20201021, 17:52  #1 
1976 Toyota Corona years forever!
"Wayne"
Nov 2006
Saskatchewan, Canada
10712_{8} Posts 
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 
20201021, 18:25  #2 
"Viliam Furík"
Jul 2018
Martin, Slovakia
2×193 Posts 
The whole process is here.
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. 
20201021, 18:48  #3  
1976 Toyota Corona years forever!
"Wayne"
Nov 2006
Saskatchewan, Canada
2×3^{2}×11×23 Posts 
Quote:
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. Last fiddled with by petrw1 on 20201021 at 18:54 

20201021, 19:26  #4  
"Viliam Furík"
Jul 2018
Martin, Slovakia
110000010_{2} Posts 
Quote:
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:
 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^{2}  1 = (a1)(a+1) Last fiddled with by Viliam Furik on 20201021 at 19:36 

20201022, 07:43  #5 
Romulan Interpreter
Jun 2011
Thailand
5^{2}×7×53 Posts 
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.

20201022, 11:37  #6 
Feb 2017
Nowhere
2·3^{2}·241 Posts 
Yes, substituting x for 7! does make things much easier to handle. The obvious regrouping of the first part of the expression gives
The "obvious" multiplication then gives . I note that things can go wrong for complex values of "x" that aren't positive real numbers. 
20201022, 16:45  #7 
1976 Toyota Corona years forever!
"Wayne"
Nov 2006
Saskatchewan, Canada
1000111001010_{2} Posts 
Thanks all

20201027, 17:45  #8 
1976 Toyota Corona years forever!
"Wayne"
Nov 2006
Saskatchewan, Canada
2×3^{2}×11×23 Posts 
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 
20201028, 04:54  #9 
Romulan Interpreter
Jun 2011
Thailand
10010000111011_{2} Posts 
Do you mean like engineeers112?.
(scroll through the ppt presentation there) 
20201028, 05:42  #10  
1976 Toyota Corona years forever!
"Wayne"
Nov 2006
Saskatchewan, Canada
4554_{10} Posts 
Quote:


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