If n is an integer greater than 10, then the expression (n^2 - 2n)(n +

Question

If n is an integer greater than 10, then the expression (n^2 - 2n)(n + 1)(n - 1) MUST be divisible by which of the following?

I. 4
II. 6
III. 18

A. I only
B. II only
C. I and II only
D. II and III only
E. I, II, and III

pushpitkc · Accepted Answer

The expression ( n^2 -2n)(n+1)(n-1) can be expressed as n(n-2)(n-1)(n+1),
which is the product of 4 consecutive integers. It has also been given that the number n>10

We know that the product of 4 consecutive numbers is always divisible by  24(2^3*3) 
 Rule : The product of any n consecutive integers will be always divisible by n!
 
Hence, the expression must be divisible by  4(2^2)  and  6(2*3)  but not  18(2*3^2) 
The answer is  Option C(i and ii only)

ehsan090 · Answer

pushpitkc  can you please share a link related to shared rule please 
TIA

generis · Answer

ehsan090  , the rule is under  Consecutive Integers , here: 
 
 https://gmatclub.com/forum/math-number-theory-88376.html

Dkingdom · Answer

I took two values, i.e. 11 and 12 and applied them to the given formula. Only 4 and 6 satisfied both. While 18 satisfied only when n=11.

BrentGMATPrepNow · Answer

-----ASIDE---------------------
There's a nice rule says:   The product of k consecutive integers is divisible by k, k-1, k-2,...,2, and 1  
So, for example, the product of any 5 consecutive integers will be divisible by 5, 4, 3, 2 and 1
Likewise, the product of any 11 consecutive integers will be divisible by 11, 10, 9, . . . 3, 2 and 1
NOTE: the product may be divisible by other numbers as well, but these divisors are guaranteed.
---------------------------------

The trick here is the see that (n² - 2n)(n + 1)(n - 1) is actually  the product of 4 consecutive integers 
Since we can factor (n² - 2n) as n(n - 2), we get: (n² - 2n)(n + 1)(n - 1) = (n)(n - 2)(n + 1)(n - 1)

Now rearrange the terms to get: (n - 2)(n - 1)(n)(n + 1) 
Notice that n-2, n-1, n and n+1 represent 4 consecutive integers

According to the above   rule  , the product must be divisible by 4, 3, 2 and 1

So, the product is definitely  divisible by 4

Since we know that the product is divisible by 2 AND 3, we know that it's  divisible by 6

What about 18? Let's test a values of n.
If n = 13, then the product (n - 2)(n - 1)(n)(n + 1) becomes (11)(12)(13)(14)
Is that divisible by 18? It's hard to tell. Let's find the prime factorization of each value in the product. 
 (11)(12)(13)(14) = (11)(2)(2)(3)(13)(2)(7)
We can see that this is NOT divisible by 18 (since the product does NOT include two 3's and one 2)

Answer: C

Cheers,
Brent

steffe440 · Answer

Factor the expression:

(n^2 - 2n)(n + 1)(n - 1)

= n(n - 2)(n + 1)(n - 1)

Reorder the factors:

(n - 2)(n - 1)n(n + 1)

This is the product of 4 consecutive integers.

The product of 4 consecutive integers must be divisible by 4!, which is 24.

Since 24 is divisible by both 4 and 6, the expression must be divisible by 4 and by 6.

So statements I and II must be true.

Now check statement III: must the expression be divisible by 18?

Divisible by 18 requires the product to contain 2 × 3 × 3, meaning it needs two factors of 3.

But 4 consecutive integers do not always contain two factors of 3.

For example, let n = 13. Then the expression becomes:

(11)(12)(13)(14)

This product has only one factor of 3, coming from 12. So it is not divisible by 18.

Therefore III does not have to be true.

The correct answer is C: I and II only.

If n is an integer greater than 10, then the expression (n^2 - 2n)(n +

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If n is an integer greater than 10, then the expression (n^2 - 2n)(n +

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

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My Rewards