What is the smallest positive integer n for which √n*7! is an integer?

Question

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

BrentGMATPrepNow · Accepted Answer

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

nailin16 · Answer

B?
Coz root (n*7!)
For this to b integer take pairs which can come out of root..we wl b left wd 5 and 7 ...which need another 5 n 7 to form sqr and come out

Sent from my SM-G600FY using  GMAT Club Forum mobile app

Abhishek009 · Answer

= \sqrt{n*7*6*5*4*3*2}

= \sqrt{n*7*2*3*5*2*2*3*2}

= \sqrt{n*7*5*3^2*2^4}

= 2^2*3\sqrt{n*7*5}

Thus, n must be 35, answer must be (B) 35

Nunuboy1994 · Answer

We can solve this question if we break down the components of 7!

7 x 6 x 5 x 4 x 3 x 2 x 1 
  7 x (3 x 2) x 5 x (2 x 2) x 3 x 2 x 1
  5 x 7 x 3^ 2 x 2^4 x n

However, the exponents of the factors a perfect square MUST be multiples of 2- therefore the factors of N must necessarily contain factors of 5 and 7 so as to make the powers of abundance or powers of 5'2 and 7's in the number squared multiples of 2.

If we analyze 14-
Factors: 1, 2, 7 , 14 - cannot be the answer

If we analyze 35-
Factors: 1, 5, 7 , 35

So if N is 35 then

5^2 x 7^2 x 3^ 2 x 2^4

Now, because the exponents are multiples of two this constitutes a perfect square.

Nunuboy1994 · Answer

We can simply break this apart and apply a basic property of square roots

\sqrt{7 x 6 x 5 x 4 x 3 x 2 x 1}

\sqrt{7 x 6 x 5 x 4 x 6}

\sqrt{7 x 6^2 x 5 x 4^2} - now a basic property of any perfect square root is that the number of odd integers must be even for example 36 = \sqrt{3^2 x 2^2}

If we apply 35 which is 7 and 5 then our answer is

\sqrt{7^2 x 6^2 x 5^2 x 4^2}

Thus
 "B"

Abhishek009 · Answer

7! = 7*6*5*4*3*2

Or,  7! = 7*(2*3)*5*2^2*3*2

Or,  7! = 7*5*3^2*2^4

Thus, the minimum value of  \sqrt{n*7!}  must be 35, answer will be (B)

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What is the smallest positive integer n for which √n*7! is an integer?

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What is the smallest positive integer n for which √n*7! is an integer?

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