(basically, each digit maps to itself, except (2, 8) and (3, 7) which map to the other one in the pair).
So, the last digit of the cubic root of 148877 is therefore 3.
Now, erase the last 3 digits. You are left with 148. To figure out the first digit of the answer, you have to memorize the cubes of 1 to 10, you probably already know most of them:
1, 8, 27, 64, 125, 216, 343, 512, 729, 1000. (You probably know 343 from Halo series, and 729 is 1729 (http://en.wikipedia.org/wiki/1729_%28number%29) without one, so the only hard thing here is 216).
So, 148 is between 125 and 216. Take the lesser one, 125 is 5^3. So the first digit is 5.
The answer is therefore 53. It only works with two-digit numbers, but with some practice you can do it in seconds.
Had a classmate who was a prodigy. He did the same cubic roots trick, except he could work with non-integers.
e.g. By your method, if I ask for the cube root 270119, you'd say 69, but the answer is is closer to 64.64. He could usually give the right answer to within four sig digs.
The way I understood his explanation, he'd do all the calculations in log space, so he could pull out high order roots, but the answers were less accurate.
Convert to binary, take (approximate) log2 of the number, divide by three, take 2 to the power of ans?
(Binary both because most programmers know powers of two, and because you are more accurate when you know the closest power of two to something than the closest power of 10)
Even just approximating log2(270119) as 18 gives you an answer of 64.
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.
"he carried into adulthood the concomitant problems in relating to others—and very pertinently to women—ensuring the prerequisite internal pool of frustrations essential for lateral thinking"
I don't think this follows at all, honestly. Frustration begets more frustration, not lateral thinking.
<<There are several categories of scientists in the world; those of second or third rank do their best but never get very far. Then there is the first rank, those who make important discoveries, fundamental to scientific progress. But then there are the geniuses, like Galilei and Newton. Majorana was one of these.>>
—(Enrico Fermi about Majorana, Rome 1938)
The story is just begging for a Dan Brown style book: a missing genius, a Majorana codex containing the secrets of the Universe, the power that goes with knowing those secrets, a secret Illuminati with connections to Galilei and Newton, a connection to the Cosa Nostra, throw in a Roman/Vatican connection, a WWII connection... A rip roaring novel (with screen rights) is in there somewhere!
I'll teach you. I used to do that when I was 10. Easy enough.
Let's say you want to take a cubic root from 148877. Looks scary enough?
First, look at the last digit. It's 7. Map it through the following correspondence:
1 -> 1, 2 -> 8, 3 -> 7, 4 -> 4, 5 -> 5, 6 -> 6, 7 -> 3, 8 -> 2, 9 -> 9, 0 -> 0
(basically, each digit maps to itself, except (2, 8) and (3, 7) which map to the other one in the pair).
So, the last digit of the cubic root of 148877 is therefore 3.
Now, erase the last 3 digits. You are left with 148. To figure out the first digit of the answer, you have to memorize the cubes of 1 to 10, you probably already know most of them:
1, 8, 27, 64, 125, 216, 343, 512, 729, 1000. (You probably know 343 from Halo series, and 729 is 1729 (http://en.wikipedia.org/wiki/1729_%28number%29) without one, so the only hard thing here is 216).
So, 148 is between 125 and 216. Take the lesser one, 125 is 5^3. So the first digit is 5.
The answer is therefore 53. It only works with two-digit numbers, but with some practice you can do it in seconds.
Now go impress someone.