[araver]

Mystery Operation III

We've barely started exploring our solar system when a message is received from outer space.

It looks like an encoded operation is embedded in the message. And it also looks as it is a test of some sort.

4 *&* 8 = 6

20 *&* 5 = 12

8 *&* 31 = 34

-------------

44 *&* 47 = ?

Maybe we ought to respond to it?

What do you think?

P.S. General rules are the same as in and

[araver]

54?

No, sorry.

[araver]

Another random example (since no requests so far)

4 *&* 8 = 6

20 *&* 5 = 12

8 *&* 31 = 34

15 *&* 2 = 9

80 *&* 31 = 62

71 *&* 40 = 86

83 *&* 13 = 94

50 *&* 11 = 30

40 *&* 42 = 28

21 *&* 20 = 20

-------------

44 *&* 47 = ?

[plainglazed]

if either operand is prime subtract one. if an operand is even, divide it by its highest prime factor then multiply by its lowest prime factor (if there is a lower one). if an operand is odd, first subtract its lowest prime fractor (only have two cases thus far so maybe just subtract 3) then treat as even. finally, add the two decoded operands.

for example: 80 *&* 31 = 62 --> (80*2/5) + (31-1) = 62

This method seems to work for all the examples except the final answer. For that reason am thinking I might still be close but havent gotten anything else.

EDIT: suspect the extra manipulation of the odd operands is a little too convoluted and maybe the subtracting of one from the prime operands is too and that there is a simple single rule that works for all.

Edited January 8, 2011 by plainglazed

[araver]

4 *&* 8 = 6

20 *&* 5 = 12

8 *&* 31 = 34

15 *&* 2 = 9

80 *&* 31 = 62

71 *&* 40 = 86

83 *&* 13 = 94

50 *&* 11 = 30

40 *&* 42 = 28

21 *&* 20 = 20

22 *&* 12 = 14

-------------

44 *&* 47 = ?

[Guest]

Even a hint of a solution to this continues to elude me. Does the operation affect two digit numbers differently than single digit numbers?

[araver]

Even a hint of a solution to this continues to elude me. Does the operation affect two digit numbers differently than single digit numbers?

The operation is essentially the same, regardless of the number of digits

Adding another example to the challenge:

4 *&* 8 = 6

20 *&* 5 = 12

8 *&* 31 = 34

15 *&* 2 = 9

80 *&* 31 = 62

71 *&* 40 = 86

83 *&* 13 = 94

50 *&* 11 = 30

40 *&* 42 = 28

21 *&* 20 = 20

22 *&* 12 = 14

2 *&* 36 = 13

-------------

44 *&* 47 = ?

[araver]

Adding another example to the challenge:

4 *&* 8 = 6

20 *&* 5 = 12

8 *&* 31 = 34

15 *&* 2 = 9

80 *&* 31 = 62

71 *&* 40 = 86

83 *&* 13 = 94

50 *&* 11 = 30

40 *&* 42 = 28

21 *&* 20 = 20

22 *&* 12 = 14

2 *&* 36 = 13

80 *&* 81 = 86

-------------

44 *&* 47 = ?

[araver]

Adding another example to the challenge:

4 *&* 8 = 6

20 *&* 5 = 12

8 *&* 31 = 34

15 *&* 2 = 9

80 *&* 31 = 62

71 *&* 40 = 86

83 *&* 13 = 94

50 *&* 11 = 30

40 *&* 42 = 28

21 *&* 20 = 20

22 *&* 12 = 14

2 *&* 36 = 13

80 *&* 81 = 86

10 *&* 14 = 10

-------------

44 *&* 47 = ?

[curr3nt]

Is it 66?

[araver]

Is it 66?

Care to say why? (in a spoiler of course)

[Guest]

I should have thought of looking for factor anomalies.

[araver]

Started with prime - 1

83 *&* 13 = 94 --- (83-1) + (13-1) = 94

So went with N - ? for each number

Started filling in the ones with prime numbers

20 *&* 5 = 12 --- (20-12) + (5-1) = 12

8 *&* 31 = 34 --- (8-4) + (31-1) = 34

80 *&* 31 = 62 --- (80-48) + (31-1) = 62

71 *&* 40 = 86 --- (71-1) + (40-24) = 86

50 *&* 11 = 30 --- (50-30) + (11-1) = 30

The ? shares some common denominators

Except 15 *&* 2 = 9 --- (15-7?) + (2-1) = 9

Which I pretended wasn't there...

Figured

22 *&* 12 = 14 --- (22-12) + (12-8) = 14

With what I had

44 *&* 47 = ? --- (44-?) + (47-1) =

Guessed ? was 24 from 22 *&* 12 = 14

Making

44 *&* 47 = ? --- (44-24) + (47-1) = 66

Basically... not sure how it works. I'm just making things up.

Edited because tabs do not work, neither do spaces...

curr3nt, I like the way you deduced it - it's by no means simply a lucky guess. But the operation is not solved yet.

So, adding your answer to the known examples + a new one.

And changing the challenge to another question.

4 *&* 8 = 6

20 *&* 5 = 12

8 *&* 31 = 34

15 *&* 2 = 9

80 *&* 31 = 62

71 *&* 40 = 86

83 *&* 13 = 94

50 *&* 11 = 30

40 *&* 42 = 28

21 *&* 20 = 20

22 *&* 12 = 14

2 *&* 36 = 13

80 *&* 81 = 86

10 *&* 14 = 10

44 *&* 47 = 66 (solved by curr3nt)

60 *&* 40 = 32

-------------

96 *&* 91 = ?

Edited January 11, 2011 by araver

[curr3nt]

Noticed a pattern with numbers having 2 factors (excluding 1 and itself) with one of them being 2.

N - X pairs

4 (2*2) - 2

10 (2*5) - 6

14 (2*7) - 8

22 (2*11) - 12

This leads me to believe

6 (2*3) - 4 because the number subtracted is the largest factor + 1...except for (2*2)... can we get an example with 9 or 25 as one of the numbers?

Also multiplying the number by 2 multiplies the other number by 2 if 2 is a factor in the original number

10 - 6

20 - 12

40 - 24

80 - 48

also

22 - 12

44 - 24

If 6 - 4 then

12 - 8 (known)

24 - 16

48 - 32

96 - 64

So 96 *&* 91 is (96 - 64) + (91 - ?)

Now we come to the 15 - 7 pair. In the cases above other than 2*2 it could be the two factors added together minus one. 15 would be 3*5 which would be 5+3-1 to get the 7.

91 is 7*13 so would the number subtracted be 7+13-1 or 19?

Making it (96 - 64) + (91 - 19) = 104?

[curr3nt]

The pair 42-30 is throwing me off now. It has me thinking 14-8 should be 14-10 but that makes 10-4 which throws off 20-12, 40-24 and 80-48.

Unless something else happens with unique factors. Maybe an example with 28 and one with 63?

Also, thinking I should be in bed asleep right now...

btw, this is fun araver thanks!

[araver]

OK, it's clear you can probably deduce your way around computing almost any example of the operation

So I'm not posting a third challenge. However:

there is a special relation that actually explains the operation i.e. explains the logic behind those values.

This would be the final challenge to mark this problem as solved!

[curr3nt]

Based upon the available pattern 30-22 but I do not know why... 30 is another with unique factors (2*3*5) like 42.

Also guessing that

9-3

25-5

49-7

28-16

63-27

Trying to find an easier operation but it is eluding me.

[plainglazed]

seperate each operand into its prime factors and organize by pairing the two lowest unique prime factors and listing any remaining prime factors:

11 ==> 11

15 ==> 3,5 ==> (3,5)

20 ==> 2,2,5 ==> (2,5) and 2

36 ==> 2,2,3,3 ==> (2,3) and 2,3

42 ==> 2,3,7 ==> (2,3) and 7

for each pair of unique prime factors from above take their product and subtract their sum -1

for each unpaired unique prime factor subtract 1

for each unpaired non-unique prime factor do nothing

take the product of the above results

11 ==> 11 ==> 11-1 ==> 10

15 ==> 3,5 ==> (3,5) ==> 3*5-(3+5-1) ==> 8

20 ==> 2,2,5 ==> (2,5),2 ==> [2*5-(2+5-1)] * 2 ==> 8

36 ==> 2,2,3,3 ==> (2,3),2,3 ==> [2*3-(2+3-1)] * 2 * 3 ==> 12

42 ==> 2,3,7 ==> (2,3),7 ==> [2*3-(2+3-1)] * (7-1) ==> 12

then add the two modified operands

not sure what to do in the case of four or more unique prime factors as in 210. perhaps there would need to be two pairs (or more if appropriate) instead of one pair with two seperate primes.

also, think all the above may be better simplified.

[araver]

Both of you are sort of on the right-track. Just need to find the bigger picture.

[Guest]

96 *&* 91 = 74?

[Guest]

Briefly explain, we need to do some transformation to both of the numbers.Rule 1: If it is a prime number, minus 1.Rule 2: If the number is not a prime number, after factorizing it, 2 is a factor and there are no more than 2 factors, half one of the largest factor (ignore the 0.5) to obtain a new number. E.g 80 = 2^4 *5 Halve 5 (biggest factor) = 2, new number = 2^4 * 2 = 32Another example, 36 = 2^2 *3^2, Halve one of 3 (biggest factor) = 2^2 * 3 * 1(after halving) = 12Rule 3: The preconditions to activate rule 3 is not clear as there is limited number of examples, it can be whenever there are 2 or more odd factors present, OR whenever the number is odd(not prime of cos), depending on the authors design.Minus all the factors by 1, and obtain a new number.E.g 21 = 3 * 7 Transform to (3-1) * (7-1) = 12 (There are 2 or more odd factors)15 = 3 * 5 Transform to (3-1) * (5-1) = 8Add two transformed numbers together to obtain the solution.96 *&* 9196 = 2^5 * 3 , Rule 2: 2^5 *1 = 3291 = 7*13, Rule 3: (7-1)*(13-1) = 7296*&*91=32+72 = 104

Edited January 13, 2011 by iverson2002939

[araver]

Briefly explain, we need to do some transformation to both of the numbers.Rule 1: If it is a prime number, minus 1.Rule 2: If the number is not a prime number, after factorizing it, 2 is a factor and there are no more than 2 factors, half one of the largest factor (ignore the 0.5) to obtain a new number. E.g 80 = 2^4 *5 Halve 5 (biggest factor) = 2, new number = 2^4 * 2 = 32Another example, 36 = 2^2 *3^2, Halve one of 3 (biggest factor) = 2^2 * 3 * 1(after halving) = 12Rule 3: The preconditions to activate rule 3 is not clear as there is limited number of examples, it can be whenever there are 2 or more odd factors present, OR whenever the number is odd(not prime of cos), depending on the author’s design.Minus all the factors by 1, and obtain a new number.E.g 21 = 3 * 7 Transform to (3-1) * (7-1) = 12 (There are 2 or more odd factors)15 = 3 * 5 Transform to (3-1) * (5-1) = 8Add two transformed numbers together to obtain the solution.96 *&* 9196 = 2^5 * 3 , Rule 2: 2^5 *1 = 3291 = 7*13, Rule 3: (7-1)*(13-1) = 7296*&*91=32+72 = 104

Result is correct (same as curr3nt).

But operation was not found yet.

[araver]

Description of the operation is still OPEN!

Adding another example to the challenge:

4 *&* 8 = 6

20 *&* 5 = 12

8 *&* 31 = 34

15 *&* 2 = 9

80 *&* 31 = 62

71 *&* 40 = 86

83 *&* 13 = 94

50 *&* 11 = 30

40 *&* 42 = 28

21 *&* 20 = 20

22 *&* 12 = 14

2 *&* 36 = 13

80 *&* 81 = 86

10 *&* 14 = 10

44 *&* 47 = 66 (solved by curr3nt)

60 *&* 40 = 32

96 *&* 91 = 104 (solved by curr3nt)

33 *&* 24 = 28

[plainglazed]

yeah, had thought that bit in the OP about some sort of test was meant to direct us to a more general solution. I think what we have been finding when modifying each operand, n, is the number of positive integers less than or equal to n that are coprime to n.

EDIT: reworded to make sense. *coffee time*

Edited January 15, 2011 by plainglazed

[araver]

yeah, had thought that bit in the OP about some sort of test was meant to direct us to a more general solution. I think what we have been finding when modifying each operand, n, is the number of positive integers less than or equal to n that are coprime to n.

EDIT: reworded to make sense. *coffee time*

that is the operation. It is based on the

Euler's totient function

I'm going to have to take a small break to think of another Mystery OP.