If z is a positive integer, such that z1. What is the remainder when

Question

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

AndrewN · Answer

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

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If z is a positive integer, such that z>1. What is the remainder when

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