We are looking for the remainder of n being divided by k.
1. It's useful to know that the formula for raising (a + b) to the third power:
(a + b)^3 = a^3 + 3a^2b + 3ab^2+ b^3.
This means that n will be k^3 + 3k^2 + 3k + 1. Now, you can clearly see that the remainder will be 1 here, since k^3, 3k^2 and 3k are all divisible by k.

2. is insufficient. Knowing k = 5 is not enough: we also need some info about n.

Rules:

1) (a+1)^x / a will always give a remainder of 1 where a is an integer >1

2) a^x / a+1 will always give a remainder of 1 when n is even and will always give a remainder of a when n is odd where a is an integer



Applying rule 1 to stmt I the remainder is always 1 since a=k and n = (a+1)^x i.e (a+1)^3


The other way would be to substitute numbers

k=2

n=9

remainder 1

k=3

n=64

remainder 1

k=4

n=124

remainder 1

It follows a pattern....

suff

Choose A

DanaJ wrote:We are looking for the remainder of n being divided by k.
1. It's useful to know that the formula for raising (a + b) to the third power:
(a + b)^3 = a^3 + 3a^2b + 3ab^2+ b^3.
This means that n will be k^3 + 3k^2 + 3k + 1. Now, you can clearly see that the remainder will be 1 here, since k^3, 3k^2 and 3k are all divisible by k.

2. is insufficient. Knowing k = 5 is not enough: we also need some info about n.

DanaJ, we are all very lucky to have you on board! Thanks for the great work you are doing on this forum. It is joy to read and learn from your posts. I am sure I talk for everbody on this forum.

Thank you, logitech. I'm sure everyone thinks the same about you!