3 ms·
> So, the last digit of the cubic root of 148877 is therefore 3. So your algorithm only works for numbers that are cubes of an integer, right?
by sampo 12y ago
> So, the last digit of the cubic root of 148877 is therefore 3.
So your algorithm only works for numbers that are cubes of an integer, right?
- araes 12y agoI believe the first digit trick still works with off-numbers, but the second digit trick definitely fails, as it depends on tricks of cubics. 1=>[1], 2=>[8], 3=>2[7], 4=>6[4], 5=>12[5], 6=>21[6], 7=>34[3], 8=>51[2], 9=>72[9], 10=>100[0] (or 0=>[0]) Basically, if you memorize the first 10 cubics, then you can do the next 10 cubics (and actually, there's probably a trick for doing all the integer ones recursively) If you're really good at this trick, I'm sure you can bound non-integer numbers with referenced cubics and then do something like Newtonian descent. Its kinds of like building a lookup table of lookup tables in your head for integer cubics.
- atemerev 12y agoRight. Still enough to claim that you can do "cubic roots of 6-digit numbers in your head".