n is an integer greater than 6 + DIVISIBILITY BY 3 QUESTION

[Guest]

If n is an integer greater than 6, which of the following must be divisible by 3?
A. n (n+1) (n-4)
B. n (n+2) (n-1)
C. n (n+3) (n-5)
D. n (n+4) (n-2)
E. n (n+5) (n-6)


The answer is given as D).

Stacey/Ron or others, could you please shed some light on this? Also,there are so many different types of divisibility and remainder questions in general so is there a collection of concepts/rules other than the common ones that can be applied to these type of problems?

Please advice.

[Nauman]

2 divides all evens
3 if the sum of the digits in the number is multiple of 3
4 if the last two digits are divisible by 4
5 if digit ends in 5 or 0
6 if the number is divisible by both 2 and 3
7 no rule
8 if the last three digits are divisible by 8
9 if the sum of the digits in the number is multile of 9

[RonPurewal]

first of all, you must understand this fact: every nth integer is divisible by n.
in other words, every other integer is divisible by 2; every third integer is divisible by 3; etc.
make sure you understand this basic principle first; if you don't, then essentially none of the following will make any sense to you.

for problems like this one, in which you're considering the divisibility of unknown numbers, it can help to do the following:
(1) set up a number line, with ..., n - 1, n, n + 1, n + 2, ... placed along it
(2) write out the different CASES for which ones are divisible by 3 (or divisible by whatever number you're talking about; in this problem it just happens to be 3)

here's what i mean.

in this case, there are 3 different cases, because every third integer is divisible by 3. (in general, for divisibility by n, there will be n different ways in which to draw the number line.)
here are the 3 cases. in each case, the multiples of 3 are in orange, and the non-multiples of 3 are in the default color (black on my view, although i'm not sure whether that's the default for everyone else).

case 1
... n-6 n-5 n-4 n-3 n-2 n-1 n n+1 n+2 n+3 n+4 n+5 n+6 ...

case 2
... n-6 n-5 n-4 n-3 n-2 n-1 n n+1 n+2 n+3 n+4 n+5 n+6 ...

case 3
... n-6 n-5 n-4 n-3 n-2 n-1 n n+1 n+2 n+3 n+4 n+5 n+6 ...

once you've done this, just look at the 3 cases for each of the answer choices. if any of the factors is in orange, then that particular product is a multiple of 3 (because, if there's a 3 anywhere in the factorization, then the whole product is a multiple of 3 - as with any other factor).

for choice (a), all three cases include one orange factor apiece: n is orange in case 1, n+1 is orange in case 3, and n-4 is orange in case 2. therefore, (a) is the correct answer.

for choice (b), case 3 contains no orange factors, so (b) doesn't have to be a multiple of 3.

for choice (c), case 2 contains no orange factors, so (b) doesn't have to be a multiple of 3.

for choice (d), case 2 contains no orange factors, so (b) doesn't have to be a multiple of 3.

for choice (e), case 3 contains no orange factors, so (b) doesn't have to be a multiple of 3.

note that the correct answer is a, not d.

--

note that you could also just plug in a ton of numbers and see what happens with the factorizations: just plug in 7, 8, 9, ... (per the directions) for the choices and see what the products are. don't multiply the products if you do that; just see whether the numbers in the products are divisible by 3.

[Rehannsr]

I understand case 1 because it shows that n is divisible by 3 and then every third number is divisible by 3. But how, for case 2 and 3, can n+2 and n+1 be shown in orange as divisible by 3, when we proved in case one that n+2 and n+1 result in remainders, and thus are NOT divisible by 3?

What am I missing?

[mirzank]

maybe an instructor can weigh in on this aswell, but I thought I'd give it a shot while you wait. be warned, i've never taught before so the following might be confusing but maybe it will guide you in right direction.

Since n is greater than 6, and the rule previously explained was that every nth factor is divisible by 3, since we don't know the actual value of n (is it 7, 8,9 or 10 etc) what we're trying to check is no matter what the nth term is, are there enough cases to ensure that atleast one of them is a factor of 3.

Maybe thats not clear enough so I'll digress a bit. Say you had to find out if a number n was divisible by 5, then you would have to check if in the timeline you've drawn has any factors of 5. But you'd want to check if n =5, n=6, n=7, n=8, etc. If n was greater than 5 but less than 12, you would draw multiple timeline. Say you started with n=7, then you would draw the n, n+1, n+2 timeline etc for case 1. But for case 2 you would now assume n as 8. But for the sake of consistency with the first timeline, even though you're starting at 8, you would call 8 n+1. Next would be 9 which would be n+2. Instead of 3 cases, imagine just one case, except it would get really messy and confusing to try to pinpoint starting points once with n, once with n+1, once with n+2.

Back to the multiple of 3 case, the way i followed the instructions was as follows. I only needed 3 cases really. n could be any starting number, 7, 8, 9, 200, 201, etc. (because remember, if you have 3 numbers, n, n+1 and n+2, then you know their product is a multiple of 3 regardless of where you start n because atleast one of the terms is a multiple of 3) All i needed to know was if any of the choices had one multiple of 3. And you need the timelines to see if N+1, N-6 etc are multiples.

so i started with a timeline for n-6,...n-3, n-2, ....n, n+1, n+2, ....n+6

then below n i had one timeline with n=7, below that another one for n=8, below one for n=9 (i.e. i assumed numbers rather than base it purely on the rule) (you could have 8,9,10, or 9,10,11, etc it wouldnt make a difference).

then i checked the answer choices. Like i said, and here is i believe the answer to your confusion of why n, N+1, n-1. For each answer choice, i started with the n timeline. n could be 7, 8 or 9 in the timelines below. if n=7, then n+1 = 8, and n-4 = 3. yep multiple of 3. then if n=8, n+1=9. so multiple of 3. then n = 9. sufficient, multiple of 3. No matter what value of n you start with you will find a multiple of 3 in EACH case for answer choice A.

follow this process for all choices and you will see that for B, C, D and E, they might be divisible by 3 if say n=7, but not on n=8. So answer would be maybe divisible, maybe not, since we dont know for sure what n is, maybe is not a good enough answer.

I understand case 1 because it shows that n is divisible by 3 and then every third number is divisible by 3. But how, for case 2 and 3, can n+2 and n+1 be shown in orange as divisible by 3, when we proved in case one that n+2 and n+1 result in remainders, and thus are NOT divisible by 3?

What am I missing?

[Rijul Negi]

n is an integer greater than 6

so, n can be
6+1
6+2
6+3
6+4
6+5
.
.
.
.
etc

so we can express n as 3k,3k+1 or 3k+2

substituting n in terms of k to the options

A.

for n=3k
n(n+1)(n-4) => divisible by 3
for n = 3k+1
(3k+1)(3k+2)(3k-3) => 3(3k+1)(3k+2)(k-1) => divisible by 3
for n = 3k+2
(3k+2)(3k+3)(3k-2) => 3(3k+2)(k+1)(3k-2) => divisible by 3


This shows that equation in option A will be divisible by 3 for all integers greater than 6

we dont need to check next options...

Option A is the correct option.

[tim]

Good discussion here. Let us know if you need any further clarification from an instructor..