1) My Love for you is like $L= m.C^2$...because it grows xxx with speed of xxxx.
2) My Love is like a xxxx it goes on xxxxx.
3) You must be the xxxxx xxxx of xxxxx because I feel irrational around you.
4) The derivative of my love for you is xxxx because my love for you is xxxx.
5) Honey, you are sweeter than xxxxxx.
6) I am a mental math ...
We have that
Upon editing this, and reviewing hexomino's answer, that our reasoning is pretty much identical, although not in the same order. I don't have time to write a complete explanation, especially when it's already written.
Partial answer that I'm saving for now
(For convenience, I will call $PCRON$ "the root" and $PRINCETOM$ "the square".
We can first deduce that the digit N
We can also do some quick tests to find the approximate range
Let's try that:
We can then determine that C is
!Key to solving this is rearranging terms, otherwise it becomes messy.
!$(A!+B) $- $(A+B)^2$
!+$ (C!+B)$ -$ (2C+B)^2$
!+$(D!+B)$ - $(DB)^2$ = $0$
!For 1st term, simple inspection yields A=4, B=1.
!For second term, C=3, B=1.
!For third term,
based on Factorial values
! D has to be 7
!Everything falls into place ...
Partial answer (could be used as a springboard by somebody else)
Preliminary size considerations:
If one of $A,C,D$ is 7, then
[to be completed]
If none of $A,C,D$ is 7, then
Considering each of these possibilities in turn:
So Case 2a is impossible.
[to be completed]
After wondering why people weren't doing very simple ones. I reread the rules realizing I misread the part where it says "Each contains all the digits 1 to 9 occurs only once."
after correcting my self, I came up with this:
Since no answer was posted so far, in spite of hints, I am posting an answer. In general, most of my sequences won’t be found in OEIS. They are generated using simple rules based on widely known sequences and they can be regenerated without calculators and computers. They have several practical applications in security and cryptography.
OK - here is a real solution this time without concatenation.
I have a couple sub expressions.
Now I can construct $A,B,C$.
Not sure if this is in the spirit of the puzzle since I am using the fact that $1 \div 1 = 1$ and getting a set of pan digital sub-expressions which equal to 1.
EDIT: I've added some other possibilities which don't bear as much ...
Square 1. Sum of digits 1.
Square 4. Sum of digits 4.
Square 9. Sum of digits 9.
Square 16. Sum of digits 7.
Square 25. Sum of digits 7.
Square 36. Sum of digits 9.
Square 49. Sum of digits 13.
Square 64. Sum of digits 10.
Square 81. Sum of digits 9.
Square 100. Sum of digits 1.
Square 121. Sum of digits 4.
Square 144. Sum of digits 9.
Intent of the puzzle is to create numberless sudoku system which can represent all the digits as well as a pattern which could be used to solve sudoku.
My intent was not properly conveyed.
Here is my thinking:
Letter “ T “ represents number 2. Symbol I select is dot “.” Based on defined rules, all other numbers are generated to be used in Sudoku ...
I don't think this is much of an answer because I don't think I've added anything new.
In order for the puzzle to be playable, we would need to have all 9 digits representable in our customisation.
So for the symbol, I will use + and for the character T. Here is one way to encode 9 distinct sets using 'T' and '+'.
T + TT
++ T+ +T
I got it down to
Using all the same methods as @RandAlThor, but choosing a
For double checking purposes, here's the sudoku as solved by https://sudokusolver.net/
For further improvement, it's very likely that some digit(s), particularly those that occur twice, can be replaced by adding both a 2 and a 3, giving a sudoku that has more clues but fewer ...