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05 Apr 2012, 13:04
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
45% (02:32) correct
55%
(02:18)
wrong
based on 576
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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)
(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)
05 Apr 2012, 14:17
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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.
Answer: B.
(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.
Answer: B.
General Discussion
05 Apr 2012, 20:19
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Galiya
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.
