## South African Skeptics

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# Some number fun

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Rigil Kent
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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 = ?
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GCG
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skeptical mantis is skeptical

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

i knew that was going to be higher grade....
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st0nes
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 « 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
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Mefiante
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In solidarity with rwenzori: Κοπρος φανεται

 « 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
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Faerie
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 « 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?
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Mefiante
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In solidarity with rwenzori: Κοπρος φανεται

 « 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
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Faerie
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 « 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.
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Faerie
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 « 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

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

Throw some more! I'm enjoying this!

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st0nes
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 « 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
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Mefiante
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 « Reply #9 on: March 16, 2011, 14:32:22 PM »

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.

'Luthon64
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st0nes
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 « Reply #10 on: March 16, 2011, 14:58:46 PM »

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.

'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?
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Mefiante
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 « Reply #11 on: March 16, 2011, 15:42:39 PM »

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

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
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Mefiante
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In solidarity with rwenzori: Κοπρος φανεται

 « 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
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benguela
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An infinitesimal subset of the observable universe

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

3 ○ 2 ○ 7 = 641

641 is a prime, hmmm interesting, curious facts about 641
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benguela
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An infinitesimal subset of the observable universe

 « 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

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Mefiante
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 « Reply #15 on: March 17, 2011, 16:22:59 PM »

Correct.  Just to be sure, what’s 7 ○ 4 ○ 9?

Anyone else worked it out?

'Luthon64
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benguela
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An infinitesimal subset of the observable universe

 « Reply #16 on: March 17, 2011, 16:26:29 PM »

Just to be sure, what’s 7 ○ 4 ○ 9?

863
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Deezil
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 « Reply #17 on: March 17, 2011, 16:32:00 PM »

Ag nee bliksem ... this last one is killing me!
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Mefiante
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In solidarity with rwenzori: Κοπρος φανεται

 « Reply #18 on: March 17, 2011, 16:44:08 PM »

Okay, that clinches it.  You’ve got it.

Perhaps you’d like to try your hand at one that’s rather trickier:

If
1 ○ 3 ○ 7 → 2195
2 ○ 6 ○ 9 → 7841
4 ○ 8 ○ 3 → 6129
6 ○ 7 ○ 4 → 6433
7 ○ 4 ○ 9 → 7514
then
2 ○ 4 ○ 6 → ?

'Luthon64
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Deezil
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 « Reply #19 on: March 17, 2011, 17:11:29 PM »

Please explain previous one ... PM (or spoiler tag) is fine ... I'm beginning to pull my hair out
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Mefiante
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In solidarity with rwenzori: Κοπρος φανεται

 « Reply #20 on: March 17, 2011, 17:17:11 PM »

The previous one in Reply #18 might well give you sleepless nights.   If so, try this instead for relief:

If
1 ○ 3 ○ 7 → 353
2 ○ 9 ○ 8 → 595
4 ○ 8 ○ 5 → 193
6 ○ 7 ○ 4 → 119
7 ○ 4 ○ 6 → 239
then
3 ○ 5 ○ 7 → ?

'Luthon64
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Mefiante
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In solidarity with rwenzori: Κοπρος φανεται

 « Reply #21 on: March 17, 2011, 17:33:32 PM »

PM (or spoiler tag)
I don’t think the forum software supports spoiler tags, but I may be wrong.  I sent you a PM just in case.

'Luthon64
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Lurkie
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 « Reply #22 on: March 17, 2011, 18:54:58 PM »

Quote
If
1 ○ 3 ○ 7 → 353
2 ○ 9 ○ 8 → 595
4 ○ 8 ○ 5 → 193
6 ○ 7 ○ 4 → 119
7 ○ 4 ○ 6 → 239
then
3 ○ 5 ○ 7 → ?

3 ○ 5 ○ 7 → 371
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Mefiante
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In solidarity with rwenzori: Κοπρος φανεται

 « Reply #23 on: March 17, 2011, 19:23:56 PM »

3 ○ 5 ○ 7 → 371
Correct.  Well done for getting it so quickly!

'Luthon64
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Rigil Kent
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Three men make a tiger.

 « Reply #24 on: March 17, 2011, 20:30:57 PM »

Bloody nora, Lurkie! I was planning on working on it over Easter Weekend so as to impress the socks off everyone first thing in May. Respect!

Mintaka
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Mefiante
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In solidarity with rwenzori: Κοπρος φανεται

 « Reply #25 on: March 17, 2011, 20:54:36 PM »

Well, you can do that anyway if you like.  For everyone but Lurkie, what does 6 ○ 1 ○ 9 yield in the scheme of Reply #20?

'Luthon64
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Lurkie
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 « Reply #26 on: March 18, 2011, 09:26:03 AM »

Quote
Bloody nora, Lurkie! I was planning on working on it over Easter Weekend so as to impress the socks off everyone first thing in May. Respect!

Thanks Mintaka! Do the puzzle anyway. It is mathematically elegant. I'm never mad about problems that involve stuff like modular arithmetic - don't like throwing parts of numbers away.
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Lurkie
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 « Reply #27 on: March 21, 2011, 14:48:30 PM »

Quote
1 ○ 3 ○ 7 → 2195
2 ○ 6 ○ 9 → 7841
4 ○ 8 ○ 3 → 6129
6 ○ 7 ○ 4 → 6433
7 ○ 4 ○ 9 → 7514
then
2 ○ 4 ○ 6 → ?

Mefiante, this puzzle was a SOB. I tried the usual combinations of addition, subtraction, multiplication, inverses then graduated to powers, factorials, Laplace transforms and Lagragian multipliers (ok, not the last two!)

Finally,
2 ○ 4 ○ 6 → 4148

Didn't like having to throw some numbers away ...

I sound like I'm complaining, but I'm not. Thoroughly enjoyed the chase, despite some head-bashing moments!

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Mefiante
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 « Reply #28 on: March 21, 2011, 14:54:50 PM »

Correct again, well done!  Anyone else?

'Luthon64
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Rigil Kent
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Three men make a tiger.

 « Reply #29 on: March 21, 2011, 18:06:31 PM »

Anyone else?

Nope ... I have now finally embraced the new and somewhat less agonizing passtime of sticking roasting forks in my legs.

ETA: .... but just for interest, is there an elegant way of solving these chestnuts, or is inspection and a keen eye the only hope? They look a bit like simultaneous equations, and its almost as if they beg to be solved in a similar manner ... but, of course, they are not equations at all.

Mintaka
 « Last Edit: March 21, 2011, 18:41:42 PM by Mintaka » Logged
Lurkie
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 « Reply #30 on: March 22, 2011, 13:45:26 PM »

Quote
ETA: .... but just for interest, is there an elegant way of solving these chestnuts, or is inspection and a keen eye the only hope? They look a bit like simultaneous equations, and its almost as if they beg to be solved in a similar manner ... but, of course, they are not equations at all.

I'd love to learn an elegant way to solve these rotters. Roasting fork moments almost drove me to write some code to implement a brute force search to test lots and lots of possible combinations of operators for the given number sequence. Only problem is that there are gazillions of combinations. But this will still miss the problems where only a demented human brain will pick up a non-mathematical pattern.

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Tweefo
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 « Reply #31 on: March 22, 2011, 13:56:00 PM »

Enough of these brilliant minds being wasted on games. Here is a real life number problem: My bank balance need to grow! Fast!
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Hermes
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 « Reply #32 on: March 22, 2011, 14:29:48 PM »

Enough of these brilliant minds being wasted on games. Here is a real life number problem: My bank balance need to grow! Fast!
No problem, if

\$1 ○ \$3 ○ \$7 → \$353
\$2 ○ \$9 ○ \$8 → \$595
\$4 ○ \$8 ○ \$5 → \$193
\$6 ○ \$7 ○ \$4 → \$119
\$7 ○ \$4 ○ \$6 → \$239
then
\$3 ○ \$5 ○ \$7 → ?

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Lurkie
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 « Reply #33 on: March 22, 2011, 14:47:13 PM »

Quote
No problem, if

\$1 ○ \$3 ○ \$7 → \$353
\$2 ○ \$9 ○ \$8 → \$595
\$4 ○ \$8 ○ \$5 → \$193
\$6 ○ \$7 ○ \$4 → \$119
\$7 ○ \$4 ○ \$6 → \$239
then
\$3 ○ \$5 ○ \$7 → ?

Hmm. Not good if the operators divide the numbers resulting in an infinitely diminishing balance. I suggest a new operator where ○=\$
 « Last Edit: March 23, 2011, 09:53:34 AM by Lurkie, Reason: speling misteak » Logged
Mefiante
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In solidarity with rwenzori: Κοπρος φανεται

 « Reply #34 on: March 22, 2011, 15:46:33 PM »

Mathematically speaking, these problems aren’t especially interesting or edifying.  At best, they can be used to illustrate some core mathematical ideas whose application is mostly cryptographic.  And no, there is no fixed approach to them because the variety of possible schemes is virtually limitless.

That said, for the insatiably curious, frustrated and bagshot among us, here’s…

(click to show/hide)

(click to show/hide)

(click to show/hide)

'Luthon64
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Mefiante
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In solidarity with rwenzori: Κοπρος φανεται

 « Reply #35 on: March 22, 2011, 16:36:22 PM »

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Wandapec
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100% Proud Atheist/Skeptic

 « Reply #36 on: May 31, 2011, 15:20:52 PM »

Got this in a mail today...
Quote
THIS IS THE ONLY TIME WE WILL SEE AND LIVE THIS EVENT

Calendar for July 2011.

July

Sun    Mon    Tue    Wed    Thu    Fri    Sat
1    2
3    4    5    6    7    8    9
10   11    12    13    14    15   16
17   18    19    20    21    22   23
24   25    26    27    28    29   30
31

This year, July has 5 Fridays, 5 Saturdays and 5 Sundays. This happens
once every 823 years.

This year we're going to experience four unusual dates.

1/1/11, 1/11/11, 11/1/11, 11/11/11 and that's not all...

Take the last two digits of the year in which you were born - now add
the age you will be this year,

The results will be 111 for everyone in whole world..
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