If x is a positive integer greater than 2, the product of x consecutive positive integers must be divisible by which of

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If x is a positive integer greater than 2, the product of x consecutive positive integers must be divisible by which of the following?

I. x - 1
II. 2x
III. x!

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

Answer: E
Source: Economist GMAT
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BTGModeratorVI wrote:
Sat Apr 18, 2020 9:15 am
If x is a positive integer greater than 2, the product of x consecutive positive integers must be divisible by which of the following?

I. x – 1
II. 2x
III. x!

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

Answer: E
Source: Economist GMAT
So, we have x ≥ 3.

Product of x consecutive positive integers = 1*2*3*4* ... *(x – 1)*x = x!

Let's see each statement one by one.

I. x –1: Note that x! is always divisible by (x – 1). True.

II. 2x: With x ≥ 3, the minimum value of x! = 1*2*3 = 6, which is divisible by 2x= 2*3 = 6. True.

III. x!: Already seen that it's true.

The correct answer: E

Hope this helps!

-Jay
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BTGModeratorVI wrote:
Sat Apr 18, 2020 9:15 am
If x is a positive integer greater than 2, the product of x consecutive positive integers must be divisible by which of the following?

I. x - 1
II. 2x
III. x!

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

Answer: E
Source: Economist GMAT
----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.
----------------------------------

So, the product of x consecutive integers must be divisible by x, x-1, x-2, ....3, 2, and 1
So, we can immediately see that the product will be divisible by (x-1) and x! (since x! = (x)(x-1)(x-2).....(3)(2)(1)
So statements I and III are true

What about statement II?
We already know that the product of x consecutive integers must be divisible by x, x-1, x-2, ....3, 2, and 1
Since x > 2, then we know 2 and x are DIFFERENT numbers.
So, the product of x consecutive integers must be divisible x AND by 2.
This means the product must be divisible by 2x
So statement II is true

Answer: E

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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BTGModeratorVI wrote:
Sat Apr 18, 2020 9:15 am
If x is a positive integer greater than 2, the product of x consecutive positive integers must be divisible by which of the following?

I. x - 1
II. 2x
III. x!

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

Answer: E
Source: Economist GMAT
We can adhere to the rule that the product of x consecutive positive integers is divisible by x!. Thus, III is true. Additionally since 2 and x are distinct factors of x!, x! is divisible by 2x. Lastly, since x - 1 will always be a factor of x!, x! is divisible by x - 1. So, I and II are also true.

Answer: E

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