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

Question

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

Skywalker18 · Accepted Answer

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
 Answer B

chetan2u · Answer

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

drasticgre · Answer

awaiting for the reveal.
TIA.

suravi · Answer

it seems the calculation is lengthy. waiting for official solution.

saikiransomala · Answer

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

Regards

9SidMn · Answer

B , by reducing to powers of primes . took 5 mins.

KrishnakumarKA1 · Answer

(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

akshayk · Answer

Is there a shorter way to do this?

chesstitans · Answer

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.

saukrit · Answer

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

MsInvBanker · Answer

The answer is B, 672.

saukrit · Answer

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

bumpbot · Answer

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

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