Fiery Factorials!

[Response to Challenge Master Note on Louis W's solution]

A bit of observation tells us that the following product $\mathcal{P}_M$ (for positive integer $M$ ) can be generalized and used for evaluating both the original problem and the one in the note.

$\mathcal{P}_M=\frac{\displaystyle\prod_{n=1}^{2M}n!}{\displaystyle\prod_{n=1}^{M}\bigg[(2n-1)!\bigg]^2}=\frac{\displaystyle\prod_{n=1}^M (2n)!}{\displaystyle\prod_{n=1}^M(2n-1)!}=\prod_{n=1}^M\frac{(2n)!}{(2n-1)!}=\prod_{n=1}^M (2n)\\ \implies \mathcal{P}_M=\prod_{n=1}^M (2n)=\left(\prod_{n=1}^M2\right)\left(\prod_{n=1}^Mn\right)=2^M\times M!$

Our original problem is $\mathcal{P}_5=2^5\times 5!=2^5\times 120=2^8\times 15$ .

The problem in the note is $\mathcal{P}_{50}=2^{50}\times 50!$

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We used a few trivial product identities and the following almost trivial identity:

$\prod_{i=1}^{2x} i!=\left(\prod_{i=1}^x (2i)!\right)\left(\prod_{i=1}^x (2i-1)!\right)$

First, to simplify a little: $\frac{1!\times2!\times3!\times\cdot\cdot\cdot\times10!}{(1!)^{2}(3!)^{2}(5!)^{2}(7!)^{2}(9!)^{2}}=\frac{2!\times3!\times\cdot\cdot\cdot\times10!}{(3!)^{2}(5!)^{2}(7!)^{2}(9!)^{2}}$ Now to rearrange: $=\frac{2\times\color{#3D99F6}{3!}\times4\times\color{#27D2E7}{3!}\times\color{fuchsia}{5!}\times6\times\color{#20A900}{5!}\times\color{#EC7300}{7!}\times8\times\color{#69047E}{7!}\times\color{#D61F06}{9!}\times10\times\color{teal}{9!}}{\color{#3D99F6}{3!}\times\color{#27D2E7}{3!}\times\color{fuchsia}{5!}\times\color{#20A900}{5!}\times\color{#EC7300}{7!}\times\color{#69047E}{7!}\times\color{#D61F06}{9!}\times\color{teal}{9!}}$ All of the colors cancel each other out and you are left with: $= 2\times4\times6\times8\times10=\color{#D61F06}{2}\times\color{#D61F06}{2}\times\color{#D61F06}{2}\times\color{#D61F06}{2}\times3\times\color{#D61F06}{2}\times\color{#D61F06}{2}\times\color{#D61F06}{2}\times\color{#D61F06}{2}\times5$ $=2^{\color{#D61F06}{8}}\times15\space\space\space\Box$

The numerator can be paired into an even factorial and an odd factorial (ex: 9!10!). This can be rewritten as 9!9!x10 or (9!)^2 x10. If we rewrite all the numerator pairs as such we get 1!1!x2 x 3!3!x4 x 5!5!x6 x 7!7!x8 x 9!9!x10. All the factorials cancel with the denominator leaving 2x4x6x8x10. Rewritten as primes we fet 2x(2x2)x(2x3)x(2x2x2)x2x5=15x2^8

Simplify this we get (2!×4!×6!×8!×10!)/(1!×3!×5!×7!×9!)=15×2^n Where 2!/1!=2 lly we get above step 2×4×6×8×10=15×2^n 2×4×2×8×2=2^n Then left side we can change Base 2 form 2^8=2^n Finally we got d ans n=8

On solving we will get( 2 4 6 8 10)/15. On simplifying we'll get 256 which is none other than 2^8. Therefore the required answer is 8.

(2! X 4! X 6! X 8! 10!) / (1! X 3! X 5! X 7! X 9! X 15) = 256 = 2^8

Exponent of a prime p(here 2) in a factorial of n is given by Ep(n!) = floor(n/p) + floor(n/p^2) + ......... so on since 15 in RHS does not resolve to 2 as a factor n will contain all the power of 2 of LHS. Now find power of 2 for each factorial using above formula. Ep(2!) = 1 Ep(3!) = 1 Ep(4!) = 3 Ep(5!) = floor(5/2) + floor(5/4) + floor(5/8) = 2+1+0 = 3 Ep(6!) = 4 Ep(7!) = 4 Ep(8!) = 7 Ep(9!) = 7 Ep(10!) = 8

Now calculate power of 2 in each factorial for numerator it will be (1+1+3+3+4+4+7+7+8) = 38 For denominator it will be (1+1+3+3+4+4+7+7) = 30

Ans = 38-30 = 8

jus observe and mind does all the math and gives 8 3:)

15 x 2^n = 1! 2! 3! 4! 5! 6! 7! 8! 9! 10!/ 1! 3!3! 5!5! 7!7! 9!9! = 2! 4! 6! 8! 10!/ 3! 5! 7! 9! = 2! 4 6 8 10 = 15 x 2^8