Consecutive Integers

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Consecutive Integers

by aleph777 » Wed Jan 05, 2011 5:01 am
Hello,

Quick question about DS problem 170 in the OG12:

If n is a positive integer, is n^3 - n divisible by 4?

(1) n = 2k + 1, where k is an integer.
(2) n^2 + n is divisible by 6.

OA: A

I followed through the logic of this, realizing n^3 - n was the series of consecutive integers (n - 1)n(n +1). But I interpreted this to mean that these three could only be divisible by 4 if either n was a multiple of four or if n was any odd number greater than 1.

Because statement (1) says k is an integer, even if it is 1, then n = 3, which makes the statement sufficient. However, I was reading the MGMAT OG Companion, which states "We are told that n must be a positive integer. Thus, the lowest possible value of n - 1 is zero. Even in this case, the product will be divisible by 4, since 0 is divisible by 4."

I don't follow that statement. If statement 1 didn't specify that k was an integer, and therefore n could have been 1, n + 1 could have been 2, and n - 1 could have been 0, statement 1 would have still be sufficient?

It's a bit of a hypothetical, but I want to make sure I understand in case something like that ever shows up. Thanks!
Source: — Data Sufficiency |

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by arora007 » Wed Jan 05, 2011 5:14 am
aleph777 wrote:Hello,

Quick question about DS problem 170 in the OG12:

If n is a positive integer, is n^3 - n divisible by 4?

(1) n = 2k + 1, where k is an integer.
(2) n^2 + n is divisible by 6.

OA: A

I don't follow that statement. If statement 1 didn't specify that k was an integer, and therefore n could have been 1, n + 1 could have been 2, and n - 1 could have been 0, statement 1 would have still be sufficient?
Now if k were not an integer.. you could have plugged in k=1/2 thus n would have been 2 and n^3 -n would not be divisible by 4. Thus not sufficient.... The answer choice would then have been E.
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by aleph777 » Wed Jan 05, 2011 6:45 am
Thanks, Arora.

My my confusion was with the MGMAT explanation.

It says if n were 0, it would still be divisible by 4. But I don't understand that? 0/4 = 0 r4. And I thought that's partially why statement 1 says k is an integer. To explain it is neither a fraction nor zero. I'm just trying to wrap my head around the MGMAT explanation that even if n were zero, then n^3 - n would be divisible by 4...

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by GMATGuruNY » Wed Jan 05, 2011 8:30 am
aleph777 wrote:Thanks, Arora.

My my confusion was with the MGMAT explanation.

It says if n were 0, it would still be divisible by 4. But I don't understand that? 0/4 = 0 r4. And I thought that's partially why statement 1 says k is an integer. To explain it is neither a fraction nor zero. I'm just trying to wrap my head around the MGMAT explanation that even if n were zero, then n^3 - n would be divisible by 4...
The remainder after division is not how much remains of the divisor (the number on the bottom) but how much remains of the dividend (the number on top; the number being divided into). When any integer if divided into 0, nothing of the dividend (0) remains, so there is no remainder.

In other words, 0 is divisible by every positive integer.
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by MAAJ » Thu Jan 06, 2011 2:07 pm
So MGMAT explanation is correct ;) thanks! Just love these books!
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