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I watched on tv that a guy could say the value of $\sqrt [ 3 ]{ x } $ for a given $x$, in less than ten seconds. Moreover, he could do for $\sqrt [ 13 ]{ x } $, $\sqrt [ 103 ]{ x } $, $\sqrt [ 1003 ]{ x } $. I know that he was looking to the the last 3-4 digits of $x$. For example if the last digit of x is $1$, than the last digit of $\sqrt [ 3 ]{ x } $ must be $1$. However I could not model for the last 3-4 digits. So, the question is:

If you know that $0<\sqrt [ 3 ]{ x }<99 $ , what is the value of $\sqrt [ 3 ]{ x }$ for a given $x$

For example for $x=157464$, you can say $\sqrt [ 3 ]{ x }=54 $ by looking for three or two digits of $x$

Additionally build the same model for $\sqrt [ 13 ]{ x } $,$\sqrt [ 103 ]{ x }$,$\sqrt [ 1003 ]{ x }$

Note: $\sqrt [ 3 ]{ x } $ and $x$ are integers

Note2: $x$ was given as 3684849759365252680119420227475775049492674443146905344822745288298143046106331441531798764782169420286244212297417702761555740760756048067228412045874270238715327556678211597193757451059582259503545059959722118963270933846793818460752099206747525659014794031653506061992511860134802242876788750520888820110080505679940477828522656005828478195470651831585339407491541412547532043564019267830576862266475657742655079354385864558909932489824922657536960135959364334680734490825496314934021284389706462034993094441427484468431831131329495454329359968306231964741732502732936584806687832283240524973095558123557893421615630299392659144456999160880961090629674198135946098548815074663065403716229881899998734677466563511307098758301725180878228542151982532894727010449270694673832788117758121432660157632144863623257572958865327742110214978113074412669861204102215454568565209907518122414604893305540601040443389773145449494505681564919573412773764876132208183779019683251734342466854174739320538873007486268980279129645406862537226568433208719346636825106745675952144307470309490075574156853916253370585261188823302350410947564043990554122093606148767744

He could say that $\sqrt [ 1003 ]{ x }=14$ very quickly.

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some people can execute practiced algorithms in their mind quickly. – user59671 May 12 '13 at 15:28
up vote 3 down vote accepted

Knowing that a number is a perfect cube, you can use a bit of simple logic to find the value of it quickly.

You begin by noting the number of digits in the resulting number - if $N$ is the number of digits in the perfect cube, then you know that the cube root will have $\left\lfloor\frac{N-1}3\right\rfloor+1$ digits.

Now, look at mod $10$. In this case, the sequence from 0 to 9 produces the sequence: $$ 0, 1, 8, 7, 4, 5, 6, 3, 2, 9 $$ As such, knowing that 2<->8 and 3<->7, and the others remain where they are, you can quickly identify the last digit of the number you're looking for. For instance, with the example you provided, $157464$ ends in 4 and has $N=6$ digits, so we're looking for a $\left\lfloor\frac{6-1}{3}\right\rfloor+1 = 2$ digit number, and it ends in 4.

In the case where you know that it's a 2 digit number, it's fairly easy to then use a bit of bounding to find the right value. Notice that $$ 50^3 = 125000 $$ and $$ 60^3 = 216000 $$ As such, you know that it has to be between $50$ and $60$. With the knowledge that the last digit is $4$, you quickly get $54$ as the final answer.

Where more digits are required, a more detailed method has to be used, of course. But the logic works the same in any case.

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