Why does the following 8,9,10 pattern exist (please explain or direct
me to an explanation/proof (that counts as a correct answer)) among
cubes of numbers:
2^3 = 8
3^3 = 9
4^3 = 64; 6+4 = 10
5^3 = 125; 1+2+5 = 8
6^3 = 216; 2+1+6 = 9
7^3 = 343; 3+4+3 = 10
8^3 = 512; 5+1+2 = 8
9^3 = 729; 7+2+9 = 18; 1+8 = 9
10^3 = 1000; 10+0+0 = 10
11^3 = 1331; 1+3+3+1 = 8
12^3 = 1728; 1+7+2+8 = 18; 1+8 = 9
13^3 = 2197; 2+1+9+7 = 19; 1+9 = 10
etc etc
this silly pattern kept me from having a good night's sleep last night
thanks for any help!

Interesting question, I had never seen this before. The explanation is
pretty simple, though. It is the old "casting out nines" business. The
sum of the digits of any natural number has the same remainder when
divided by nine as the number itself. Then just look at three cases
for a natural number n:

(3n)^3 = 27n^3 = 0 modulo 9
(3n+1)^3 = 9n^3+9n^2+9n+1 = 1 modulo 9
(3n-1)^3 = 9n^3-9n^2+9n-1 = -1 modulo 9

Get it?

The proposition does not seem to hold:

19^3 = 6859

6 + 8 + 5 + 9= 28
2 + 8 = 10
1 + 0 = 1

Methodoogy:

I typed

 (each succeding prime number after 13) ^ 3

 in the Google search bar and got the 6859 for 19.

I tried primes only - but not for an absolutely logical reason.

Oops! I misread the question. Scratch my answer.

On reflection, I need to clarify my answer. What I have proven is
that, if you carry out the summing of digits to its logical
conclusion, you get the following pattern for the cubes, starting with
1^3:

1,8,9,1,8,9,1,8,9,1,8,9.....

You use a funny rule in which, when you find the digits 1 and 0
together, you don't add them, and you also skip the case of 1^3. My
proof didn't show that your special rule will always work, allowing
you to replace all the ones in the above pattern with the number 10,
but it is true, except for the case of 1^3.

I tried raising the numbers to power 4. Here are the results of
performing similar operations:

1                           1
16                          7
81                          9
256                         4
625                         4
1296                        9
2401                        7
4096                        1
6561                        9
10000                       1
14641                       7
20736                       9
28561                       4
38416                       4
50625                       9
65536                       7
83521                       1
104976                      9.

Hopefully, you can notice a pattern here as well. My explanation for this is:

We have only 10 numbers. If you perform a similar operation on
consecutive numbers, you are bound to get some sort of sequence. These
sequences might appear immediately or may take thousands of operations
to emerge clearly. The bigger the operation, the longer the sequence
might be. Usually, in mathematics, we try to use this to our
advantage. It might be possible to have a proof for these sequences
but I am not quite sure about a scientific reason, unless one would
consider this reason logical enough.

Go to you library and check out the book "Numbers" by John H. Conway. 
This book covers all of what you are asking about and then some.  John
H. Conway is one of the great mathematicians of our time.  He created
the Game of Life (cellular automation) in the 70's and looks at the
fun puzzles in mathematics and number theory.

John Horton Conway is truly a genius and a very interesting speaker.
He has a habit of throwing out challenges to the audience. Once he
offered $10,000 for the solution of a particular problem. Someone in
the audience solved it several weeks later and claimed the prize.
Conway said he had not offered such a large sum. Unfortunately for
him, they had him on video tape. In the end, Conway agreed to send a
signed check for $10,000, which the other guy agreed to frame, hang up
in his office, and never cash.

Adding the (decimal) digits of a number (repeatedly) until a single
digit is reached gives a value congruent to the remainder of the
number divided by nine.

For example, 10^3 = 1000.  If the digits are added we would get 1, and
the remainder of 1000 divided by nine is 1.

Most often the step preceding getting such a remainder of 1 (by
"casting out nine's" as this process of adding digits is known) will
produce a 10.

What is behind the general pattern of 8,9,10 is the fact that:

  x = -1,0,+1 mod 3

implies respectively:

  x^3 = -1,0,+1 mod 9

Every number is congruent to either -1 or 0 or +1 "mod 3", which means
that subtracting the number and its "congruence" gives a result
divisible by 3.

Now if x - 1 is divisible by 3, then (x - 1)^3 is divisible by 27, and
in particular by 9.  Thus:

  x^3 - 3x^2 + 3x - 1 = 0 mod 9

  x^3 - 1 = 3x(x - 1) mod 9

But recalling that x - 1 is divisible by 3, the above shows that:

  x^3 - 1 = 0 mod 9

I.e. x = 1 mod 3 implies x^3 = 1 mod 9.

The other cases are similar or (for x = 0 mod 3) easier.  It remains
only to notice that for the cases that for positive x^3 = -1 or 0 mod
9, the only possible termination of "casting out nine's" will be the
digit 8 or 9, respectively, since we must have a single digit
congruent to -1 or 0 mod 9.

Besides exceptions to the pattern for x = 1 mod 3 that are powers of
10, so that the very first sum of digits drops us all the way to 1
without going through a sum of digits equal to 10 first, there's a
possibility of having a cube which is not a power of 10, but whose
digits sum up to a power of 10 (greater than the first power).  This
is true of 3333355 cubed, for example:

  3333355 ^ 3  =  37037759263953713875

whose digits sum directly to 100 (and hence to 1 in the next step).


regards, mathtalk-ga

mathtalk: Congratulations on finding the example that sums to 100 on
the first step. I was searching for just such an example earlier using
Google as my calculator, but it doesn't have enough digits of
precision. By our questioner's funny rule, he still gets to quit at 10
rather than casting out the final nine to get to 1, just as he does
for 10^3. As you can see from my first post, I also prefer to think of
this as -1, 0, +1, instead of 8,9,10.

Yes, hopefully the OP is already sleeping better!

regards, mt

Oh dear..if you start looking out such silly things then you have to
be sleepless all the night. Good night Sweet dreams.