Ask me others if u want
Obviously... =)
Your number = #
(2# + 6 )/2 -# =
#+3 - #=
=3
You could choose any number (that includes numbers like pi or square root of 2 or even a+bi (complex numbers))
Here is one a little harder.
Take any natural number (from 1 to infinity).
1. Divide this number by 2 until you get an odd number, go to step 2.
2. multiply by 3 and add 1
Repeat step 1
In the end you will always get the same number, what is it?
Obviously... =)
Your number = #
(2# + 6 )/2 -# =
#+3 - #=
=3
You could choose any number (that includes numbers like pi or square root of 2 or even a+bi (complex numbers))
Here is one a little harder.
Take any natural number (from 1 to infinity).
1. Divide this number by 2 until you get an odd number, go to step 2.
2. multiply by 3 and add 1
Repeat step 1
In the end you will always get the same number, what is it?
It is conjectured that you will always eventually get to 4, but it yet to be proven. A paticularly long example of is 27, which gets up to the thousands before reaching 4.
Pick a # between 1-10. (Don't tell me)
Double it
Add 6
Cut it into half
Subtract it from the # you started with
Is your answer 3?
I can do this all day long :)