When a positive integer, n, is divided by 24 the remainder is 18. When n is divided by 8 the remainder is:

A.0 B.1 C.2 D.4 E.6

I tried doing n/24=18 and I got 432 and when I divided it by 8 I got 54; am I misinterpreting the question? I really see no other way to do it. Can someone please help me?

The smallest value that n could be is 42 (42÷24 = 1 with a remainder of 18)

Adding 24 each time I got the sequence
42 66 90 114 138 162 ...

Now when you divide each of these by 8 you will have a remainder of 2 for each one

sorry, but I still don't get it...

ok, let's try this

let N = 24x + 18

for whole number values of x isn't
N = 42,66,90,114,138,162,... ?

24x would give you multiples of 24, that is, all numbers that are exactly divible by 24, so by adding 18 we create a remainder of 18

now try dividing each of those by 8
What did you notice about the remainder?

I get decimals, well, a whole unti and a decimal

I still don't think I'm doing this right, am I?

I just don't get how I'm supposed to divide 90 by 18 and get a remainder of 2, according to your first reply...

OK, n is divided by 24 and a remainder of 18.

if divided by 3, you will still have a "remainder"of 18, but that remainder is divisible by 3 twice, giving a final remainder of 2

According to your own question when you divide by 8 you get a remainder of 2, not when you divide by 18

so sorry, but I don't get how having a reminder works...

I'm sorry, I'm just trying to figure out a way of actually colsing the problem and nothing seems to work