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Can you guess my "nifty operator" (#) ?

+1 vote
117 views

This is a very difficult puzzle, I bet you can't solve it.

Let (#) (a circled star) be called "nifty operator" which is an operator and takes two natural numbers and produces a natural number.
Here you got some definitions:

4 (#) 3 = 1, 3 (#) 4 = 4
9 (#) 2 = 4, 2 (#) 9 = 6
5 (#) 7 = 5, 7 (#) 5 = 7
2 (#) 5 = 4
8 (#) 9 = 8
4 (#) 14 = 16

Can you tell me, how is a (#) b defined?

You maybe want some more tips:
~~~
If a is even then a (#) a = 0
If a is odd then a (#) a = a
n (#) 1 = n
1 (#) n = 1
~~~

If you can't solve it, here is a tip:
SPOILER 1!
(#) cannot be defined for complex numbers.
SPOILER 2!
a (#) b < a+b
SPOILER 3!
a (#) b = a X b Y (a Z b) where X,Y,Z are operators
SPOILER 4!
The result is always a natural number

posted Sep 5, 2019 by anonymous

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