Question from practice test I could not get

[beboppin]

If (n+1)(n-1) is divided by 24 and the remainder is r, what is r?

i) n is not divisible by 2
ii) n is not divisible by 3

[BrianSmith]

I think it's C. The problem with my answer is that it's complete butchery!!! Has anyone got a more subtle solution for this?

I went by way of elimination. If I could find two different r for each statement, it means it's not enough to reveal r.

(n+1)(n-1) is n^2-1 . If we remove the remainder r we get n^2-1-r which must be divisible by 24. I'm not sure you're supposed to do it but I skipped n=0,1,2,3,4 because then n^2-1 comes up as less than 24.

(i) n is not divisible by 2

I tried n=5: 5^2-1-r = 24-r so r=0 (24 is divisible by 24)
n=6 is divisible by 2 so it doesn't qualify
I tried n=7: 7^2-1-r = 48-r so r=0 (48 is divisible by 24)
n=8 is divisible by 2 so it doesn't qualify

So, is r always 0? I had to try one more...
n=9: 9^2-1-r=80-r ... the nearest multiple of 24 is 72 so 80-r=72 or r=8.

(i) is insufficient.

(ii) n is not divisible by 3

we already know that if n=5 and n=7, then r=0
n=9 is no good this time so I try n=8 (not divisible by 3)
n=8: 8^2-1-r=63-r nearest 24 multiple is 48 so 63-r=48 or r=15

(ii) is insufficient

(i)+(ii) I already know n=5 or n=7 means r=0
n=8,9,10 is not good (divisible by 2 or 3)
n=11: 11^2-1-r=120-r 120 is divisible by 24 so r=0

so are both together sufficient? try two more
n=12,14,15,16 are all divisible by either 2 or 3
n=13: 13^2-1-r=168-r 168 is divisible by 24 so r=0
n=17: 17^2-1-r=288-r 288 is divisible by 24 so r=0

for n's which aren't divisible by 2 or 3, r is always 0.
so (i)+(ii) is sufficient and it's C.

[tohellandback]

IMO C

1) numbers are 1,3,5,7,9,11,13
take 3- 2*4=8. remainder 8
take 5- 4*6=24. remainder 0
NOT SUFF

2) numbers are
1 2 4 5 7 8 10 11
take 2- 1*3=3, remainder 3
take 5- 4*6 remainder 0
combined
take 5,7,11,13
remainder is 0
SUFF

[ranell]

Answer is definitely C
1 -insufficient
take 1 and 3 and you will get two various remainders
2 - insufficient
take 2 and 4 and you will get two various remainders

if you combine the date from 1 and 2, you will use 1, 5, 7, 11. All the figures will be primes, except for 1. The remainder will be the same - 0