If n is a positive integer and the remainder is r when (n-1)

[ambikasrinivas]

From GMAT Prep Exam

If n is a positive integer and the remainder is r when (n-1)(n+1) is divided by 24, what is the value of r?

(1) 2 is not a factor of n
(2) 3 is not a factor of n

Answer C

I can see that from the statements, n is an odd integer not divisible by 3. This means that n-1 and n+1 are even integers, and n must be either prime, or a multiple of 7, 5, 11, etc. but not 3. But either n-1 or n+1 must be divisible by 3. I am not really sure how to go from there.

Thanks!

[george.kourdin]

i got this far

2 is not a factor of n so n is odd, thus n-1 and n+1 are both even so (n-1)(n+1) has to be divisible by 4. actually by 8 because (n-1) is some even number that is two apart form (n+1) so there is another 2 in there which gets us to 8. if someone has a more elegant way of showing this, please reply.

with 3 not being a factor of n i am kind of stumped. i feel like its something trivial where we will eventually show that (n-1)(n+1) is indeed divisible by 3 because 8 and 3 are factors of 24.

i hate gmat......

[george.kourdin]

meh....okay

n isn't divisible by 3 but n is between (n-1), n, (n+1). out of any 3 consecutive integers one must be divisible by 3 so if n isn't then either n-1 or n+1. we got our factor of 3.

this is so much fun

[ambikasrinivas]

thanks - I was having a difficult time getting to the 'divisible by 8' but now that makes sense!

[RonPurewal]

like most problems involving divisibility, this one is easier to do by just plugging in numbers.

statement 1: pick n's such that 2 is not a factor of n.
n = 1 --> (n - 1)(n + 1) = 0 --> remainder = 0
n = 3 --> (n - 1)(n + 1) = 8 --> remainder = 8
2 different answers; insufficient.
(if you don't like remainders that involve 0, then just use n = 5 instead; that one also gives a remainder of 0.)

statement 2: pick n's such that 3 is not a factor of n.
n = 1 --> (n - 1)(n + 1) = 0 --> remainder = 0
n = 2 --> (n - 1)(n + 1) = 3 --> remainder = 3
2 different answers; insufficient.
(if you don't like remainders that involve 0, then just use n = 4 instead; that one gives a remainder of 15.)

together: pick n's such that neither 2 nor 3 is a factor of n.
n = 1 --> (n - 1)(n + 1) = 0 --> remainder = 0
n = 5 --> (n - 1)(n + 1) = 24 --> remainder = 0
n = 7 --> (n - 1)(n + 1) = 48 --> remainder = 0
n = 11 --> (n - 1)(n + 1) = 120 --> remainder = 0
n = 13 --> (n - 1)(n + 1) = 168 --> remainder = 0
by this point it's clear that you're always going to get 0; sufficient.

(c) it is.

--

if you see divisibility by particular numbers, then ALWAYS have plugging available as a backup method; on a *huge* number of problems (like this one), it's way easier than using theory.

[taniafconca]

Maybe I dont see it... but when you are picking numbers isnt it
(n-1)(n+1) / 24 -> ie: n=2 8/24?
How do you conclude the remainder is 8?

Thanks!

[RonPurewal]

Maybe I dont see it... but when you are picking numbers isnt it
(n-1)(n+1) / 24 -> ie: n=2 8/24?
How do you conclude the remainder is 8?

Thanks!

the remainder when you divide 8 by 24 is 8.

in fact, if "x" is smaller than "y", then the remainder from dividing x/y is just x. (basically, x goes into y zero times, leaving the whole x as a remainder.)

as i mentioned in that post, you can avoid this whole issue by simply plugging in slightly higher numbers.