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happypuppy
GMAT 1: 720 Q50 V38 (Online)
Posts: 207
23 Jun 2022, 11:17
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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:
65%
(hard)
Question Stats:
61% (02:45) correct
39%
(02:29)
wrong
based on 90
sessions
History
Date
Time
Result
Not Attempted Yet
If z is a positive integer, such that z > 1. What is the remainder when (z - 1)(z + 1) is divided by 24?
(1) 5z is not a multiple of 2
(2) 6 + z is not divisible by 3
(1) 5z is not a multiple of 2
(2) 6 + z is not divisible by 3
23 Jun 2022, 13:08
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Quote:
I enjoyed this question, and I thought I would take a moment to explain my approach in case it may prove helpful to someone else.
Statement (1) tells us that z is odd, since any even number would make 5z divisible by 2. I started with something I knew would be divisible by 24—e.g., if z = 5, (z - 1)(z + 1) will be 24. Then I explored another option. What if z = 3? (z - 1)(z + 1) would now be 8, which would not be divisible by 24. Thus, Statement (1) is NOT sufficient.
Statement (2) tells us that z itself is not divisible by 3, since adding 6 to anything divisible by 3 would produce a number divisible by 3. Once again, we know that z = 5 is a valid input that will lead to a product of 24; likewise, we can appreciate that a number such as 2—remember, nothing prohibits such a value from this statement alone—would yield a product that was not divisible by 24: (2 - 1)(2 + 1) = 3. Thus, Statement (2) is NOT sufficient.
Together, the statements allow for odd integers greater than 1 but not divisible by 3, so 5, 7, 11, 13, and so on would be valid inputs for z. Notice how all of these integers are +/- 1 from an integer that is divisible by 3, and the product will be large enough to be divisible by 24. I tested the first few integers and then selected (C):
5: (5 - 1)(5 + 1) = (4)(6) = 24 √
7: (7 - 1)(7 + 1) = (6)(8) = 48 √
11: (11 - 1)(11 + 1) = (10)(12) = (5)(2)(12) = (5)(24) √
13: (13 - 1)(13 + 1) = (12)(14) = (12)(2)(7) = (24)(7) √
I know that this solution can be expressed algebraically, but I enjoy a numerical approach sometimes. I thought this question was well thought out. What is the source, Roy867?
- Andrew
21 Oct 2023, 04:13
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Hello from the GMAT Club BumpBot!
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Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).
Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
