(E) (k – 3)(k + 1)(k + 4)(k + 6)

Yes agree, one needs to add 6 to both 'k-4' and 'k-2' and we will end up with

(k+2)(k+3)(k+4)(k+5) four consecutive integers we have 2 even, 2 odd and of course they are all divisible by 3 (Actually by 4!)

Hope it helps
Cheers
J

Quote:
Hi Bunuel,
I am not sure if "jlgdr" still responds to this post.
Can you please elaborate on any reason why "6" is considered to be added to both "k-4" and "k-2" ?

Regards,
Yosita
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Joined: 10 Jun 2016
Posts: 45
Schools: IIM-A"19
Re: If k is a positive integer, which of the following must be  [#permalink]

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Bunuel wrote:
If k is a positive integer, which of the following must be divisible by 24?

(A) (k – 4)(k)(k + 3)(k + 7)
(B) (k – 4)(k – 2)(k + 3)(k + 5)
(C) (k – 2)(k + 3)(k + 5)(k + 6)
(D) (k + 1)(k + 3)(k + 5)(k + 7)
(E) (k – 3)(k + 1)(k + 4)(k + 6)

24=8*3.

Note that the product of two consecutive even integers is always divisible by 8 (since one of them is divisible by 4 and another by 2). Only option B offers two consecutive even numbers for any integer value of k: k-4 and k-2, if k=even or k+3 and k+5 if k=odd.

Also from the following 3 consecutive integers: (k-4), (k-3), (k-2) one must be divisible by 3, if it's not k-4 or k-2 then it must be k-3 (if it's k-4 or k-2 option B is divisible by 3 right away). But if it's k-3 then (k-3)+6=k+3 must also be divisible by 3.

So, option B: (k – 4)(k – 2)(k + 3)(k + 5) is divisible by 8 and 3 in any case.

Bunnel,

For option B the (k – 4)(k – 2)(k + 3)(k + 5) can be written as (k-1) (k-2) k (k+2). Consider k = odd number say 3 then the sequence is 1,2,3,5 which is not divisible by 24. Not getting whats the issue in my thinking. Cant we rewrite the option B as(k-1) (k-2) k (k+2). Appreciate if anyone can correct me.
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Re: If k is a positive integer, which of the following must be  [#permalink]

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coolkl wrote:
Bunuel wrote:
If k is a positive integer, which of the following must be divisible by 24?

(A) (k – 4)(k)(k + 3)(k + 7)
(B) (k – 4)(k – 2)(k + 3)(k + 5)
(C) (k – 2)(k + 3)(k + 5)(k + 6)
(D) (k + 1)(k + 3)(k + 5)(k + 7)
(E) (k – 3)(k + 1)(k + 4)(k + 6)

24=8*3.

Note that the product of two consecutive even integers is always divisible by 8 (since one of them is divisible by 4 and another by 2). Only option B offers two consecutive even numbers for any integer value of k: k-4 and k-2, if k=even or k+3 and k+5 if k=odd.

Also from the following 3 consecutive integers: (k-4), (k-3), (k-2) one must be divisible by 3, if it's not k-4 or k-2 then it must be k-3 (if it's k-4 or k-2 option B is divisible by 3 right away). But if it's k-3 then (k-3)+6=k+3 must also be divisible by 3.

So, option B: (k – 4)(k – 2)(k + 3)(k + 5) is divisible by 8 and 3 in any case.

Bunnel,

For option B the (k – 4)(k – 2)(k + 3)(k + 5) can be written as (k-1) (k-2) k (k+2). Consider k = odd number say 3 then the sequence is 1,2,3,5 which is not divisible by 24. Not getting whats the issue in my thinking. Cant we rewrite the option B as(k-1) (k-2) k (k+2). Appreciate if anyone can correct me.

Why do you think that (k-1) (k-2) k (k+2) is the same as (k – 4)(k – 2)(k + 3)(k + 5)?

Also, if k=3, then option B gives -48 which is divisible by 24.
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I am struggling between B &C. Can someone give me a test value that eliminates C? Re: If k is a positive integer, which of the following must be   [#permalink] 13 Oct 2019, 15:03
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