question from Remainders Study Hall

[gmatwork]

There are six consecutive positive integers in Set S. What is the value of positive integer, n?

1) When each integer in S is divided by n the sum of remainders is 11?
2) When each integer in S is divided by n, the remainders include five different values?

Can you please post the solution for this one?

OA: B

[tim]

Before we help with this question, we need you to show some effort of your own. What did you try on this question? Where did you get stuck?

[kapilnitt]

Hi tim,

Will you allow me to reply the post?

Priyanka: I'm just giving you the hint, until you follow the instruction of tim.

6 numbers in the set are in sequence so the reminder will also come in sequence and if out of 6 reminders 5 are same, it is only possible when count goes from 0 to 5.

Thanks
Kapil

[RonPurewal]

6 numbers in the set are in sequence so the reminder will also come in sequence and if out of 6 reminders 5 are same, it is only possible when count goes from 0 to 5.

Thanks
Kapil

right idea. not quite the right numbers.
if you had remainders 0 through 5, that would be six, not five, different remainders (count them: 0, 1, 2, 3, 4, 5).

the only way to get exactly five different remainders is if n = 5, so that the remainders (from dividing consecutive integers) repeat in the order 0, 1, 2, 3, 4, 0, 1, 2, 3, 4, 0, 1, 2, 3, 4, ...
this way, if you divide six consecutive integers by 5, you'll see five different remainders (0, 1, 2, 3, and 4). you'll see exactly one of them twice, but that's irrelevant for the purpose of counting how many different remainders show up.

if n is less than 5, then there aren't even five remainders possible in the first place.
if n is greater than 5, all six of the remainders will be different. (try it yourself.)

[RonPurewal]

and, in case you don't know why statement 1 is insufficient -- you basically have to figure that one out by trial and error.

if n = 5, then the remainders repeat in this sequence:
0, 1, 2, 3, 4, 0, 1, 2, 3, 4, 0, 1, 2, 3, 4, ...
with these remainders, if you take six numbers that leave the remainders 1, 2, 3, 4, 0, 1, then those add up to eleven. so, n could be 5.

if n = 4, then the remainders repeat in this sequence:
0, 1, 2, 3, 0, 1, 2, 3, 0, 1, 2, 3, ...
here, if you take six numbers that leave the remainders 2, 3, 0, 1, 2, 3, then those also add up to eleven. so, n could be 4 as well.

that's two different values of n, so, insufficient.

i don't know if there's any algebraic way to prove that this statement is insufficient.