Bunuel wrote:

What is the greatest prime factor of (11! × 10! + 10! × 9!)/111?

A. 2

B. 3

C. 5

D. 7

E. 11

We are trying to find the greatest prime factor of the given value.

Since we can only find the greatest prime factor of an INTEGER, we can conclude that

the numerator must be divisible by 111, so that it cancels out with the denominator.

How does this happen?

First notice that we can factor 10! out of the given numerator

That is, 11! × 10! + 10! × 9! = 10!(11! + 9!)

Next we should recognize that there's also a 9! "hiding" within 11!

11! = (11)(10)

(9)(8)(7)(6)(5)(4)(3)(2)(1)= (11)(10)(

9!)

This means (11! + 9!) = 9![(11)(10) + 1]

So, 11! × 10! + 10! × 9! = 10!(11! + 9!)

= (10!)(9!)[(11)(10) + 1]

= (10!)(9!)[110 + 1]

= (10!)(9!)[111]

So, (11! × 10! + 10! × 9!)/111 = (10!)(9!)[111]/111

= (10!)(9!)

= [(10)(9)(8)(7)(6)(5)(4)(3)(2)(1)][(9)(8)(7)(6)(5)(4)(3)(2)(1)]

= [(2)(5)(3)(3)(2)(2)(2)(

7)(3)(2)(5)(2)(2)(3)(2)(1)][(3)(3)(2)(2)(2)(7)(3)(2)(5)(2)(2)(3)(2)(1)]

The greatest prime factor is

7Answer: D

Cheers,

Brent

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Brent Hanneson – GMATPrepNow.com

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