AustinKL wrote:
What is the smallest positive integer n for which \(\sqrt{n * 7!}\) is an integer?
A. 14
B. 35
C. 70
D. 105
E. 210
IMPORTANT CONCEPTS:
#1) In order for √K to be an integer, K must be a perfect square (e.g., 1, 4, 9, 16, 25, 36, 49, etc)
#2) The prime factorization of a perfect square will have an even number of each prime
For example: 400 is a perfect square.
400 = 2x2x2x2x5x5. Here, we have four 2's and two 5's
This should make sense, because the even numbers allow us to split the primes into two EQUAL groups to demonstrate that the number is a square.
For example: 400 = 2x2x2x2x5x5 = (2x2x5)(2x2x5) = (2x2x5)²
Likewise, 576 is a perfect square.
576 = 2X2X2X2X2X2X3X3 = (2X2X2X3)(2X2X2X3) = (2X2X2X3)²---ONTO THE QUESTION!!----------------------------------------
For the original question, we must recognize that n * 7! must be a perfect square.
n * 7! = (n)(7)(6)(5)(4)(3)(2)(1)
= (n)(7)(3)(2)(5)(2)(2)(3)(2)
= (n)(2)(2)(2)(2)(3)(3)(5)(7)
As we can see, we already have an EVEN number of 2's and 3's.
So, to get an even number of all of the primes (and thus create a perfect square), we need to add a 5 and a 7 to the prime factorization of n * 7!
So, let
n = (5)(7) = 35
This means n * 7! = (2)(2)(2)(2)(3)(3)(5)
(5)(7)
(7), in which case it is a perfect square AND \(\sqrt{n * 7!}\) is an integer
Answer: B
Cheers,
Brent
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