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 follo - The Beat The GMAT Forum

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 follo

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BTGModeratorVI: Legendary Member; Posts: 1223; Joined: Sat Feb 15, 2020 2:23 pmFollowed by: 1 member

by BTGModeratorVI » Sat Apr 18, 2020 9:22 am

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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

Answer: C
Source: Magoosh

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Re: 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 f

by Brent@GMATPrepNow » Sun Apr 19, 2020 6:14 am

BTGModeratorVI wrote: ↑
Sat Apr 18, 2020 9:22 am
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

Answer: C
Source: Magoosh

-----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

Brent Hanneson - Creator of GMATPrepNow.com

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Scott@TargetTestPrep: GMAT Instructor; Posts: 8156; Joined: Sat Apr 25, 2015 10:56 amFollowed by: 29 members; Location: Los Angeles, CA; Thanked: 43 times

Re: 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 f

by Scott@TargetTestPrep » Thu Apr 23, 2020 3:11 am

BTGModeratorVI wrote: ↑
Sat Apr 18, 2020 9:22 am
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

Answer: C
Source: Magoosh

Simplifying, we have:

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

So we have the product of 4 consecutive integers which must be divisible by 4! = 24. Thus, (n^2 - 2n)(n + 1)(n - 1) MUST be divisible by 4 and 6. We can easily see that the product need not be divisible by 18 by letting n = 12.

Answer: C

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