remainder

[cans]

What is the remainder when the positive integer n is divided by the positive integer k, where
k>1?
(1) n=(k+1)^3 (i.e. cube of (k+1))
(2) k=5

[krishnasty]

What is the remainder when the positive integer n is divided by the positive integer k, where
k>1?
(1) n=(k+1)^3 (i.e. cube of (k+1))
(2) k=5

K has to be greater than 1
hence, let k = 2
1 ) k = 2 and n = 3^3 = 27 --> 27/2 = 1(remainder)
let k = 3 and n = 4^3 = 64 --> 64/3 = 1 (remainder)

Hence, sufficient

2) k = 5
no idea of n.
Hence, insufficient

IMO, A

[galaxian]

IMO A, from the same substitution method explained above.

[phanideepak]

1 : (k+1)^3/k = [k^3 + 1 +3k(1+k)]/k so in this only 1 is left out with out a k so the remainder is 1.

Sufficient.

2 : K=5 but we do not know n so insuff

answer is A

[magicalhat]

A, imo.

What is the remainder when the positive integer n is divided by the positive integer k, where
k>1?
(1) n=(k+1)^3 (i.e. cube of (k+1))
(2) k=5

[Ian Stewart]

What is the remainder when the positive integer n is divided by the positive integer k, where
k>1?
(1) n=(k+1)^3 (i.e. cube of (k+1))
(2) k=5

When you expand (k+1)^3, every term you get will be a multiple of k except for the 1^3 = 1 term at the end. So (k+1)^3 will be 1 greater than a multiple of k, and thus the remainder will be 1 when it is divided by k. So the answer is A, since Statement 2 alone is of no help.

In fact, for the same reason, if k > 1, then (k+1)^n will always give you a remainder of 1 when you divide it by k, assuming n and k are positive integers.

[cans]

OA A

[najeeb775]

Another way of looking at this..
If Reminder(K/n) = r
then Reminder(K^3/n) = Reminder(r^3/n)

Therefore:
Since Reminder((K+1)/K) = 1
Hence Reminder ((K+1)^3/K) = 1^3/K = 1

Hence A alone is sufficient to answer it.

B gives no idea about n.
IMO A

[Sanjay2706]

A is the answer.

[Sanjay2706]

A is the answer.