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Author Topic:

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
Yes.  No. (Answered in order.)
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Mefiante
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In solidarity with rwenzori: Κοπρος φανεται

 « 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.

'Luthon64
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st0nes
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 « 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.

'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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In solidarity with rwenzori: Κοπρος φανεται

 « 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