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

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

n^3-n=n(n^2-1)=(n-1)n(n+1), so we are asked whether the product of 3 consecutive integers is divisible by 4.

(1) n = 2k + 1, where k is an integer --> n=odd --> as n is odd then both n-1 and n+1 are even hence (n-1)n(n+1) is divisible by 4. Sufficient.

(2) n^2 + n is divisible by 6 --> if n=2 then n^3-n=6 and the answer is NO but if n=3 then n^3-n=24 and the answer is YES. Not sufficient.

Answer: A.
_________________

Although this question is on finding out divisibility by 4, using the divisibility rule of 4 may not fetch the desired result here. Instead, we need to work on identifying the type of numbers that are part of the given expression.

The given expression, \(n^3\) – n can be broken down into (n-1) * n * (n+1). This is the product of 3 consecutive integers. The product of any 3 consecutive integers is always divisible by 3! i.e. 6.

Note: The product of any n consecutive integers is always divisible by n! and exactly one of these numbers will be exactly divisible by n.


For the product to be divisible by 4, two of the numbers should be even. Any data that helps us establish this will be sufficient data.

From statement I, we know that n = 2k + 1, where k is an integer. This is nothing but the definition of an odd number. So, n is odd; this means that (n-1) and (n+1) is even. Since there are 2 even numbers in the product, the number has to have a four and hence be divisible by 4.

Statement I alone is sufficient. Answer options B, C and E can be ruled out, possible answer options are A or D.

From statement II, \(n^2\) + n is divisible by 6. \(n^2\) + n can be broken down as n*(n+1). This is the product of 2 numbers.
So, as per the statement, the product of 2 numbers is divisible by 6. This means that at least one of these two numbers is even and the other is a multiple of 3. But, we do not know anything about the number (n-1).

For example, if n = 2, (n-1) = 1 and (n+1) = 3. The product of (n-1) * n * (n+1) is not divisible by 4.

On the other hand, if n = 3, (n-1) = 2 and (n+1) = 4. The product of (n-1) * n * (n+1) is divisible by 4.

The data given in statement II is insufficient. Answer option D can be eliminated.
The correct answer option is A.

Hope this helps!
_________________

to solve this problem:
1) substitute the constraint 2k+1 in the equation

A: (2k+1)(2k+2)(2k)/4 = (k+1)(k)(2k+1) DIVISIBLE by 4.
B: (2k+1)(2k+2)/6 = (2k+1)(k+1)/3. But we don't get any clue about N itself. So, how can we say that n^3-n is divisible by 4. If anything from looking at both equations we know only that 2k+2 is divisible by 2. To be divisible by 4 we need to two sets of 2's which is lacking from the information given in B.

To answer your doubt:
1) n^2 + n is div by 6 then n^3-n is div by 4

Let's try: n = 2 it satisfies n^2 + n (6) is div by 6; BUT n^3-n(6) is NOT div by 4 (because 6 IS NOT divisible by 4).
Let's try: n = 3 it satisfies n^2 + n(12) is div by 6; BUT n^3-n(24) IS div by 4 (because 24 IS divisible by 4).

So, since n= 2 doesnt work and n = 3 works. It's insufficient.

if n=1 though then the result is zero, is zero considered to be divisible by 4 ? so can we say zero is also a multiple of every integer ? if not , A is not sufficient by itself then.

Zero is divisible by EVERY integer except zero itself, since 0/integer=integer (or, which is the same, zero is a multiple of every integer).

Check Number Theory chapter of Math Book for more hints/tips/rules on this subject: https://gmatclub.com/forum/math-number-theory-88376.html

Hope it helps.
_________________

In the statement one when we express n=2k+1, why cant we take k=0, in that case n=1 and the product will be zero...Can we say then zero is divisible by 4?

Zero is a divisible by every integer, except zero itself. Integer \(a\) is a multiple of integer \(b\) means that \(a\) is "evenly divisible" by \(b\), i.e., divisible by \(b\) without a remainder. Now, since zero/integer=integer then zero is a multiple of every integer (except zero itself).

Check Number Theory chapter of Math Book for more: https://gmatclub.com/forum/math-number-theory-88376.html

Hope it helps.
_________________

Sorry Bunuel, just to make this clear

if k = 0 then n = 1 (2 * 0 + 1)

this means that (n-1)(n)(n+1) is 0 * 1 * 2

which is 0 (zero) and therefore divisible by 4 because zero is divisible by 4 ... right?

Exactly so: if n=1 then (n-1)(n)(n+1)=0*1*2=0 and 0 is divisible by every positive integer including 4.
_________________

(n^3 - n) mod 4 = 0 ?
n(n^2 - 1) mod 4 = 0 ?
(n-1)(n)(n+1) mod 4 = 0 ?

1)n = 2k+1
=> 4k(2k+1)(k+1)
this is div by 4. hence, sufficient.

2) n^2 + n mod 6 = 0
n(n+1) mod 6 = 0
so either n or n+1 must be divisible by 2
if n is div by 2 then original expression is not div by 4
if n+1 is div by 2 then original expression is div by 4
hence, insufficient.

A.
_________________

Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.


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

-> When you modify the condition and problem, n^3-n=4t?(t is a positive integer) --> (n-1)n(n+1)=4t? and n-1=even?. Since n-1=even and n+1=even, it becomes yes.
In 1), n-1=2k and it is sufficient.
In 2), from n(n+1)=6m(m is a positve integer), n=3 yes n=2 no, which is not sufficient. Therefore, the answer is A.


-> For cases where we need 1 more equation, such as original conditions with “1 variable”, or “2 variables and 1 equation”, or “3 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 59 % chance that D is the answer, while A or B has 38% chance and C or E has 3% chance. Since D is most likely to be the answer using 1) and 2) separately according to DS definition. Obviously there may be cases where the answer is A, B, C or E.

n^3-n= (n-1)*n*(n+1) [product of 3 consecutive integers]

1) n is an odd number; hence, n-1 and n+1 are even numbers. The product of n-1 and n+1 is always divisible by 4.
Sufficient

2) n(n+1) is divisible by 6.
If n is odd multiple of 2, then (n-1)*n*(n+1) is not divisible by 4
In all other cases (n-1)*n*(n+1) is divisible by 4.

Insufficient.

Good question!

St 2 according to me is easier; therefore I start with an analysis of statement 2. Let us test two scenarios (1,2,3) and (2,3,4). In the former case, we get a "no" answer and in the latter case, we get a yes answer. Hence, not sufficient. We can remove option B and D

St 1 directly says that n is an odd number. This would mean that n^3 – n or (n-1) * n * (n+1) is definitely divisible by (2*2 or 4). Hence st 1 is sufficient.

Option A is the correct answer.
_________________

Hello,
ArvindCrackVerbal

If n=5; n+1=6 then n(n+1)=30. Here, 5*6=30. One (6) is multiple of 3 and the other (5) is not EVEN
Could you explain this scenario, please?
Thanks__
_________________

Hello Asad,

It would be my pleasure to.

In my statement, I have mentioned that at least one of them should be a multiple of 2 and the other should be a multiple of 3. I see that you have recognized an obvious flaw in the language I used, isn’t it? Thanks for correcting my language here.

I should have probably said “ One of them should be a multiple of 2 and one of them should be a multiple of 3.” This would take care of the case that you have taken up i.e. the same number being a multiple of both 2 and 3.

Regardless of that, I don’t see how it changes the answer. Because if n = 5 \(n^3\)-n = 120 which is divisible by 4. But we have already seen that for other values of n, like n = 3, \(n^3\)-n is not divisible by 4 although n(n+1) is divisible by 6. So, essentially, you have just given me a YES case whereas there’s already a NO case to prove that Statement II alone is insufficient.
Hope that answers your query.

Thanks!
_________________

n^3 -n = n (n^2 -1) = (n-1) n (n+1)....so its the product of 3 consecutive integers. Will be divisible by 4 if there is two even integers or has a multiple of 4.

1) since the lowest value of n is 3. So, n^3 -n will contain two even integers and always is divisible by 4. sufficient

2) n (n +1)= 6k,when k =1, n =2 and n + 1 =3, not divisible by 4. But when n = 3, n+1 =4, n63-n is divisible by 4. not sufficient.
A is the answer.

Solution:

We need to determine whether n^3 - n is divisible by 4. Notice that n^3 - n can be factored as n(n^2 - 1) = n(n - 1)(n + 1), which can be expressed as a product of three consecutive integers: (n - 1)(n)(n + 1).

Statement One Alone:

n = 2k + 1, where k is an integer

Statement one tells us that n is an odd integer. Thus, (n - 1) and (n + 1) are both even. We have (n - 1)(n)(n + 1) as even x odd x even, and so each even factor contributes at least one 2 to the product. Therefore, n^3 – n is divisible by 4.

(Note that even if n = 1, resulting in a product of 0, statement one still holds because zero is divisible by 4.)

Statement one alone is sufficient.

Statement Two Alone:

n^2 + n is divisible by 6

If n = 2, we see that n^2 + n = 4 + 2 = 6 is divisible by 6. However, n(n - 1)(n + 1) = 2(1)(3) = 6 is not divisible by 4.

On the other hand, if n = 3, then n^2 + n = 9 + 3 = 12 is also divisible by 6. In this case, n(n - 1)(n + 1) = 3(2)(4) IS divisible by 4.

Statement two alone is not sufficient.

Answer: A
_________________

My solution was exactly the same, but why are we able to firmly conclude that (n-1)n(n+1) is divisible by 4 without plugging in numbers?

(n-1)n(n+1) does turn out to be even, but which even number?

For example, if you plug in n= 1 then you get 0 / 4 so NO, but n = 3 YES
_________________

CEdward, 0 is divisible by 4.