To distribute:
(k+1)^3 = (k+1)(k+1)(k+1)
let's take the first two sets of parentheses first:
(k+1)(k+1) = k^2 + 2k + 1
now, multiple in our third (k+1):
(k+1)(k^2 + 2k + 1) =
k^3 + 2k^2 + k + k^2 + 2k + 1) =
k^3 + 3k^2 + 3k + 1 = n
They ask about n/k, so:
(k^3 + 3k^2 + 3k + 1)
k
= k^2 + 3k + 3 + 1/k. In this algebraic representation of division, the numerator of the fraction is the remainder. So the remainder is always 1.
You can also try some real numbers and see if there's a pattern. We know they're both positive integers and k>1.
Try k = 2. Then n = (2+1)^3 = 27. 27/2 = 13 R1
Try k = 3. Then n = (3+1)^3 = 64. 64/3 = 21 R1
So the first thing I notice is that, when k is even, n is odd, and vice versa. The second thing I notice is that I've gotten R1 for both. Interesting.
k = 4. Then n = (4+1)^3 = 125. 125/4 = 31 R1. R1 again.
k = 5. Then n = (5+1)^3 = 216. 216/5 = 43 R1. R1 yet again!
At this point I'm feeling pretty comfortable that there's a pattern. I'm going to get R1 no matter what I try for k.
Please note: I do not use the Private Messaging system! I will not see any PMs that you send to me!!
Stacey Koprince
GMAT Instructor
Director of Online Community
Manhattan GMAT
Contributor to Beat The GMAT!
Learn more about me
