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17 Jun 2019, 01:07
Kudos
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
15%
(low)
Question Stats:
78% (01:43) correct
22%
(01:53)
wrong
based on 171
sessions
History
Date
Time
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Not Attempted Yet
If p is any odd integer greater than 1, which of the following must be a multiple of 24?
A. \(p(p − 1)(p − 2)\)
B. \(p(p + 1)\)
C. \(p(p^2− 1)\)
D. \(p^3 − p^2\)
E. \(p(p + 3)(p + 2)\)
A. \(p(p − 1)(p − 2)\)
B. \(p(p + 1)\)
C. \(p(p^2− 1)\)
D. \(p^3 − p^2\)
E. \(p(p + 3)(p + 2)\)
firas92
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Joined: 16 Jan 2019
Last visit: 02 Dec 2024
Posts: 616
Given Kudos: 142
Location: India
Concentration: General Management
Schools: Tepper '23 (M$) Foster '23 (D)
GMAT 1: 740 Q50 V40
WE:Sales (Other)
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17 Jun 2019, 01:18
Kudos
Bookmarks
Assign any odd number greater than 1 for p
Say, p=3
In this case, only (C) is a multiple of 24
Answer is (C)
Hit Kudos if this helped!
Say, p=3
In this case, only (C) is a multiple of 24
Answer is (C)
Hit Kudos if this helped!
17 Jun 2019, 10:58
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Bunuel
Given p = odd (o)
24 = 8*3
A. p(p − 1)(p − 2) = odd*even*odd --> Need not be a multiple of 8 always - NO
B. p(p + 1) = odd*even --> Need not be a multiple of 8 always - NO
C. p(p^2− 1) = (p - 1)p(p + 1) = even*odd*even - YES
Note:
(1) In the product of any 3 consecutive integers, there will ALWAYS be a multiple of 3 {(1,2,3), (2,3,4), (5,6,7) etc}
(2) Product of any 2 consecutive even numbers will ALWAYS be a multiple of 8 {(2*4), (4*6), (10*12) etc}
D. p^3 − p^2 = p^2(p - 1) = odd*odd*even --> Need not be a multiple of 8 always - NO
E. p(p + 3)(p + 2) = odd*even*odd --> Need not be a multiple of 8 always - NO
IMO Option C
Pls Hit Kudos if you like the solution
