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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Brent Hanneson – Founder of gmatprepnow.com