4 ms·
https://math.stackexchange.com/questions/1142767/why-does-this-cube-root-trick-work https://math.stackexchange.com/questions/1142767/why-does-th... Also, I hav
by thrdbndndn 4y ago
https://math.stackexchange.com/questions/1142767/why-does-this-cube-root-trick-work https://math.stackexchange.com/questions/1142767/why-does-th...
Also, I have an additional question.
Obviously this trick relies on the fact that "the last digit of the cube of all numbers from 0 to 9 is distinct".
Is this fact just coincidence, or is it bound to happen?
- rk06 4y ago0^3 = 0 1^3 = 1 2^3 = 8 3^3 = 27 4^3 = 64 5^3 = 125 6^3 = 216 7^3 = 343 8^3 = 512 9^3 = 729 As you can see, for decimal system, last digit of cubes is unique for all digits. As last digit of cube depends on last digit of cube root only (same for squares and so on), it is bound to be true for all numbers
- thrdbndndn 4y agoYou misunderstand my question. I was asking if this (cubes of number 0-9, the one you list, all have unique last digit in 0-9) a coincidence or not. (I know a number's cube's last digit is only affected by its own last digit.)
- archi42 4y agoThe question is in which case(s) `f(x) = x^n mod m` is a bijection for `0 <= x < m`. My intuition is to guess that `gcd(n,m) = 1` is at least a sufficient condition (correct math term? In german it's "hinreichende Bedingung"). Maybe it's the sole necessary condition. But I did only think about this for 5 to 10s and did not spend too much time on looking for a counter example, let alone a proof strategy (my number theory is a bit rusty -- ah, but on second thought, if the theory is correct, then I bet the Chinese remainder theorem is involved).
- deleted 4y ago[deleted]
- sbaiddn 4y agoI think it only works if you know its a perfect cube from the get go. Which is useless for me in my day to day unless Im talking to my number theory brother in law. But he only likes to talk about acting and scrabble.
- cdelsolar 4y agoIs he a rated tournament player? There are dozens of us!
- deleted 4y ago[deleted]