2 ms·
0^3 = 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
by rk06 4y ago
0^3 = 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).