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phil1882 added an answer to a question recursive number addition
alright thanks good to know.

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phil1882 added an answer to a question recursive number addition
the strings <<>><><>, <><><<>> are considered equivalent, though for the proposes of uniqueness, i put the factors in order from lowest to highest.

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phil1882 added an answer to a question recursive number addition
correct.
one of my questions is about addition.
basically my question is, given a number system that uses multiplcation as its basis, (multiplying two numbers in recursive format is just concatenation) an you come up with a method for addition?
ill do 7 as an example
7 is prime.
< 7 >
7 is the 4th prime so replace the 7 with a 4.
< 4 >
4 is composite, its factors are 2 and 2.
< 2 2 >
2 and 2 are prime.
<<2><2>>
the number two is the first prime so cant be reduced further. replace with blank.
<<><>>

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phil1882 added a question in New Logic/Math Puzzles
recursive number additionif n is composite, break it into its prime factors, if n is prime, place < > around n and replace n with the nth prime its is. the first 20 numbers then are...
1
<> 2
<<>> 3
<><> 4
<<<>>> 5
<><<>> 6
<<><>> 7
<><><> 8
<<>><<>> 9
<><<<>>> 10
<<<<>>>> 11
<><><<>> 12
<<><<>>> 13
<><<><>>14
<<>><<<>>> 15
<><><><> 16
<<<><>>> 17
<><<>><<>> 18
<<><><>> 19
<><><<<>>> 20
hypothesis:
1) there is no general method for adding recursive numbers.
2) numbers that differ by 1 wont differ in recursive representation by more than 2 brackets
3) symmetric recursive numbers, recursive numbers that can be represented in such a way that they are a mirror image at the middle, grow at a logarithmic rate somewhat similar to the primes.
can you confirm or disprove any of these?
 7 replies
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phil1882 added an answer to a question Prove that you solved sudoku
it seems to me what you really need is an encryption method that can be reversed if necessary but would take longer to reverse than to solve the sudoku yourself and get the same encryption. what do you think of this idea? it may not even require a computer, depending on how difficult the encryption method is.

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phil1882 added an answer to a question 11 Letters
doorenownly is the best i could com up with.

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phil1882 added an answer to a question The Android unlock pattern
i'd say its

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phil1882 added an answer to a question Folding a paper into a sphere
assuming you also had scissors...

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phil1882 added an answer to a question Equilateral triangle on a grid

phil1882 added an answer to a question Triangle with a given perimeter
no clue on this one, you can divide a single line segment to any measurement, but showing any point m to construct a triangle of any parameter? much much harder i think.

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phil1882 added an answer to a question numbers by the book
alright sorry, i tried to make it such that there was only 1 solution but looks like i failed. thanks for finding all 3.

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phil1882 added an answer to a question if a,b,c, and d are natural numbers revisited.
yeap, or more specifically a power of 3.

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phil1882 added an answer to a question if a,b,c, and d are natural numbers revisited.
i'm not going to list how to make each weight, however

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phil1882 added an answer to a question numbers by the book
this solution doesn't work, as tim and fred both end in 2.
the numbers 15 should appear once in each digit place between men, as hint #2 tries to explain.

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phil1882 added a question in New Logic/Math Puzzles
numbers by the bookFive men, one of which is Greg each selected 5 unique 3 digit numbers from a hat. given the clues, can you tell what the numbers are and the men who selected them?
1. Each digit for the number chosen by each man is unique. (that is, if the number begins with 1, 1 wont appear anywhere else in the number.)
2. Each digit for each place between men is unique. (that is, if a number begins with 1, no other number begins with 1.)
3. Only the digits 15 are used.
4. Andy has the smallest number, and it's evenly divisible by 5.
5. Both Bob's and Fred's numbers are evenly divisible by 3.
6. Bob's number is smaller than Fred's.
7. Tim's number is evenly divisible by 2, and a prime.
8. Tim's number is greater than Bob's.
9. Fred's middle most number is prime.
10. Bob's middle and last number are odd.
11. the middle digit of Tim's number is twice as big as Andy's.
good luck and enjoy!
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