Challenge Question Dispute

[Guest]

09/30/02

Question

Given that n is an integer, is n "” 1 divisible by 3?

1) n^2 + n is not divisible by 3
2) 3n+5>= k+8, where k is a positive multiple of 3


The above is one of the problems from the challenge archives that I have bought......The explanation states that (1) is not sufficient....................Here is the explanation given by Manhattan..

"Statement (1) gives us information about n^2 + n , which can be rewritten as the product of two consecutive integers as follows:n^2 +n = n(n+1)
Since the question asks us about n "” 1, we can see that we are dealing with three consecutive integers: n "” 1, n, and n + 1 .
By definition, the product of consecutive nonzero integers is divisible by the number of terms. Thus the product of three consecutive nonzero integers must be divisible by 3.
Since we are told in Statement (1) that the product n^2 +n is not divisible by 3, we know that neither n nor n + 1 is divisible by 3. Therefore it seems that n "” 1 must be divisible by 3.
However, this only holds if the integers in the consecutive set are nonzero integers. Since Statement (1) does not tell us this, it is not sufficient. "


However, by definition zero is a multiple of all numbers and hence (1) should be sufficient.
May I request Manhattan tutor to either confirm or clarify the above statement..
Thanks in advance

[JadranLee]

You're absolutely right! On p. 108 of the Official Guide, we are told that:

If x and y are integers, and x is not equal to zero, then x is a divisor (factor) of y provided that y=xn for some integer n.

This implies that every nonzero integer x is a factor of zero, because if y=0 and n=0 in the above equation x can be any nonzero integer.

The Official Guide clearly indicates that zero is divisible by 2, for example, when it defines an even integer as any integer that is divisible by two.

So you're right, we need to change our solution to that challenge problem. (I'll take care of it soon.)

-Jad

[Guest]

Thank you Jad for the quick response and clarification!!!

[christiancryan]

Thank you, Guest, for flagging the error -- we've fixed it in the archived problem. I'm not sure what source the writer was using for the quoted rule, nor how the problem got past our censors, but we certainly appreciate your bringing the issue to our attention! Good luck on your continued preparation.

[Guest]

Thank you for the quick implementation!