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If k is a positive integer, which of the following must be
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05 Apr 2012, 12:04
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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)
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Re: If k is a positive integer, which of the following must be
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05 Apr 2012, 13:17
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: k4 and k2, if k=even or k+3 and k+5 if k=odd. Also from the following 3 consecutive integers: (k4), (k3), (k2) one must be divisible by 3, if it's not k4 or k2 then it must be k3 (if it's k4 or k2 option B is divisible by 3 right away). But if it's k3 then (k3)+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. Answer: B.
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Re: If k is a positive integer, which of the following must be
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05 Apr 2012, 19:19
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
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06 Apr 2012, 11:15
Bunuel, Karishma thank you so much for help!



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Re: If k is a positive integer, which of the following must be
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15 Apr 2012, 02:52
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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Re: If k is a positive integer, which of the following must be
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15 Apr 2012, 02:59
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
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Re: If k is a positive integer, which of the following must be
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31 Jan 2014, 08:00
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 'k4' and 'k2' 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



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If k is a positive integer, which of the following must be
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17 Jul 2016, 06:41
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 'k4' and 'k2' 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 "k4" and "k2" ?
Also, please note that in your solution, you have added that "6" to "k3". Thanks in advance.
Regards, Yosita



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Re: If k is a positive integer, which of the following must be
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08 Mar 2017, 11:42
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: k4 and k2, if k=even or k+3 and k+5 if k=odd.
Also from the following 3 consecutive integers: (k4), (k3), (k2) one must be divisible by 3, if it's not k4 or k2 then it must be k3 (if it's k4 or k2 option B is divisible by 3 right away). But if it's k3 then (k3)+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.
Answer: B. Bunnel, For option B the (k – 4)(k – 2)(k + 3)(k + 5) can be written as (k1) (k2) 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(k1) (k2) 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
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08 Mar 2017, 22:51
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: k4 and k2, if k=even or k+3 and k+5 if k=odd.
Also from the following 3 consecutive integers: (k4), (k3), (k2) one must be divisible by 3, if it's not k4 or k2 then it must be k3 (if it's k4 or k2 option B is divisible by 3 right away). But if it's k3 then (k3)+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.
Answer: B. Bunnel, For option B the (k – 4)(k – 2)(k + 3)(k + 5) can be written as (k1) (k2) 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(k1) (k2) k (k+2). Appreciate if anyone can correct me. Why do you think that (k1) (k2) 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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Re: If k is a positive integer, which of the following must be
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