How Can the Product of Factorials from 1 to 15 Be Expressed as \(A^2B!\)?

[anemone]

Here is this week's POTW:

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Find $A$ and $B$ such that the product $$\prod_{n=1}^{15} (n+1)!$$ can be written in the form $A^2B!$, where $A,\,B$ are positive integers. -----

Remember to read the http://www.mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to http://www.mathhelpboards.com/forms.php?do=form&fid=2!

 

[anemone]

Congratulations to the following members for their correct solution:

1. kaliprasad
2. greg1313 Solution from kaliprasad:

Spoiler

Because $(n+1)! = n! * (n+1)$
we have
$ \prod_{n=1}^{15}(n+1)!=2! * (3!)^2 * 4 * (5!)^2 * 6 * (7!)^2 * 8 * (9!)^2 * 10 * (11!)^2 * 12 * (13!)^2 * 14 * (15!)^2 * 16$
= $2! * (3! * 5! * 7!*9!*11!*13!*15!)^2 * 4 * 6 * 8 * 10 * 12 *14 * 16$
= $2! * (3! * 5! * 7!*9!*11!*13!*15!)^2 * 2^2 * 2* 3 * 2^3 * 2* 5 * 2^2 * 3 * 2* 7 * 2^4$
= $2! * (3! * 5! * 7!*9!*11!*13!*15!)^2 * 2^{14}* 3^2 * 5 * 7$
(there is an isolated 7 and an isolated 5 on the expression so this should get absorbed in factorial)
= $3! * (3! * 5! * 7!*9!*11!*13!*15!)^2 * 2^{14}* 3 * 5 * 7$
= $4 ! * (3! * 5! * 7!*9!*11!*13!*15!)^2 * 2^{12}* 3 * 5 * 7$
= $5! * (3! * 5! * 7!*9!*11!*13!*15!)^2 * 2^{12}* 3 * 7$
= $6! * (3! * 5! * 7!*9!*11!*13!*15!)^2 * 2^{11} * 7$
= $7! * (3! * 5! * 7!*9!*11!*13!*15!)^2 * 2^{11}$

(Now we have 2^11 is not a power of 2 but 2^3 =8 so we get)

= $8! * (3! * 5! * 7!*9!*11!*13!*15!)^2 * 2^8$ solution (1)

as $3^3 = 9$ so we get also

= $9! * (2 * 5! * 7!*9!*11!*13!*15!)^2 * 2^8$ solution (2)

so there are 2 solutions
solution 1

A = 3! * 5! * 7!*9!*11!*13!*15! * 2^4 and B = 8

solution 2

A = 32 * 5! * 7!*9!*11!*13!*15! and B = 9