Consecutive Intergers "n is an integer"

[andrew]

In the Challenge Archive, 2002 JUL-SEP, September 30th "Neighborly Numbers"

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

(1) n² + n, is not divisible by 3

(2) 3n+5 ≥ k+8, where k is a positive multiple of 3

I know that using Statement (1) (n-1)n(n+1) is three consecutive integers and that three consecutive integers are divisible by 3, but if n is 1 than (n-1) would equal 0 and this would not be, right? unless 0 is supposed to be divisible by 3???
Statement (2) break down this and it tells you that n ≥ 2, not sufficient

I was sure that this answer was C because the question does not specify that n is an integer greater than 1. So using both statements n ≥ 2 and the product of any three consecutive integers is divisible by 3.

But the answer is A???

Anyone???

[fahadmuhammad86]

Try plugging in numbers of statement 1. I used 1,-3,6,-6 and got No's for all of them i.e n-1 is not divisible by 3, thus answer is A. You are right abt the 2nd statement.

I hope this helps

[andrew]

Sorry if you look at the answer to the question in the Challenge Problem Archives it says n - 1 is divisible by 3...

I'm a little confused on your plugging in numbers theory, Statement (1) says n² + n is not divisible by 3... therefore you cannot plug in numbers like 6 & -6... if you do plug in numbers like 4, 4² + 4 = 20 (not divisible by 3) and n - 1 is divisible by 3 as it is 3 itslef...

So I know n-1 is divisible by 3, but I really thought that you had to know that n was greater than 1, right?

Thank you so much for the reply though... if you want check out the answer that is posted in the challenge archive, 2002 JUL-SEP, September 30th "Neighborly Numbers"

[fahadmuhammad86]

Sorry about my last post, U are right n-1 is divisible by 3. Here's how i did it (The plug-in technique is something i learned at Princeton, so i will try to explain it as much as i can).

Statement 1 says that n^2+n is not divisible by 3.
Now they way i do it is by plugging in numbers. Let try the following numbers for n 1,4,-2,-5,0. Now keep in mind that the number i choose have to satisfy the Statement 1 and generate a number that is not divisible by 3. You have to satisfy statement 1 in order to answer the original question. Think of it as you trying to get into a club, but u have to pass the bouncer 1st in order to get in. If u choose a number that doesn't satisfy statement 1, you cant go on to answering the question i.e. is n-1 divisible by 3. So the numbers i mentioned earlier satisfy statement 1 and thus show that n-1 is divisible by 3. Im not a student for manhattan course so i cant really check the answer to the question

Hope this helps, if ur getting confused on my method, just ignore it since its not something that is a part of your course/curriculum.

[andrew]

if n is 1 than n-1 is 0... not divisible by 3??? that's what i'm confused on... 1 satisfies statement 1 but n-1 would be zero and isn't divisible by 3...


and the answer given is A (and says n-1 is divisible by 3)

thank you for trying to help

is there a way to get an instructor to look at this???

[vnr1995]

n^2 + n = product of two consecutive numbers.

i. 2.3
ii. 3.4
iii 4.5
iv 5.6

look at the pattern

For (1) to be true, n = 4, which is nothing but n = 1 (modulo 3). This is sufficient.

if 3 divides n - 4, it divides n - 1 as well.

[Ben Ku]

The problem states:

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

(1) n² + n, is not divisible by 3

(2) 3n+5 ≥ k+8, where k is a positive multiple of 3

If we make a short list:
n-4, n-3, n-2, n-1, n, n+1, n+2, n+3

The question is asking, is n-1 divisible by 3? In other words, is n-1 a multiple of 3? This means that n-4, n-1, n+2 must all be multiples of 3 as well. Still another rephrase might be "is n one larger than a multiple of 3?"

I know that using Statement (1) (n-1)n(n+1) is three consecutive integers and that three consecutive integers are divisible by 3, but if n is 1 than (n-1) would equal 0 and this would not be, right? unless 0 is supposed to be divisible by 3???

If we simplify (1) we get n^2 + n = n(n + 1). If n(n + 1) is NOT divisible by 3, then it tells us that neither n nor n + 1 are multiples of 3. Since these two are not multiples of 3, then n - 1 MUST be a multiple of 3. So statement (1) is sufficient.

Statement (2) break down this and it tells you that n ≥ 2, not sufficient

Your restatement of (2) is almost correct. If 3n+5 ≥ k+8
That means 3n ≥ k + 3. If k is a positive multiple of 3, then we can express k as 3p, where p = 1, 2, 3, .... Now, 3n ≥ 3p + 3, so n ≥ p + 1. We don't exactly know what k is, so we don't know exactly what p is. But since p can be any number, then we can't determine whether n - 1 is a multiple of 3.

(A) is the answer. Hope that helps.

[epatmaloney]

I'm with Andrew on this one.

I get the math and logic behind it, but shouldn't the question state something like: "Given that N>1......."?

I may be wrong, but I don't think Andrew's question has been answered.

[messi10]

Hi,

I am a bit confused by this whole post. There seems to be a lot of replies but no one is saying that 0 is a multiple of 3, because I am sure it is.

Regards

Sunil

[jnelson0612]

Hi,

I am a bit confused by this whole post. There seems to be a lot of replies but no one is saying that 0 is a multiple of 3, because I am sure it is.

Regards

Sunil

Hi Sunil,
0 is considered a multiple of all integers. Hope this helps!