What is the reminder when the positive integer n is divided

Question

What is the reminder when the positive integer  n  is divided by the postivie integer  k , where k>1?

(1) n=(k+1)^3

(2) k=5

OA= A

Please, any explanation?

Thank you!

piper · Answer

For statement 1, if K > 1, k could be 2, 3, 4, 5, 6, 7, etc. With k being any of those numbers, when you put K, in the formula and you find n, you can have different remainders for n/k. Therefore not sufficient.

For statement 2, k is 5 we can't find n. Therefore not sufficient.

For both statements, if k is 5, you will get one answer for n. Therefore you can find the remainder for n/k.

Hope that helps.

JohnLewis1980 · Answer

Sorry Piper, I put the wrong OA.

The right one is  A  and not C.

I edit the original post

(C is what I chose)

Cheers

FN · Answer

I dont think OA can be A..

n=(K+1)^3, suppose K=2 then 3^3=27/3 remainder is 0..but if K=4, then 5^3/3 has a non-zero remainder..

I think C is the correct answer..

JohnLewis1980 · Answer

mmm, it could be possible that the GMAT PrepSoft were wrong?

I attached an screen with the OA

Cheers

JohnLewis1980 · Answer

could you please post your explanation?

Thank you

FN · Answer

OOPs...OA is Correct..

n=(K+1)^3, suppose n=3^3 then k=2 then remainder is 1..

you will always note that remainder is 1..

A is correct

JohnLewis1980 · Answer

yes, now it's clearer

This kind of problem is easier to solve using a couple of examples

Cheers

scthakur · Answer

From stmt1: n = k^2 + 2k + 1 and since the first two terms are divisible by k, the remainder will be 1. Hence, stmt 1 is sufficient.

prasun84 · Answer

hi,
n=(k+1)3 and not (k+1)2 to break it into k^2 +2k+1.
however, (k+1)3 is  k^3+1+3k^2+3k --

the remainder will always be 1.

bigtreezl · Answer

scthakur...how did you arrive at  n = k^2 + 2k + 1  from n = (k+1)^3 ?

anuprajan5 · Answer

Answer is A. The remanider is always 1. Reason being that (k+1)^3 = K^3+3k^2+2k+1. Hence k is common in all variable except 1. Hence 1 will always be the remainder.

scthakur · Answer

My fault. I wrongly read as (k+1)^2. But, even with k+1)^3, all the terms but the last will have k and hence the remainder will always be 1.

What is the reminder when the positive integer n is divided

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