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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? [#permalink]

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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
[Reveal] Spoiler: OA

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Re: What is the smallest positive integer n for which √n*7! is an integer? [#permalink]

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New post 22 Feb 2017, 00:07
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

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Re: What is the smallest positive integer n for which √n*7! is an integer? [#permalink]

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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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Re: What is the smallest positive integer n for which √n*7! is an integer? [#permalink]

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New post 22 Feb 2017, 10:24
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


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

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

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

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

Thus, n must be 35, answer must be (B) 35
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Re: What is the smallest positive integer n for which √n*7! is an integer? [#permalink]

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New post 15 Apr 2017, 17:27
ziyuen 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


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

[square_root] 7 x 6 x 5 x 4 x 3 x 2 x 1
[square_root] 7 x (3 x 2) x 5 x (2 x 2) x 3 x 2 x 1
[square_root] 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

[square_root] 5^2 x 7^2 x 3^ 2 x 2^4

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

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Re: What is the smallest positive integer n for which √n*7! is an integer? [#permalink]

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New post 19 Jun 2017, 08:51
hazelnut 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


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"

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Re: What is the smallest positive integer n for which √n*7! is an integer? [#permalink]

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New post 19 Jun 2017, 09:59
hazelnut 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


\(7! = 7*6*5*4*3*2\)

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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Re: What is the smallest positive integer n for which √n*7! is an integer?   [#permalink] 19 Jun 2017, 09:59
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