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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