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-   -   Can someone help me understand this... (https://www.mersenneforum.org/showthread.php?t=26111)

petrw1 2020-10-21 17:52

Can someone help me understand this...
 
Or simplify it:

How does such a seemingly odd string of irrational numbers equate to exactly 142?
Better yet how would someone have come up with this --- mathematically; not via trial and error?

Interestingly, the first half is almost exactly twice the second half.

(√√7! + √√7!) * (√(√7! + (√7!)/7!)) = 142

Thanks

Viliam Furik 2020-10-21 18:25

The whole process is [URL="https://drive.google.com/file/d/1Wtm5pVcE8RUBqG0tSqycenKk9mffLie5/view?usp=sharing"]here[/URL].

If we say the 7 factorial is x, then we can generalize it for x, and by shifting things around, we find that expression of the form (√√x + √√x) * √(√x + (√x)/x) is a whole number whenever x+1 is a square.

petrw1 2020-10-21 18:48

[QUOTE=Viliam Furik;560579]The whole process is [URL="https://drive.google.com/file/d/1Wtm5pVcE8RUBqG0tSqycenKk9mffLie5/view?usp=sharing"]here[/URL].

If we say the 7 factorial is x, then we can generalize it for x, and by shifting things around, we find that expression of the form (√√x + √√x) * √(√x + (√x)/x) is a whole number whenever x+1 is a square.[/QUOTE]

Thanks a lot....no big deal, but you dropped he 2x on line 2 of your napkin.

So If I knew what I was doing I could reverse this process and get other whole numbers starting with √(x!+1) ??
x can be 4, 5, or 7.

Viliam Furik 2020-10-21 19:26

[QUOTE=petrw1;560586]Thanks a lot....no big deal, but you dropped he 2x on line 2 of your napkin.[/QUOTE]Yes, I know, thus the implication arrow afterwards, instead of an equality sign. Removing the 2 doesn't change the rest. If I were to get some fraction in the end, I could simply put it back. But it doesn't change the rationality, which is the whole question.

And, it isn't a napkin. (But I think you know that) It is my notebook I use for maths in the school. (BTW, I am graduating in May 2021, at least that's what I thought a year ago. Who knows what else might Covid take - we are learning online since the last Monday, and it's possible it will be until Christmas)

[QUOTE=petrw1;560586]
So If I knew what I was doing I could reverse this process and get other whole numbers starting with √(x!+1) ??
x can be 4, 5, or 7.[/QUOTE]
Yes! Absolutely. You can shove in any number for x, even primorials, perfect powers, and also Riesel primes with even powers of base and square ks (based on few look-ups, they might not exist), but sadly enough, no Mersenne primes except M2, as 2[SUP]p[/SUP] - 1 + 1 = 2[SUP]p[/SUP], which is not a square if the p is odd.

----
EDIT:
Silly me. Of course there can't be a Riesel prime with k being square and n being even, because of the almighty algebraic factors of a[SUP]2[/SUP] - 1 = (a-1)(a+1)

LaurV 2020-10-22 07:43

[QUOTE=petrw1;560573]
(√√7! + √√7!) * (√(√7! + (√7!)/7!)) = 142
[/QUOTE]
Let \(\alpha=\sqrt{7!}\). You have \(2\sqrt\alpha\cdot\sqrt{\alpha+\frac{\alpha}{\alpha^2}}\). Which, when multiply the radicals and simplify the fraction under it, becomes \(2\sqrt{\alpha^2+1}\). Now substitute back the \(\alpha\), you have \(2\sqrt{7!+1}\), or \(2\sqrt{5041}\), which is 2*71.

Dr Sardonicus 2020-10-22 11:37

Yes, substituting x for 7! does make things much easier to handle. The obvious regrouping of the first part of the expression gives

[tex]2x^{\frac{1}{4}}\cdot\(x^{\frac{1}{2}}\;+\;x^{\frac{-1}{2}}\)^{\frac{1}{2}}[/tex]

The "obvious" multiplication then gives [tex]2\sqrt{x+1}[/tex].

I note that things can go wrong for complex values of "x" that aren't positive real numbers.

petrw1 2020-10-22 16:45

Thanks all

petrw1 2020-10-27 17:45

I can follow the simplification that gets one to 2 x SQRT(7!+1) = 142

Can someone explain how you would start at 2 x SQRT(7!+1) = 142
and given that the final equation must contain only 7's, get to the original equation I supplied in post 1.
For example the 2 at the front could become something like (7+7)/7 OR <any function with a 7> + <same> iff that function can be divided out in another part of the entire formula to equal 2.
....confused??? Me too.

(My son has a game/puzzle app when he must find a formula using the least number of each digit from 1 to 9 to get a number)

He is limited to +. -, x, /, SQRT and concatenation. ie 77 or 777

LaurV 2020-10-28 04:54

Do you mean like [URL="https://www.slideshare.net/ByronWillems/an-engineer-112"]engineeers112[/URL]?.:razz:
(scroll through the ppt presentation there)

petrw1 2020-10-28 05:42

[QUOTE=LaurV;561311]Do you mean like [URL="https://www.slideshare.net/ByronWillems/an-engineer-112"]engineeers112[/URL]?.:razz:
(scroll through the ppt presentation there)[/QUOTE]

Exactly ... that is sooooo much clearer!!!!


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