An old member returns...

[Thomas Radtke]

(02-22-2015 11:47 AM)J-F Garnier Wrote:  Let's take all this with humour... :-)

Tried and failed. I'm here to learn and talk about calculators and some math. I'd buy a TV set if I would like to watch Tom & Jerry instead.

[Gerson W. Barbosa]

(02-22-2015 11:47 AM)J-F Garnier Wrote:  

(02-20-2015 02:51 PM)Thomas Radtke Wrote:  The signature already told the story ... if you know V. and a little physics.

(02-20-2015 09:27 PM)Gerson W. Barbosa Wrote:  Actually VA equals W only when var equals zero [ VA = sqrt(W^2 + var^2) ]. This has nothing to do with VA's little riddle though. The square root of VA might be W for a quite different reason, if I got him right. Also, VA as we know him is a real entity, not an apparent one. :-)
P.S.: I thought you were referring to another kind of VA. Yes, your VA does equal W. Sorry!

As an electric engineer and knowing Valentin, I can only share your point :-)

In one of his famous S&SMC , Valentin asked to find a solution on the HP-71B to the equation:
Abs(Ln(x*x)-Ln(x^2)) > 4.012007 (BTW, I found a solution ...)
In the same way, is it possible to have square of VA significantly different from the product of VA by VA?

Let's take all this with humour... :-)

J-F

I've made a great mistake (please see my updated reply to Valentín). So one possible answer to the square root of VA is 45, NOT W, if my line of thought matches Valentín's.

I am sorry for having misled all of you from the solution. Again, my apologies!

This and the answer to the first problem might give a hint to both solutions, in case no one has figured them out yet.

Best regards,

Gerson.

[walter b]

(02-20-2015 02:51 PM)Thomas Radtke Wrote:  The signature already told the story ... if you know V. and a little physics.

Still waiting for an answer telling me what the claim SQRT(VA) = W should have to do with physics. TIA for enlightenment.

d:-?

<<<< Walter; No one here owes you an explanation for anything, especially since you wouldn't give up the name of who you were calling illiterate - when you were asked. I'm still waiting for an answer for that. Additionally; You should stop following Thomas around our forum and trolling him. It's gone on for too long. Just because he answers member's questions about "your" 34s, it doesn't give you the right to badger him.>>>>

[Gerson W. Barbosa]

(02-22-2015 04:50 PM)walter b Wrote:  

(02-20-2015 02:51 PM)Thomas Radtke Wrote:  The signature already told the story ... if you know V. and a little physics.

Still waiting for an answer telling me what the claim SQRT(VA) = W should have to do with physics. TIA for enlightenment.

d:-?

All my fault because of putting 1 instead of 10 in line 07 and thus getting 32, that is, "W", instead of 45... Again, apologies!

Code:


00 { 34-Byte Prgm }
01>LBL "VA"
02 32                        ; start with base 32, since V = 31 (A = 10, B = 11, ...)
03>LBL 00
04 RCL ST X
05 31
06 ×
07 10
08 +                       
09 SQRT
10 FP                        ; check whether 31*base + 10 is a perfect square 
11 X=0?                                   
12 GTO 01                    ; if ok go show square root and base (stack registers X and Y, respectively)
13 X<>Y
14 1                         ; else
15 +                         ;     try next base
16 GTO 00
17>LBL 01
18 X<>Y
19 LASTX
20 .END.  [code]

BTW, by the same reasoning the square root of JFG is 330646 and the square root of GWB is 14944740. I haven't been able to find the square roots of WB and TR, but perhaps I haven't searched long enough.

Gerson.

Edited to add comments to the code

P.S.: Sqrt(GWB) above is wrong, but at least Sqrt(JFG) is correct :-)
P.P.S: I've just found out a member here has Sqrt = O(24) that is, 24 in base 34. Someone else with an even lower square root? :-)

[Thomas Klemm]

(02-22-2015 04:41 PM)Gerson W. Barbosa Wrote:  So one possible answer to the square root of VA is 45, NOT W, if my line of thought matches Valentín's.

The integer solutions to \(31n+10=m^2\) are:
\(n=31k^2+28k+6; m=31k+14; k \in \mathbb{Z}\)
\(n=31k^2+34k+9; m=31k+17; k \in \mathbb{Z}\)

Thus e. g. 48₇₄ is another solution for the square root of VA₇₄.

Cheers
Thomas

[Valentin Albillo]

.
Hi, all:

Thanks for your interest and kind replies. I'm sorry for the delay in posting my own reply but this is about the faster I can manage. The answer to my explicit mini-challenge, i.e.:

" If 9x9=29, what's the square root of 69 ? "

was correctly given by Gerson W. Barbosa and it's "F". First one would need to realize that for the 9x9=29 equality to hold the numbers given can't be in base 10, then compute the actual base B, like this:

9x9 = 81 (base 10) = 29 (base B) = 2*B+9 => B=(81-9)/2 = 36

now, knowing the base B, we have that 69 (base 36) = 6*36+9 = 225 (base 10) and finally:

sqrt(69 (base 36)) = sqrt(225 (base 10)) = 15 (base 10 ) = F (base 36)

which is the answer. I think that the statement:

" If 9x9 = 29 then sqrt(69) = F "

is pretty nice and seemingly nonsensical at first sight.As for the second, implicit mini-challenge, namely:

" ... if VA is square, what's the square root of me ? "

you first need to realize that, again, VA is a number in some base which has V and A as digits and which happens to be a square, and me is also a number in the same base, digits m and e this time, which is also a perfect square.

Don't be surprised by the first number being expressed with uppercase letters for its digits while the second number uses lowercase letters, this is akin to "23AF46E5" and "23af446e5" being equally valid and used representations in base 16, the case of the letters is irrelevant.

The second mini-challenge then is to find numbers VA and ME (or va and me, if you prefer) that are simultaneously perfect squares in some base B that allows for those digits. The programming isn't complicated at all and I'll give just the answers for you to check your results:

First solution:

VA (base 65) = 2025 (base 10) => sqrt(2025 (base 10)) = 45 (base 10)
ME (base 65) = 1444 (base 10) => sqrt(1444 (base 10)) = 38 (base 10)

Second solution:

VA (base 539431265) = 16722369225 (base 10), whose sqrt is 129315 (base 10)
ME (base 539431265) = 11867487844 (base 10), whose sqrt is 108938 (base 10)

and there are no other solutions up to base 194,576,544,937 (almost 200 billion)

PS: I need to know how to properly format math expressions in a post, most obviously subscripts and tables.

Best regards.
V.
.

[Thomas Klemm]

(02-23-2015 06:59 PM)Valentin Albillo Wrote:  PS: I need to know how to properly format math expressions in a post, most obviously subscripts and tables.

You can use LaTeX within ( ) or [ ]: you just have to escape them with \.
There are plenty of examples in the Test forum.

HTH
Thomas

[Valentin Albillo]

(02-23-2015 07:10 PM)Thomas Klemm Wrote:  

(02-23-2015 06:59 PM)Valentin Albillo Wrote:  PS: I need to know how to properly format math expressions in a post, most obviously subscripts and tables.

You can use LaTeX within ( ) or [ ]: you just have to escape them with \.
There are plenty of examples in the Test forum.

HTH
Thomas

Thanks a lot, Thomas, that's all the info I needed.

Regards.
V.

[Jeff_Kearns]

Welcome back Valentin! I have enjoyed reading your many posts over the years and those from the archives, in addition to your excellent articles (keepsakes all). Whatever happened to your lycos website?

Jeff Kearns

[Valentin Albillo]

(02-24-2015 02:17 AM)Jeff_Kearns Wrote:  Welcome back Valentin! I have enjoyed reading your many posts over the years and those from the archives, in addition to your excellent articles (keepsakes all). Whatever happened to your lycos website?

Thanks a lot for your warm welcome and kind words, Jeff.

I have no idea what happened to the Lycos website, I found myself unable to access it many years ago so I could never update its contents, add to it or edit it in any way.

I think it was completely infected with spam websites and contents to the point that the server's storage space was completely filled up and further, the moment someone deleted some content spam would immediately gobble up the freed storage, rendering the site unusable for everyone else. This was pre-Captchas, of course.

That, together with tons and tons of adds and popups pestering and annoying every visitor utterly killed it for me. For a time I would send my PDF articles to HPCC's Datafile for publication on my two regular sections there but that ended in 2007 (together with my subscription to HPCC) for reasons known to everyone who frequented the old MoHP at the time and which I won't repeat here.

That left me with a very sour feeling and a number of unpublished PDF articles, most of them almost ready for publication, some 100 pages in all, including my cherished "50th Birthday Special" one, which I was about to send to Datafile for publication in the March 2008 issue but alas, it wasn't to be.

Now that the sourness has mostly disappeared and I've joined this new incarnation of MoHP I might be tempted to finish the PDF articles, probably I should, but I don't know where to publish them and I need to have some suitable place available before I commit myself to do it. On the other hand, Challenges and mini-challenges are easier to come by and in due time they'll surely appear here for everyone to have a chance at solving them.

Thanks again for your interest and best regards.
V.

[Jeff_Kearns]

Valentin,

Regarding your pdf articles online: I was able to locate most of them through the WayBackMachine.

This was shared with the forum members in a thread at the beginning of September of 2014 called: "Where to find DatafileVA005.pdf"


Jeff

[Thomas Klemm]

Link for the lazy: Where to find DatafileVA005.pdf