[Prime]

In the 6th grade my math teacher showed the class a procedure to approximate square root. I asked whether there was a nice easy way to extract roots of higher powers. She said -- “No.” Few days later I came up to the teacher and asked her to raise any two-digit number to the third power and tell me the result. She obliged. Whereupon, it took me just a couple of seconds to tell her the original number she cubed. I asked to try it again with another number. She was annoyed, as there were no calculators back then, and raising a two-digit number into the third power was taking much longer than for me to take the root.

How is it done?

Can this trick be used with other powers? (Other than cube.)

Greater than 2-digit numbers?

[Guest]

In the 6th grade my math teacher showed the class a procedure to approximate square root. I asked whether there was a nice easy way to extract roots of higher powers. She said -- “No.” Few days later I came up to the teacher and asked her to raise any two-digit number to the third power and tell me the result. She obliged. Whereupon, it took me just a couple of seconds to tell her the original number she cubed. I asked to try it again with another number. She was annoyed, as there were no calculators back then, and raising a two-digit number into the third power was taking much longer than for me to take the root.

How is it done?

Can this trick be used with other powers? (Other than cube.)

Greater than 2-digit numbers?

I hope I can explain this well enough. It is not really as complicated as I think I make it sound, easy to come up with in your head once you understand. Thanks alot for making me think about this one Prime. I had never heard of this before and enjoyed figuring something out by myself for once, even if it was pretty obvious once I knew it was possible.

Here is how it is done:

To determine the first digit in the two digit number you need to remember cubed numbers. Any number that is greater than or equal to 1000(n^3), but less than 1000((n+1)^3) indicates that the first digit equals n. For Example...

If X is > or = 125000 (a.k.a. 1000(5^3)) but X is < 216000 (a.k.a. 1000(6^3)) Than the first digit of X must be 5.

To determine the second digit in the two digit number, look at the last digit in the cubed number.

0=0, 1=1, 2=8, 3=7, 4=4, 5=5, 6=6, 7=3, 8=2, and 9=9

For Example, X=551368 than the last digit of the cubed root of X must be 2 (because the last digit of X is 8) and the first digit of X must be 8 (because X is greater than 1000(8^3) but less than 1000(9^3). Thus, the cubed root of X=82

This trick can be continued for greater than 2 digit cubed roots, however it becomes more difficult to remember the cubed numbers to determine the first two digits

Depends on the power. Powers of 4 will always end in 1 or 6 so it would be a bit of a guess to determine the last number(except when the number being multiplied ends in 5, in this case the last digit will be five) but when the power is five, the last digit in the number will be the same as the last digit in the number being multiplied.

[Prime]

I hope I can explain this well enough. It is not really as complicated as I think I make it sound, easy to come up with in your head once you understand. ...

Here is how it is done:

To determine the first digit in the two digit number you need to remember cubed numbers. Any number that is greater than or equal to 1000(n^3), but less than 1000((n+1)^3) indicates that the first digit equals n. For Example...

If X is > or = 125000 (a.k.a. 1000(5^3)) but X is < 216000 (a.k.a. 1000(6^3)) Than the first digit of X must be 5.

To determine the second digit in the two digit number, look at the last digit in the cubed number.

0=0, 1=1, 2=8, 3=7, 4=4, 5=5, 6=6, 7=3, 8=2, and 9=9

For Example, X=551368 than the last digit of the cubed root of X must be 2 (because the last digit of X is 8) and the first digit of X must be 8 (because X is greater than 1000(8^3) but less than 1000(9^3). Thus, the cubed root of X=82

This trick can be continued for greater than 2 digit cubed roots, however it becomes more difficult to remember the cubed numbers to determine the first two digits

Depends on the power. Powers of 4 will always end in 1 or 6 so it would be a bit of a guess to determine the last number(except when the number being multiplied ends in 5, in this case the last digit will be five) but when the power is five, the last digit in the number will be the same as the last digit in the number being multiplied.

Yes. That's how the trick was performed.

To rephrase the method for guessing cubic root:

You must memorize 9 cubes:

10³=1,000

20³=8,000

30³=27,000

40³=64,000

50³=125,000

60³=216,000

70³=343,000

80³=512,000

90³=729,000

That helps you to guess the first digit (tens) of the number. Whereas, the second digit yields a unique ending: 0-->0; 1-->1; 2-->8; 3-->7; 4-->4; 5-->5; 6-->6; 7-->3; 8-->2; 9-->9.

The ending is going to be unique for any odd power. E.g., the 5th power will end with the same digit as the original number. In general, any 4n+1 power will end with the same digit as the number raised, and any 4n+3 power has the same pattern as the cube shown above. (Integer n>=0.)

But how to memorize those large numbers? Imagine you show your trick with a power of 13 to a group of people, and then someone goes: "here I raised something into the power of 7..."

So more work is needed for other odd powers.

As for the 3-digit numbers, I wouldn't want to memorize 100 different values. Need some other trick there. Frankly, I don't know what it is, but I have an idea...