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
OA C
Source: Magoosh
If n is an integer greater than 10, then the expression (n^2
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-----ASIDE---------------------BTGmoderatorDC wrote:If n is an integer greater than 10, then the expression (n² - 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
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
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Brent
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Simplifying we have:BTGmoderatorDC wrote: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
OA C
Source: Magoosh
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 taking n = 3.
Answer: C
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The question says n is greater than 10. Then, how can we take n = 3?
Scott@TargetTestPrep wrote:Simplifying we have:BTGmoderatorDC wrote: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
OA C
Source: Magoosh
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 taking n = 3.
Answer: C
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Simplifying, we have:BTGmoderatorDC wrote: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
OA C
Source: Magoosh
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
Scott Woodbury-Stewart
Founder and CEO
[email protected]
See why Target Test Prep is rated 5 out of 5 stars on BEAT the GMAT. Read our reviews