Perfect square divisors

Let $N = 1!2!3!4!5!6!7!8!9!$

we took out the product of each number

$N = 1^9 \times 2^8 \times 3^7 \times 4^6 \times 5^5 \times 6^4 \times 7^3 \times 8^2 \times 9^1$

Then we do prime factorization, hence

$N = 2^{30} \times 3^{13} \times 5^{5} \times 7^{3}$

now, we need to find how many perfect squares that divide $N$ . This is similar to finding the number of divisor of a positive integer N, but we need the power of primes to be even (so the resulting divisor would be a perfect square)

Thus

from $0$ to $30$ there are 16 even numbers

from $0$ to $13$ there are 7 even numbers

from $0$ to $5$ there are 3 even numbers

from $0$ to $3$ there are 2 even numbers

then, the answer is $16 \times 7 \times 3 \times 2 = \boxed{672}$

N = $1! 2! 3! 4! 5! 6! 7! 8! 9!$ =

$2*$

$2*3*$

$2*3*4*$

$2*3*4*5*$

$2*3*4*5*6*$

$2*3*4*5*6*7*$

$2*3*4*5*6*7*8*$

$2*3*4*5*6*7*8*9$

looking down those vertical columns. That's equal to:

$2^8*3^7*4^6*5^5*6^4*7^3*8^2*9^1$

Break the composites into primes:

$2^8*3^7*(2^2)^6*5^5*(2*3)^4*7^3*(2^3)^2*3^2$ =

$2^8*3^7*2^{12}*5^5*2^4*3^4*7^3*2^6*3^2$ =

$2^{30}*3^{13}*5^5*7^3$

Every divisor of N is of the form $2^p*3^q*5^r*7^s$ , where

$0\leq p\leq 30$

$0\leq q\leq 13$

$0\leq r\leq 5$

$0\leq s\leq 3$

The divisors which are perfect squares have even exponents, (including $0$ ).

$0\leq p\leq 30$ contains $16$ even exponents and $15$ odd exponents

$0\leq q\leq 13$ contains $7$ even exponents and $7$ odd exponents

$0\leq r\leq 5$ contains $3$ even exponents and $3$ odd exponents

$0\leq s\leq 3$ contains $2$ even exponents and $2$ odd exponents

Therefore, there are $16$ choices for $p$ , $7$ choices for $q$ , $3$ choices for $r$ , and $2$ choices for $s$ .

So, the answer is $16*7*3*2 = \boxed{672}$ perfect square divisors of N.

1!2!3!4!5!6!7!8!9! = 2^30 * 3 ^13 * 5^5 * 7^3

From 2^30 we can have 16 perfect squares. (Including 1)

From 3^13 we can have 7 perfect squares.

From 5^5 we can have 3 perfect squares.

From 7^3 we can have 2 perfect squares.

Total perfect squares divisors = 16 * 7 * 3 * 2 = 672

How many perfect squares are divisors of the product

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

2* 2 3 2 3 4* 2 3 4 5 2 3 4 5 6* 2 3 4 5 6 7 2 3 4 5 6 7 8* 2 3 4 5 6 7 8*9

looking down those vertical columns that's equal to:

(2^8)(3^7)(4^6)(5^5)(6^4)(7^3)(8^2)(9^1)

Break the composites into primes:

(2^8)(3^7)((2^2)^6)(5^5)((2*3)^4)(7^3)((2^3)^2)(3^2) =

(2^8)(3^7)(2^12)(5^5)(2^4*3^4)(7^3)(2^6)(3^2) =

(2^30)(3^13)(5^5)(7^3)

Every divisor of N is of the form (2^p)(3^q)(5^r)(7^s), where

0 <= p <= 30 0 <= q <= 13 0 <= r <= 5 0 <= s <= 3

The divisors which are perfect squares have even exponents, (including 0).

0 <= p <= 30 contains 16 even exponents and 15 odd exponents 0 <= q <= 13 contains 7 even exponents and 7 odd exponents 0 <= r <= 5 contains 3 even exponents and 3 odd exponents
0 <= s <= 3 contains 2 even exponents and 2 odd exponents

Therefore for divisor (2^p)(3^q)(5^r)(7^s),

there are 16 choices for p, 7 choices for q, 3 choices for r, and 2 choices for s.

Answer: 16 7 3*2 = 672 perfect square divisors of N.

$$1!2!3!4!5!6!7!8!9!=2^{30}3^{13}5^5 7^3$$

and the answer is $$( \lfloor 30/2 \rfloor +1)(\lfloor 13/2 \rfloor+1)(\lfloor 5/2 \rfloor + 1)(\lfloor 3/2 \rfloor +1)=672 $$

1!2!3!4!5!6!7!8!9!=2^8 * 3^7 * 4^6 * 5^5 * 6^4 * 7^3 * 8^2 * 9=2^30 * 3^13 * 5^5 * 7^3. Now to get perfect squares we need to take combinations of even powers of all terms including 0. So,in our factor we can take product of 0 or 2 or 4 or.........28 or 30 number of 2s.Clearly 16 such ways. Similarly for 3:7 ways,for 5:3 ways,and for 7:2 ways to gets combinations of even powers. As all the events are related we multiply.Leading us to the result 16 7 3*2=672.

Notice that (1!2!3!4!5!6!7!8!9!) could be rewrite as: 9.(8²).(7³).(6^4).(5^5).(4^6).(3^7).(2^8).(1^9)

Next step: Put this in prime factors: (3²).(2^6).(7³).(2^4.3^4).(5^5).(2^12).(3^7).(2^8).1 = (2^30).(3^13).(5^5).(7^3)

Divisors of (1!2!3!4!5!6!7!8!9!): 2^a . 3^b . 5^c . 7^d

Perfect square is a number that your prime power is even. Then: 2's power: 0<=a<=30, -> 30/2 +1 divisors: 16. 3's power: 0<=b<13, -> 12/2 + 1 divisors: 7. 5's power: 0<= c < 5, -> 4/2 + 1 divisors: 3. 7's power: 0<=d < 3, 2/2 + 1 divisors: 2.

By Rule of Product: 17.7.3.2 divisors are perfect square here.

Nice problem :)

the product can be written as $2^{30}3^{13}5^57^3$ and we can write all even powers of these primes like this:

$2^0, 2^2, 2^4 ... 2^{30}$

$3^0, 3^2, 3^4 ... 3^{12}$

$5^0, 5^2, 5^4$

$7^0, 7^2$

from the first set we have 16 perfect squares, from the second we have 7, from third we have 3 and from the last we have 2. Since a perfect square can be a product of 2 (or more) perfect squares ( $5^4\times 2^8$ ) we need to check for all the possible values witch is $16\times 7\times 3\times 2=\boxed{672}$

It's easy see that this product is equal: 2^30 * 3^13 * 5^5 * 7^3 , => by basic counting the answer is 16 7 3*2=672.

Let $n=1!2!3!4!5!6!7!8!9!$ . $n=1^{9} \cdot 2^8 \cdot 3^7 \cdot 4^6 \cdot 5^5 \cdot 6^4 \cdot 7^3 \cdot 8^2 \cdot 9=2^{30} \cdot 3^{13} \cdot 5^5 \cdot 7^3$ . A perfect square divisor of $n$ is of the form $N=2^{2r_2} \cdot 3^{2r_3} \cdot 5^{2r_5} \cdot 7^{2r_7}$ where $r_2,r_3, r_5,$ and $r_7$ are integers such that $0\leq r_2 \leq 15$ , $0 \leq r_3 \leq 6$ , $0 \leq r_5 \leq 2$ and $0\leq r_7 \leq 1$ . There are $16 \cdot 7 \cdot 3 \cdot 2$ possibilities for $N$ .

We must know if 1!2!3!4!5!6!7!8!9! = 2^30 x 3^13 x 5^5 x 7^3. Then, we also know if perfect squares have exponent 2n, so.. 2^30 x 3^13 x 5^5 x 7^3 that a perfect squares are 2^30 x 3^12 x 5^4 x 7^2 ---> {2^15 x 3^6 x 5^2 x 7^1}^2 ---> Now, We have many exponent 15,6,2,1. To find many divisors of it product is (exponent1 + 1)...(exponent n + 1) ---> (15+1)x(6+1)x(2+1)x(1+1) = 16 x 7 x 3 x 2 = 672. Answer : 672. HAPPY CHRISTMAS AND NEW YEAR DAY 2014 ^ ^

=2^30x3^13x5^5x7^3 therefore you have a choice of 16x7x3x2 perfect squares =672