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

Missinga · 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

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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