BrainDen.com - Brain Teasers

## Question

In a future not so far way, Earth archaeologists find on a far away planet a fragment from a long lost civilization.

This fragment involves an unknown operation *|*.

Unlocking its secrets may lead to a breakthrough in understanding their civilization.

Can you do it?

If

`21 *|* 7 = 11`
and
`17924 *|* 10751 = 851`
then how much is:
`1982 *|* 2010 = ?`

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• 0
• Solution

Was in the rush to post it before somebody else does and messed up 1982 *|* 2010 = 1960

Here are all the given clues and the result converted to the base-3 system.

```
21	*|*	7	=	11				0000000210	*|*	0000000021	=	0000000102

17924	*|*	10751	=	851				0220120212	*|*	0112202012	=	0001011112

1089	*|*	121	=	2069				0001111100	*|*	0000011111	=	0002211122

10	*|*	23	=	6				0000000101	*|*	0000000212	=	0000000020

1	*|*	0	=	2				0000000001	*|*	0000000000	=	0000000002

999	*|*	909	=	1359				0001101000	*|*	0001020200	=	0001212100

14	*|*	14	=	14				0000000112	*|*	0000000112	=	0000000112

2	*|*	1	=	0				0000000002	*|*	0000000001	=	0000000000

1	*|*	1	=	1				0000000001	*|*	0000000001	=	0000000001

3	*|*	0	=	6				0000000010	*|*	0000000000	=	0000000020

13	*|*	8	=	18				0000000111	*|*	0000000022	=	0000000200

17924	*|*	5	=	9329				0220120212	*|*	0000000012	=	0110210112

16	*|*	0	=	23				0000000121	*|*	0000000000	=	0000000212

19	*|*	1	=	10				0000000201	*|*	0000000001	=	0000000101

1234	*|*	49	=	1780				0001200201	*|*	0000001211	=	0002102221

1	*|*	9	=	20				0000000001	*|*	0000000100	=	0000000202

2	*|*	9	=	19				0000000002	*|*	0000000100	=	0000000201

1982	*|*	2010	=	1960				0002201102	*|*	0002202110	=	0002200121

```

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• 0 ahhh..if we only had the first one..it is so easy.( If a *|* b = c then c = a-b-(a/b) )The second one mix everything up!

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ahhh..if we only had the first one..it is so easy.( If a *|* b = c then c = a-b-(a/b) )The second one mix everything up! let's just say they were a little more advanced and the operation is not that simple.

Nice thought though, I did not see that one (just chose a small example at random).

Well, here's a small hint to what the operation is. It is commutative.

`a *|* b = b *|* a`
And a second hint to what the operation is not. It is not associative.
`(a *|* b) *|* c does not equal a *|* (b *|* c) all the time`

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• 0

23 - by taking the absolute value of the difference of the two given numbers and then taking the highest prime factor of that difference and adding to that the product of the sum of the remaining prime factors and the product of the remaining prime factors -

21 - 7 = 14 --> prime factors = 2,7 so 7 + (2) * (2) = 11

17924 - 10751 = 7173 --> prime factors = 3,3,797 so 797 + (3+3)*(3*3) = 851

2010 - 1982 = 28 --> prime factors = 2,2,7 so 7 + (2+2)*(2*2) = 23

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23 - by taking the absolute value of the difference of the two given numbers and then taking the highest prime factor of that difference and adding to that the product of the sum of the remaining prime factors and the product of the remaining prime factors -

21 - 7 = 14 --> prime factors = 2,7 so 7 + (2) * (2) = 11

17924 - 10751 = 7173 --> prime factors = 3,3,797 so 797 + (3+3)*(3*3) = 851

2010 - 1982 = 28 --> prime factors = 2,2,7 so 7 + (2+2)*(2*2) = 23

Indeed, your operation works for the given examples. I like it It's not however the *|* that I had in mind (it gives a different result to the question).

Nonetheless, it deserves a prize as a valid hypothesis.

Perhaps a new example can re-launch the challenge?

Give me a pair (a,b) and I'll compute a*|*b. (i.e. assume another tablet has been found on the planet, etc, to maintain the "cover story" ).

Edited by araver
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• 0

Knew it was a little too convoluted. Would have thought it much easier to find several cases that satisfied the operation given only two data points. And tried many many different approaches. A tribute to a well thought out challenge. Will be looking for another tablet as well as continuing analyzing what we've found thus far.

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Indeed, your operation works for the given examples. I like it It's not however the *|* that I had in mind (it gives a different result to the question).

Nonetheless, it deserves a prize as a valid hypothesis.

Perhaps a new example can re-launch the challenge?

Give me a pair (a,b) and I'll compute a*|*b. (i.e. assume another tablet has been found on the planet, etc, to maintain the "cover story" ).

What is 1 *|* 0 ?

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What is 1 *|* 0 ? That is a big hint `1 *|* 0 = 2`

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• 0 ##### Share on other sites
• 0 This puzzle seems very hard, at least to my limited brain .

You gotta give more clues. What is 1 *|*1 and 1 *|* 2? You can use spoilers so as not to spoil it for other members.

Thank you.

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This puzzle seems very hard, at least to my limited brain .

You gotta give more clues. What is 1 *|*1 and 1 *|* 2? You can use spoilers so as not to spoil it for other members.

Thank you.

Here's one solution that fits the existing 6 clues. Plenty more where this comes from =)

a *|* b = ( 1700 + 1309 * a1 + 182 * a2 + 985* a3 + 1440* b1 + 1398 b2 ) mod 2087

Edited by bushindo
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• 0 Here's one solution that fits the existing 6 clues. Plenty more where this comes from =)

a *|* b = ( 1700 + 1309 * a1 + 182 * a2 + 985* a3 + 1440* b1 + 1398 b2 ) mod 2087

It surely doesn't fit 1 *|* 0 =2. Also your operation is clearly not commutative (a *|* b <> b *|* a) while OP mentioned *|* is commutative.

Also, can you please let me know how did you come up with such a complex formula? Do you have some standard method for these kind of problems or you are using some computer algorithm?

Edited by amateur
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• 0

It surely doesn't fit 1 *|* 0 =2. Also your operation is clearly not commutative (a *|* b <> b *|* a) while OP mentioned *|* is commutative.

Also, can you please let me know how did you come up with such a complex formula? Do you have some standard method for these kind of problems or you are using some computer algorithm?

Actually, the example above does work for 1 *|* 0; ( 1700 + 1309*(1) + 182*(1) + 985*(1) ) = 4176, which is equal to 2 modulo 2087. I did not notice the post that required the operation to be commutative. That can be easily fixed. This is one example that fits all 6 clues and is commutative

a *|* b = ( 421 + 1136* (a+b)1 + 1882* (a+b)2 + 1093* (a+b)3 + 1634* (a+b)4 + 97* (a+b)5 ) mod 2087.

It is straightforward to find these coefficients using linear algebra. A new tablet may show up that shows this example to be incorrect, however, it is relatively easy to construct another example that fits the new information.

Edited by bushindo
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• 0 Actually, the example above does work for 1 *|* 0...

I used 1390 instead of 1309 while copying your formula. Yes it does seem to work for the examples (1,0) , (10,23), (21,7)... I did not check for others.

And thanks for pointing me to linear algebra. It has been ages since I touched a maths text book ##### Share on other sites
• 0

a *|* b = ( 421 + 1136* (a+b)1 + 1882* (a+b)2 + 1093* (a+b)3 + 1634* (a+b)4 + 97* (a+b)5 ) mod 2087.

It is straightforward to find these coefficients using linear algebra. A new tablet may show up that shows this example to be incorrect, however, it is relatively easy to construct another example that fits the new information.

I like your thinking very much , but its

... finite. For every linear-algebra generated function (sort of interpolation) you can think of, I am almost sure I can find a pair that doesn't satisfy it. I'm pretty confident (although I haven't tried a formal proof), that in the long run, this operation cannot be expressed with a finite formula as above.

So, technically, we're pretty much down to an iterative no-cheating version of the classic game: "Tell me the number I am thinking of".

You cannot win (probably, as I said, I don't have a formal proof, just a hunch) and I cannot win (since the game never ends).

However, in theory, at infinity, I win.

In the original setting of the problem:

"In a future not so far way, Earth archaeologists find on a far away planet a fragment from a long lost civilization. This fragment involves an unknown operation *|*. Unlocking its secrets may lead to a breakthrough in understanding their civilization. "

Please allow me to state (and add to the OP) 2 informal assumptions I made when posting the original setting of the problem:

1) We're talking about aliens so they *might* possess different insights than humans.

2) It's a fairly used operation on their world, so one *might* expect the symmetry /simplicity / "beauty" of a human operation (e.g. addition/ multiplication, etc.)

Disclaimer: I am not implying the fact that an alien operation is bound to be hard to grasp by humans, just that in worst-case scenarios it might be difficult to translate in human-operations. This may or may not be the case here. Just sayin'.

P.S. I'm not sure how sane I am at this hour so if I don't make much sense, feel free to skip these assumptions. Other than the slightest hint possible, there's no real value / necessity in these assumptions.

EDIT: grammar Edited by araver
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• 0

I like your thinking very much , but its

... finite. For every linear-algebra generated function (sort of interpolation) you can think of, I am almost sure I can find a pair that doesn't satisfy it. I'm pretty confident (although I haven't tried a formal proof), that in the long run, this operation cannot be expressed with a finite formula as above.

So, technically, we're pretty much down to an iterative no-cheating version of the classic game: "Tell me the number I am thinking of".

You cannot win (probably, as I said, I don't have a formal proof, just a hunch) and I cannot win (since the game never ends).

However, in theory, at infinity, I win.

In the original setting of the problem:

"In a future not so far way, Earth archaeologists find on a far away planet a fragment from a long lost civilization. This fragment involves an unknown operation *|*. Unlocking its secrets may lead to a breakthrough in understanding their civilization. "

Please allow me to state (and add to the OP) 2 informal assumptions I made when posting the original setting of the problem:

1) We're talking about aliens so they *might* possess different insights than humans.

2) It's a fairly used operation on their world, so one *might* expect the symmetry /simplicity / "beauty" of a human operation (e.g. addition/ multiplication, etc.)

Disclaimer: I am not implying the fact that an alien operation is bound to be hard to grasp by humans, just that in worst-case scenarios it might be difficult to translate in human-operations. This may or may not be the case here. Just sayin'.

P.S. I'm not sure how sane I am at this hour so if I don't make much sense, feel free to skip these assumptions. Other than the slightest hint possible, there's no real value / necessity in these assumptions.

EDIT: grammar I was just being facetious. I know I didn't have the correct function, but wanted to point out the fact that

For any set of N clues for a *|* b, there are an infinite number of functions f(a, b) that fit the N clues.

The challenge is to find one that has 'symmetry /simplicity / "beauty"'. I'm not really good at this type of problem, so I'll wait for the correct answer from the brain denizens. Nice puzzle. It's rare for a puzzle on brainden to be unsolved this long.

Edited by bushindo
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• 0

I was just being facetious. I know I didn't have the correct function, but wanted to point out the fact that

For any set of N clues for a *|* b, there are an infinite number of functions f(a, b) that fits the N clues.

The challenge is to find one that has 'symmetry /simplicity / "beauty"'. I'm not really good at this type of problem, so I'll wait for the correct answer from the brain denizens. Nice puzzle. It's rare for a puzzle on brainden to be unsolved this long.

You are right and also, I know that you are very good (from this and other posts you made in other topics) with linear algebra. Also, I forgot to say that you've probably uncovered

on what the operation is not. More to the point, it should depend on the least number of (arbitrary) constants as possible.

Because basic constants deduced from a finite set of rules are most likely to be contradicted when uncovering extra rules (without breaking the consistency of course). Only true-invariants are bound to be forever lasting. ( I think I'm slightly more riddle-inclined right now)

However even if the answer is not obtained on this particular path, this doesn't imply that you cannot reach it. As I know from other topics, you're also a very good programmer, so you have very good chances to approach this problem (successfully) from another angle. I believe there are more ways to approach and solve it - either from mathematical specific viewpoints or theoretical computer-science specific viewpoints.

OFF-TOPIC Thank you for the (implied) compliment. I was worried at first that no one would try to solve it.

Edited by araver
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• 0

This will be the final example, as I feel it's enough info already.

Recap:

```
21 *|* 7 = 11

17924 *|* 10751 = 851

1089 *|* 121 = 2069

10 *|* 23 = 6

1 *|* 0 = 2

999 *|* 909 = 1359

14 *|* 14 = 14

2 *|* 1 = 0

```
and
```
a *|* b = b *|* a

```

Edited by araver
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This will be the final example, as I feel it's enough info already.

Recap:

```
21 *|* 7 = 11
17924 *|* 10751 = 851
1089 *|* 121 = 2069
10 *|* 23 = 6
1 *|* 0 = 2
999 *|* 909 = 1359
14 *|* 14 = 14
2 *|* 1 = 0
[/code]

and

[code]
a *|* b = b *|* a
```

Oh...OK. I guess I'm done. Thanks for the puzzle!

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