Thread: Perfect Square of a Product of Factorials Proof

Sorry about the double post if some of you are reading both sections but I was unsure where this could go, so I decided to ask in both.

Here is an interesting proof (in my opinion) maybe some of you will have some insight. I have managed to prove it but in a long and non efficient way. Here's the proof:

We have 1!*2!*3!*4!*5!*6!...*100!

Prove that you can eliminate one factor so that the remaining number is a perfect square.

What I have done is using the help of a computer prime factor each number 1! through 100!, then added their powers and from that I saw that by eliminating 50! the number 1!*2!*3!*4!*5!*6!...*100! will have its prime factorization all to an even power, resulting in a perfect square. Now I am sure there should be a more efficient way in doing this, any help?

Hello, ride12!

given: .

Prove that you can eliminate one factor so that the remaining number is a perfect square.

Consider the frequency of each factor in the number



The odd factors {1, 3, 5, ... , 99} all appear an even number of times.
. . They will comprise a square number.

The even factors {2, 4, 6, ..., 100} all appear an odd number of times.
. . If we eliminate one of each, the rest will comprise a square number.

But

So we can eliminate just
. . and leave the , a square.


Then: . . . . a square

Brilliant!
Thank you very much for the help. I was just considering the odd and even factors when I refreshed the forum page, but you pointed me in the right direction

Thank again, and continue the great work