[ *September 1, 2003* ]

Recurring last digit multiplication numbers.

by Allan Jenks (email)

The below has piqued Allan's idle curiosity for many years,

with no-one able, or more likely, willing, to explain the reason

for the numerical phenomena. I wonder if YOU would take the time

to explain it, in as simple terms as possible.

Take each number from 2 to 9

and multiply each successive sum by that same number:

E.g. :

2 x 2 = 4

2 x 4 = 8

2 x 8 = 16

2 x 16 = 32

2 x 32 = 64

2 x 64 = 128

2 x 128 = 256

2 x 256 = 512

and so on

It will be noted that the final digit of each sum forms

a four-digit recurring pattern.

In the case of 2, the pattern is, as shown in teal above,

4862, 4862, and so on.

It will be seen that the numbers 2 to 9 elicit the recurring

last digit patterns shown below:

2 4862

3 9713

4 6464

5 5555

6 6666

7 9317

8 4268

9 1919

The sum of EACH of the above 4 digit patterns is 20,

- EXCEPT for 6, which pattern equals 24.

QUESTION:

(1) WHY do all the numbers (except 6) produce the sum of 20 ?

or

(2) WHY is 6 the only one that does not do so ?

Allan would GREATLY appreciate it if you would be willing to take up

some of your valuable time and give me what will no doubt be

the very simple explanation!

I forwarded Allan Jenks' intriguing problem to Jean Claude Rosa

- someone with more mathematical expertise than myself -

and who gave already a few elements for a possible answer (?).

First he put the __last digit__ data in a nice table as follows

x | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 |

x^{2} | 0 | 1 | 4 | 9 | 6 | 5 | 6 | 9 | 4 | 1 |

x^{3} | 0 | 1 | 8 | 7 | 4 | 5 | 6 | 3 | 2 | 9 |

x^{4} | 0 | 1 | 6 | 1 | 6 | 5 | 6 | 1 | 6 | 1 |

total | 0 | 4 | 20 | 20 | 20 | 20 | 24 | 20 | 20 | 20 |

Then Jean Claude resorted to modulo 5 with the above table giving

x | 0 | 1 | 2 | 3 | 4 | 0 | 1 | 2 | 3 | 4 |

x^{2} | 0 | 1 | 4 | 4 | 1 | 0 | 1 | 4 | 4 | 1 |

x^{3} | 0 | 1 | 3 | 2 | 4 | 0 | 1 | 3 | 2 | 4 |

x^{4} | 0 | 1 | 1 | 1 | 1 | 0 | 1 | 1 | 1 | 1 |

total modulo 5 | 0 | 4 | 0 | 0 | 0 | 0 | 4 | 0 | 0 | 0 |

Jean Claude thinks that this second table makes apparent

what the first table hides (particularly why one finds 24

for the case x = 6 and 20 for the others

(with exception for cases 0 & 1)).

Are these tables sufficient to make a case ?

It is up to you to tell me... or to Allan...

[*July 21, 2004* ]

Phil Carmody (email) contributes the following to the topic :

Regarding: Recurring last digit multiplication numbers.

All cycles length must divide 4, as phi(10) = phi(2)*phi(5) = 1*4 = 4.

All cycles of length 4 must have x and -x (or 10-x) two positions apart

as -1, or 9 is the only square root of 1 that isn't 1. (As 10 is twice a

prime.) Therefore their cycle's sum is x+y+10-x+10-y = 20.

All the cycles of length 2 must have x and -x (or 10-x) in consecutive

positions (for the same reason as the 4-cycles).

Hence they'll sum to x+10-x+x+10-x = 20.

All that remain are the cycles formed by 0,1,5,6. These on the whole

don't sum to 20, except for 5. 5 works because it is half 10.

If you were to repeat the investigation with base 2.q for arbitrary odd

prime q, you'd see exactly the same pattern. Only 0, 1, and q+1 would

fail to sum to 4q.