South Africa Flag logo

South African Skeptics

November 21, 2019, 11:07:05 AM
Welcome, Guest. Please login or register.
Did you miss your activation email?

Login with username, password and session length
Go to mobile page.
News: Please read the posting guidelines before posting.
   
   Skeptic Forum Board Index   Help Forum Rules Search GoogleTagged Login Register Chat Blogroll  
Pages: [1] 2 3  All   Go Down
  Print  
Author Topic:

Some number fun

 (Read 14569 times)
0 Members and 1 Guest are viewing this topic.
Rigil Kent
Clotting Factor
Hero Member
*****

Skeptical ability: +19/-3
Offline Offline

Posts: 2463


Three men make a tiger.


« on: March 16, 2011, 10:03:07 AM »

IF

5+3+2 = 151022
 
9+2+4 = 183652
 
8+6+3 = 482466
 
5+4+5 = 202541
 
THEN
 
7+2+5 = ?
Logged
GCG
Hero Member
*****

Skeptical ability: +8/-4
Offline Offline

Posts: 1829


skeptical mantis is skeptical


adele horn
WWW
« Reply #1 on: March 16, 2011, 10:06:42 AM »

i knew that was going to be higher grade....  Sad
Logged
st0nes
Hero Member
*****

Skeptical ability: +10/-1
Offline Offline

Posts: 942



mark.widdicombe1
WWW
« Reply #2 on: March 16, 2011, 10:39:58 AM »

IF

5+3+2 = 151022
 
9+2+4 = 183652
 
8+6+3 = 482466
 
5+4+5 = 202541
 
THEN
 
7+2+5 = ?
143547
Logged
Mefiante
Defollyant Iconoclast
Hero Member
*****

Skeptical ability: +61/-9
Offline Offline

Posts: 3756


In solidarity with rwenzori: Κοπρος φανεται


WWW
« Reply #3 on: March 16, 2011, 10:55:11 AM »

Correct, 143547.  Now the real question where the real fun begins:  Is 7+2+5 unique?  That is, can 143547 be generated by the same scheme but using a different triplet?

'Luthon64
Logged
Faerie
Hero Member
*****

Skeptical ability: +10/-2
Offline Offline

Posts: 2114



« Reply #4 on: March 16, 2011, 11:06:26 AM »

IF

5+3+2 = 151022
 
9+2+4 = 183652
 
8+6+3 = 482466
 
5+4+5 = 202541
 
THEN
 
7+2+5 = ?
143547

Ok, so I can figure out the first bit, and I can get to the last two numbers, but then I neuk off the bus with the middle two figures. What am I so obviously missing?
Logged
Mefiante
Defollyant Iconoclast
Hero Member
*****

Skeptical ability: +61/-9
Offline Offline

Posts: 3756


In solidarity with rwenzori: Κοπρος φανεται


WWW
« Reply #5 on: March 16, 2011, 11:15:44 AM »

If you can get the last two numbers, you must necessarily be able to get the middle two.

'Luthon64
Logged
Faerie
Hero Member
*****

Skeptical ability: +10/-2
Offline Offline

Posts: 2114



« Reply #6 on: March 16, 2011, 11:21:28 AM »

If you can get the last two numbers, you must necessarily be able to get the middle two.

'Luthon64

Huh? Wait, lemme go figure my shit out. Dont tell me. I'll get this figured out.
Logged
Faerie
Hero Member
*****

Skeptical ability: +10/-2
Offline Offline

Posts: 2114



« Reply #7 on: March 16, 2011, 11:44:58 AM »

Flippin basic maths getting me down - sorry about the hijack, but this is what I've got:

7*2 = 14
7*5 = 35
14+35-2 = 47

 Grin

Heh, std 7 maths in action!! (and I failed that got a grand 18% for it)

Throw some more! I'm enjoying this!



Logged
st0nes
Hero Member
*****

Skeptical ability: +10/-1
Offline Offline

Posts: 942



mark.widdicombe1
WWW
« Reply #8 on: March 16, 2011, 13:02:46 PM »

Correct, 143547.  Now the real question where the real fun begins:  Is 7+2+5 unique?  That is, can 143547 be generated by the same scheme but using a different triplet?

'Luthon64
Yes.  No. (Answered in order.)
Logged
Mefiante
Defollyant Iconoclast
Hero Member
*****

Skeptical ability: +61/-9
Offline Offline

Posts: 3756


In solidarity with rwenzori: Κοπρος φανεται


WWW
« Reply #9 on: March 16, 2011, 14:32:22 PM »

Yes.  No. (Answered in order.)
Correct.  Final question:  Can you prove the general case that a given valid result from applying this scheme uniquely fixes the triplet?

(For the interested reader, the original problem belongs to a branch of mathematics called “group theory.”)

Later today I’ll see if I can dig up a few more number problems, preferably ones that Google knows nothing of. Evil

'Luthon64
Logged
st0nes
Hero Member
*****

Skeptical ability: +10/-1
Offline Offline

Posts: 942



mark.widdicombe1
WWW
« Reply #10 on: March 16, 2011, 14:58:46 PM »

Yes.  No. (Answered in order.)
Correct.  Final question:  Can you prove the general case that a given valid result from applying this scheme uniquely fixes the triplet?

(For the interested reader, the original problem belongs to a branch of mathematics called “group theory.”)

Later today I’ll see if I can dig up a few more number problems, preferably ones that Google knows nothing of. Evil

'Luthon64
Not formally (IANAM), but let's give it a bash.  BTW, I'm not doing your homework, am I?  Taking the last triplet as an example, we have
$xy=14$,
$xz=35$, and
$xy+xz-y=47$
All of these are linear equations--we have three variables, three equations with a single solution for each variable.  Therefore a valid result can only be arrived at by one triplet?
Logged
Mefiante
Defollyant Iconoclast
Hero Member
*****

Skeptical ability: +61/-9
Offline Offline

Posts: 3756


In solidarity with rwenzori: Κοπρος φανεται


WWW
« Reply #11 on: March 16, 2011, 15:42:39 PM »

Again correct in essence – and, no, you’re not doing my homework. Tongue

In the general case, we have the relation

        a + b + c = r1r2r3r4r5r6.

(Note the use of “+” vs. “+” to distinguish the operation from usual arithmetical addition.)

Given r1r2r3r4r5r6, you know that

        r5r6 = r1r2 + r3r4b,

whence

        b = r1r2 + r3r4r5r6.

The above clearly fixes the value of b uniquely.

Once b is known, a is uniquely determined by a × b = r1r2.  Finally, knowing a, c is uniquely determined by a × c = r3r4.  Thus, a, b and c are unique to a given r1r2r3r4r5r6, and the desired result is proved.

'Luthon64
Logged
Mefiante
Defollyant Iconoclast
Hero Member
*****

Skeptical ability: +61/-9
Offline Offline

Posts: 3756


In solidarity with rwenzori: Κοπρος φανεται


WWW
« Reply #12 on: March 16, 2011, 22:45:37 PM »

If
        2 ○ 6 ○ 9 = 248
        3 ○ 2 ○ 7 = 641
        4 ○ 8 ○ 3 = 242
        6 ○ 7 ○ 3 = 218
        8 ○ 3 ○ 6 = 488

then

        5 ○ 5 ○ 5 = ?

'Luthon64
Logged
benguela
Full Member
***

Skeptical ability: +3/-0
Offline Offline

Posts: 223


An infinitesimal subset of the observable universe


benguela
WWW
« Reply #13 on: March 17, 2011, 10:04:15 AM »

        3 ○ 2 ○ 7 = 641


641 is a prime, hmmm interesting, curious facts about 641
Logged
benguela
Full Member
***

Skeptical ability: +3/-0
Offline Offline

Posts: 223


An infinitesimal subset of the observable universe


benguela
WWW
« Reply #14 on: March 17, 2011, 15:52:45 PM »

If
        2 ○ 6 ○ 9 = 248
        3 ○ 2 ○ 7 = 641
        4 ○ 8 ○ 3 = 242
        6 ○ 7 ○ 3 = 218
        8 ○ 3 ○ 6 = 488

then

        5 ○ 5 ○ 5 = ?

'Luthon64

Answer: 555
Logged
Pages: [1] 2 3  All   Go Up
  Print  
GoogleTagged: mss then google must reply skeptic


 
Jump to:  

Powered by SMF 1.1.11 | SMF © 2006-2009, Simple Machines LLC
Page created in 0.851 seconds with 23 sceptic queries.
Google visited last this page February 26, 2019, 11:04:31 AM
Privacy Policy