BTGmoderatorDC wrote:For integers x and y, 1 < x < y. Is x^y a factor of 11!?

(1) x > 3
(2) x is prime

OA A

Source: Veritas Prep

Given: 1 < x < y; where x and y are positive integers

We have to determine whether x^y is a factor of 11!.

Let's take each statement one by one.

(1) x > 3

Case 1: Say x = Very big number; thus, y is also a very big number. So, can conclude that x^y would not be a factor of 11!. The answer is No,

Let's see at smaller values of x and y, do we have x^y a factor of 11!?

Case 2: Say x = 4, smallest possible number; thus, y = 5, smallest possible value for y.

x^y = 4^5 = 2^10

Note that 11! = 1.2.3.4.5.6.7.8.9.10.11. There are eight 2's in 11!. Since the exponent of 2^10 is 10 and is greater than 8, x^y cannot be a factor of 11!. The answer is No,

There is no need to explore further such as x^y = 5^6 since greater numbers would obviously be insufficient in 11!.

Unique answer. Sufficient.

(2) x is prime

Case 1: Say = 2 and y = 3; thus, x^y = 2^3. We already saw that 11! has eight 2's, more than sufficient for 2^3 to be its factor. The answer is Yes.
Case 2: Say x = Very big prime number; thus, y is also a very big number. So, can conclude that x^y would not be a factor of 11!. The answer is No,

No unique answer. Insufficient.

The correct answer: A

Hope this helps!

-Jay
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