If n is a positive integer, what must be true of n^3 n?

Question

If  n  is a positive integer, what must be true of  n^3 – n ?

(A) It is divisible by  4 .
(B) It is odd.
(C) It is a multiple of  6 .
(D) It is a prime number.
(E) It has, at most, two distinct prime factors.

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Sketchitesh · Answer

n = +ve integer 
 n^3 – n  = (n-1)(n)(n+1) => 3 consecutive integers

0*1*2
1*2*3
2*3*4
4*5*6
5*6*7 ....

1. Divisible by 4 (Not necessary 1*2*3)  Incorrect  
2. Odd (Will always be even, since each term will definitely have 2 as factor  Incorrect 
3. is multiple of 6 (All terms will have always 2 and 3 as factor, consecutive integers, except first which is 0)  Correct 
4. It is prime (Not necessary 2*3*4 = 24)  Incorrect  
5. At most, two distinct prime factor (5*6*7 have 2,3,5,7 as factors)  Incorrect

RDM42 · Answer

(n^3)-n 
n(n-1)(n+1) 
Let, n= 1 ; (1)*(0)*(2)=0
n= 2 ; (2)*(1)*(3)= 6 
n= 3 ; (3)*(2)*(4)= 24
n= 4 ; (4)*(3)*(5)= 60 
......and so on 
As, we can see from above only option C (It is a multiple of 6) holds true

Answer: C

NilayMaheshwari · Answer

n^3-n = n(n^2-1) = n(n+1)(n-1) = (n-1)n(n+1)
which is also the product of 3 consecutive integers.

And any product of "n" consecutive integers is always divisible by n!
Thus, above expression will always be divisble by 3! that is 6.
Thus, C

Dereno · Answer

The question asked is : If  n  is a positive integer, what must be true of  n^3 – n ?

let’s expand  n^3 – n

n (n^2– 1)

n (n+1)(n-1)

this can be rearranged as :    (n-1)* n *(n+1)   . Which is nothing but   product of three Consecutive   terms.

Options:

(A) It is divisible by  4 .

Case: n=2 , the equation becomes 1*2*3 =6 not divisible by 4. Wrong

(B) It is odd.

Product Of three consecutive integers, at least one term is even. The output will be even. Hence, wrong.

(C) It is a multiple of  6 .

This is the answer.  THE PRODUCT OF THREE CONSECUTIVE POSITIVE INTEGERS IS A MULTIPLE OF 6. 
 
(D) It is a prime number.

Case: n=2 , the equation becomes 1*2*3 =6. The number 6 is Not a prime number.

(E) It has, at most, two distinct prime factors.

Case : if n =6 , the equation becomes = 5*6*7 . The prime factors are 2,3,5,7 ( total 4 factors). Hence, wrong.

OPTION C

ManifestDreamMBA · Answer

n^3 – n 
 n(n^2 – 1)  
 (n-1)n(n+1)

We know 3 consecutive integers are divisible by 3! = 6

Answer C

WarpCode · Answer

Should it be stated in the question that it's a positive integer greater than 1? If I plug in 1, C is not true.

Bunuel · Answer

No. For n = 1, we ger n^3 - n = 0, which  is  a multiple of 6.

ZERO:

1. Zero is an  INTEGER .

2. Zero is an  EVEN  integer.

3. Zero is  neither positive nor negative  (the only one of this kind)

4. Zero is  divisible by EVERY integer except 0 itself  ( \frac{0}{x} = 0 , so 0 is a divisible by every number, x).

5. Zero is a  multiple of EVERY integer  ( x*0 = 0 , so 0 is a multiple of any number, x)

6. Zero is  NOT a prime number  (neither is 1 by the way; the smallest prime number is 2).

7. Division by zero is NOT allowed:  anything/0 is undefined .

8. Any non-zero number  to the power of 0 equals 1  ( x^0 = 1 )

9.  0^0  case is  NOT tested on the GMAT .

10. If the exponent n is positive (n > 0),  0^n = 0 .

11. If the exponent n is negative (n < 0),  0^n  is undefined, because  0^{negative}=0^n=\frac{1}{0^{(-n)}} = \frac{1}{0} , which is undefined.  You CANNOT take 0 to the negative power .

12.  0! = 1! = 1 .

Hope it helps.

egmat · Answer

Hi WarpCode,

Good eye for stress-testing the answer with the smallest case, but here's the piece that resolves it: when you plug in  n = 1 , you get  (0)(1)(2) = 0 , and  0  is a multiple of  6 .

That's the whole knot. A number is a "multiple of 6" if it can be written as  6 × (some integer) . Since  0 = 6 × 0 , zero qualifies - it's a multiple of  6  (and of every integer).

So C still holds at n = 1. Nothing is missing from the question, and there's no need to restrict it to integers greater than  1 .

Why 0 behaves this way

Divisibility asks whether the division comes out to a whole number with no remainder.  0 ÷ 6 = 0 , exactly, no remainder - so  6  divides  0  cleanly. In fact,  0  is divisible by  any  nonzero integer for the same reason.

That's why the posted solutions using  (n-1)·n·(n+1)  are safe even at the n = 1 end: the product being  0  doesn't break the rule, it satisfies it.

Quick check to lock it in:

- Is  0  a multiple of  4 ? (0 = 4 × 0 - yes)
- Is  0  a multiple of  7 ? (0 = 7 × 0 - yes)

Same idea every time: zero is a multiple of everything, so it never trips up a "must be a multiple of..." question.

Answer: C

alamin8 · Answer

I believe you have seen bunuel's zero related all the possible facts in GMAT! Thanks  Bunuel  for reminding those.

If n is a positive integer, what must be true of n^3 n?

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If n is a positive integer, what must be true of n^3 n?

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