If n is a positive integer, then n(n+1)(n+2) is

(E) divisible by 4 whenever n is even

If n is even => even x odd x even (Prod of two even numbers always divisible by 2x2)

Manhattan NP covers these well.

Will be of the form

Odd, even, odd = even
Even, odd, even = even

Can quickly rule out all but E

Posted from my mobile device

We can generalize:

If n is even, then n + 2 is also even and n and n + 2 are consecutive even numbers.

2 * 2 = 4, so any two even numbers multiplied together will yield a multiple of 4.

Therefore, any two consecutive even numbers multiplied together will yield a multiple of 4.

THEREFORE, if n is even, n(n + 2) is always a multiple of 4.

But actually, we can go a step further (this may be useful for some future problems):

Every second even number is a multiple of 4. Therefore, given any two consecutive even numbers, one of them will always be a multiple of 4.

4 * 2 is 8, so any multiple of 4 multiplied by another even number will yield a multiple of 8.

Therefore, any two consecutive even numbers multiplied together will yield a multiple of 8.

THEREFORE, if n is even, n(n + 2) is always a multiple of 8.

n(n+1)(n+2) is the product of three consecutive integers because n is an integer.

0,1,2
-200,-199,-198
100,101,102
-1,0,1

In any set of three consecutive numbers, there must be at least one odd and one even.

odd,even,odd
OR
even,odd,even

A)even only when n is even
The product of three or more consecutive integers will always be EVEN. To make the product even, we just need one even. It really doesn't matter whether n is even or n+1.

If n is even, say 0
0,1,2. product=0=even

If n is odd, say -1
-1,0,1. product=0=even

Saying that n(n+1)(n+2) will be even ONLY if n=even is NOT correct.

B)even only when n is odd

We just saw that the product will always be even irrespective of whether n is even or odd.

C)odd whenever n is odd

Product will never be odd.

D)divisible by 3 only when n is odd
Rule: Product of n consecutive number will always be divisible be n!

{1,2}: Two numbers. n=2
1*2 will be divisible by 2!=2

{45,46,47,48,49,50}: Six numbers. n=6
45*46*47*48*49*50 will be divisible by 6!=720

Similarly,
3 consecutive numbers: {1,2,3}
1*2*3 will be divisible by 3!=6
If the product is divisible by 6, it must be divisible by its factor, which is 3.

Thus, "n" can be even/odd.
FALSE.

E)divisible by 4 whenever n is even
n=2
2,3,4. Product=24

TRUE.

Ans: "E"

Hi All,

This question can be solved by TESTing VALUES (although you'll need to TEST at least 2 options to properly eliminate all of the wrong answers.

IF...
N = 1, then (N)(N+1)(N+2) = (1)(2)(3) = 6
N = 2, then (N)(N+1)(N+2) = (2)(3)(4) = 24

Answer A is eliminated by the 1st option
Answer B is eliminated by the 2nd option
Answer C is eliminated by the 1st option
Answer D is eliminated by the 2nd option
There's only one option remaining...

Final Answer:

GMAT assassins aren't born, they're made,
Rich

Solution

Given
In this question, we are given that

• The number n is a positive integer

To find
We need to determine

• With respect to the given expression n (n + 1) (n + 2), which option is correct

Approach and Working out
Option A: even only when n is even

• If we take n = 3 (odd), then the expression becomes odd x even x odd = even

Hence, option A is not true

Option B: even only when n is odd

• If we take n = 4, then the expression becomes even x odd x even = even

Hence, option B is not true

Option C: odd whenever n is odd

• If we take n = 3, then the expression becomes odd x even x odd = even

Hence, option C is not true

Option D: divisible by 3 only when n is odd

• If we take n = 6, then the expression becomes even x odd x even = even and divisible by 6

Hence, option D is not true

By default, answer choice becomes E
Option E: divisible by 12 whenever n is even

• If we take n = 4, then the expression becomes 4 x 5 x 6 = 12 x 2 x 5

Hence, option E is true

Thus, option E is the correct answer.

Correct Answer: Option E

The problem is easiest to solve by substituting numbers for n. We'll try an odd number first and then an even number.

For an odd number, let’s let n = 1: 1(1+1)(1+2) = 1(2)(3) = 6.
We see than choices A and C can’t be the correct choices. Choice A is false because, while the product is even, n is not even. Choice C is false because, while n is odd, the product is even.

For an even number, let’s let n = 2: 2(2+1)(2+2) = 2(3)(4) = 24.

We see than choices B and D can’t be the correct choices. Choice B is false because, while the product is even, n is not odd. Choice D is false because, while the product is divisible by 3, n is even.

Therefore, the only correct answer is choice E. When n is even, the product is indeed divisible by 4.

The answer is E.

Three remarks here:

1. Zero is neither positive, nor negative integer.

2. Division by zero is not allowed (number/0 is undefined) but we can divide 0 by any non-zero number (0/number=0).

3. Zero is divisible by EVERY integer except zero itself, since 0/integer=0=integer (or, which is the same, zero is a multiple of every integer). Thus even if we were told that n is integer (so if n could be a negative integer, zero or a positive integer) n(n + 1)(n + 2) would still be divisible by 4 for any even n.

Hope it's clear.

Here, n, (n+1), and (n+2) are 3 consecutive integers

The product of 3 consecutive positive integers is always divisible by 3!

Note: The product of k consecutive positive integers is always divisible by k!

Example:
The product of 2 consecutive positive integers is always divisible by 2!
The product of 3 consecutive positive integers is always divisible by 3!
The product of 4 consecutive positive integers is always divisible by 4! and so on

Why?
Considering positive integers:
# When we have 2 consecutive integers, one will be even and the other odd => Their product is even - divisible by 2 = 2!
# When we have 3 consecutive integers, one number will always be a multiple of 3. Also, 3 consecutive integers will include 2 consecutive integers.Thus, the product will be divisible by 2 x 3 => Their product is divisible by 6 = 3!
# When we have 4 consecutive integers, one number will always be a multiple of 4. Also, 4 consecutive integers will include 3 consecutive integers.Thus, the product will be divisible by 3! x 4 => Their product is divisible by 4!


In the above question:
n(n + 1)(n + 2) must be divisible by 3! = 6 (since one number must be a multiple of 3 and at least one number is even)
Thus, n(n + 1)(n + 2) is always even and always a multiple of 3
Thus, Options A, B, C, and D are incorrect.

Now, if n is even, (n + 2) is also even
Thus, n(n + 1)(n + 2) must be divisible by 3 x 2 x 2 = 12

Answer E

n(n+1)(n+2) will always be even as n is a +ve integer so that rules out A, B & C. Atleast one of n, n+1 & n+2 will be even as they are consecutive integers.

even * even is always even e.g 2*4 = 8 or 6*10 = 60 always even
even * odd is always even e.g 2*3 = 6 or 5 * 8 = 40 always even

Either of n, n+1 & n+2 will always be divisible by 3 till the time n is a +ve integer and they are consecutive integers.

Hence that leaves us with E as answer.

Also we can prove it like this way also,
First +ve even integer is 2 and not 0 (0 is neither +ve nor -ve).

so n*n+1*n+2 = 2*3*4 divisible by 4.
or if n=6 then 6*7*8 again divisible by 4.

so is E.

can we consider 0 as positive integer. If yes then whole number will become 0 and we can not divide 0 by 4.

Please clarify
Thanks.

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