+622 How many digits ($0$ to $9$) never appear as the last digit of a perfect cube?
Every digit will apear as the last digit of a perfect cube:
Examples: 10^3 ==1000
11^3==1331
8^3==512
7^3==343
14^3==2744
5^3==125
16^3=4096
3^3==27
12^3==1728
9^3 ==729
+36916 All of them can appear .....just cube the numbers 0 - 9 to find:
0^3 = 0
1^3 = 1
8^3 = 512
7^^3 = 343
4^3 = 64
5^3 = 125
6^3 = 216
3^3 = 27
2^3 = 8
9^3 = 729
