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# How many perfect squares are divisors of the product (1!)(2!)(3!)(4!)

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Math Expert
Joined: 02 Sep 2009
Posts: 51215
How many perfect squares are divisors of the product (1!)(2!)(3!)(4!)  [#permalink]

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01 Nov 2015, 23:06
4
26
00:00

Difficulty:

95% (hard)

Question Stats:

41% (02:06) correct 59% (02:40) wrong based on 167 sessions

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GMAT Challenge Problems for Students Targeting a High Score:

How many perfect squares are divisors of the product (1!)(2!)(3!)(4!)(5!)(6!)(7!)(8!)(9!)?

(A) 504
(B) 672
(C) 864
(D) 936
(E) 1008

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Posts: 7106
How many perfect squares are divisors of the product (1!)(2!)(3!)(4!)  [#permalink]

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02 Nov 2015, 00:40
2
2
Bunuel wrote:

GMAT Challenge Problems for Students Targeting a High Score:

How many perfect squares are divisors of the product (1!)(2!)(3!)(4!)(5!)(6!)(7!)(8!)(9!)?

(A) 504
(B) 672
(C) 864
(D) 936
(E) 1008

hi,
I can think of way is..
get them in form of power of perfect square and then find the factor...
$$(1!)(2!)(3!)(4!)(5!)(6!)(7!)(8!)(9!)=1*2^8*3^7*4^6*5^5*6^4*7^3*8^2*9^1.. =1*2^8*3^7*2^{12}*5^5*2^4*3^4*7^3*2^6*3^2$$....
$$=1*2^{8+12+4+6}*3^{7+4+2}*5^5*7^3.. =1*2^{30}*3^{13}*5^5*7^3$$
or=$$4^{15}*9^6*3*25^2*5*49*7...$$
separate out the perfect squares..
$$4^{15}*9^6*25^2*49..$$
no of factors= 16*7*3*2=672
ans B
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How many perfect squares are divisors of the product (1!)(2!)(3!)(4!)  [#permalink]

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Updated on: 02 Nov 2015, 00:44
7
N=(1!)(2!)(3!)(4!)(5!)(6!)(7!)(8!)(9!)

1!=1
2!=1*2
3!=1*2*3
4!=1*2*3*4
5!=1*2*3*4*5
6!=1*2*3*4*5*6
7!=1*2*3*4*5*6*7
8!=1*2*3*4*5*6*7*8
9!=1*2*3*4*5*6*7*8*9

N=(2^8)(3^7)(4^6)(5^5)(6^4)(7^3)(8^2)(9^1)
=(2^30)(3^13)(5^5)(7^3)

To find the number of perfect squares each exponent in the prime factorization must be even .
Comparing to the form (2^p)(3^q)(5^r)(7^s).
In this case,
p can take values from 0 to 30
q can take values from 0 to 13
r can take values from 0 to 5
s can take values from 0 to 3

Number of even exponents
p=16
q=7
r=3
s=2

Number of perfect squares= 16*7*3*2 = 672
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Originally posted by Skywalker18 on 02 Nov 2015, 00:38.
Last edited by Skywalker18 on 02 Nov 2015, 00:44, edited 1 time in total.
##### General Discussion
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Joined: 08 Mar 2015
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Re: How many perfect squares are divisors of the product (1!)(2!)(3!)(4!)  [#permalink]

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01 Nov 2015, 23:18
awaiting for the reveal.
TIA.
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Re: How many perfect squares are divisors of the product (1!)(2!)(3!)(4!)  [#permalink]

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02 Nov 2015, 00:13
it seems the calculation is lengthy. waiting for official solution.
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Re: How many perfect squares are divisors of the product (1!)(2!)(3!)(4!)  [#permalink]

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02 Nov 2015, 06:42
Hi,

The value of the expression is 2^30 * 3^ 13 * 5^ 5 * 7^ 3

For deducing the perfect squares that can divide the expression, remove the square roots of each prime factors.

We arrive at 2^ 15 * 3^ 6 * 5^ 2 * 7^ 1

So the number of factors will be (15+1)*(6+1)*(2+1)*(1+1) = 16*7*3*2 = 672

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Re: How many perfect squares are divisors of the product (1!)(2!)(3!)(4!)  [#permalink]

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06 Nov 2015, 12:16
Bunuel wrote:

GMAT Challenge Problems for Students Targeting a High Score:

How many perfect squares are divisors of the product (1!)(2!)(3!)(4!)(5!)(6!)(7!)(8!)(9!)?

(A) 504
(B) 672
(C) 864
(D) 936
(E) 1008

B , by reducing to powers of primes . took 5 mins.
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Re: How many perfect squares are divisors of the product (1!)(2!)(3!)(4!)  [#permalink]

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21 Mar 2017, 01:59
2
(1!)(2!)(3!)(4!)(5!)(6!)(7!)(8!)(9!)

the prime factors are 2,3,5,7
for 2 the power will be 0+1+1+3+3+4+4+7+7 = 30
for 3 the power will be 0+0+1+1+1+2+2+2+4 = 13
for 5 the power will be 0+0+0+0+1+1+1+1+1 = 5
for 7 the power will be 0+0+0+0+0+0+1+1+1 = 3

therefore the number can be expressed as 2^30 . 3^13 . 5^5 . 7^3 or 4^15. 9^6 . 25^2 . 49^1 . 3. 5.7

therefore the total number of square divisor = (15+1)(6+1)(2+1)(1+1) = 16.7.3.2 = 672
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Re: How many perfect squares are divisors of the product (1!)(2!)(3!)(4!)  [#permalink]

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26 Mar 2017, 09:28
Bunuel wrote:

GMAT Challenge Problems for Students Targeting a High Score:

How many perfect squares are divisors of the product (1!)(2!)(3!)(4!)(5!)(6!)(7!)(8!)(9!)?

(A) 504
(B) 672
(C) 864
(D) 936
(E) 1008

Is there a shorter way to do this?
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Re: How many perfect squares are divisors of the product (1!)(2!)(3!)(4!)  [#permalink]

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18 Apr 2017, 18:17
akshayk wrote:
Bunuel wrote:

GMAT Challenge Problems for Students Targeting a High Score:

How many perfect squares are divisors of the product (1!)(2!)(3!)(4!)(5!)(6!)(7!)(8!)(9!)?

(A) 504
(B) 672
(C) 864
(D) 936
(E) 1008

Is there a shorter way to do this?

this is the shortest way to come up with the answer for the question. At the first glance, the question seems complex, but it turns out to be nothing other than a problem of finding the number of factors.
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Re: How many perfect squares are divisors of the product (1!)(2!)(3!)(4!)  [#permalink]

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13 Oct 2018, 04:17
Quote:
therefore the total number of square divisor = (15+1)(6+1)(2+1)(1+1) = 16.7.3.2 = 672

Why are we doing the exponent +1 for each prime factor??
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Re: How many perfect squares are divisors of the product (1!)(2!)(3!)(4!)  [#permalink]

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13 Oct 2018, 05:41
1
The answer is B, 672.
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Re: How many perfect squares are divisors of the product (1!)(2!)(3!)(4!)  [#permalink]

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13 Oct 2018, 06:30
saurabh1921 wrote:
Quote:
therefore the total number of square divisor = (15+1)(6+1)(2+1)(1+1) = 16.7.3.2 = 672

Why are we doing the exponent +1 for each prime factor??

Well, I got the answer to my problem.
For "2", the exponent is 30, it has 16 even parts: 0, 2, 4, 6, 8, 10, 12, ..., 30.
For "3", the exponent is 13, it has 7 even parts: 0, 2, 4, 6, 8, 10, 12.
For "5", the exponent is 5, it has 3 even parts: 0, 2, 4.
For "7", the exponent is 3, it has 2 even parts: 0, 2.

So basically the +1 is for the additional 0 as the exponent for each even part. Reference for this Explanation
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Re: How many perfect squares are divisors of the product (1!)(2!)(3!)(4!) &nbs [#permalink] 13 Oct 2018, 06:30
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# How many perfect squares are divisors of the product (1!)(2!)(3!)(4!)

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