1

If I count in 3's like $(3,6,9,12,15, ...)$ - then the last digit forms a repeating sequence $[3, 6, 9 / 2, 5, 8 / 1, 4, 7 / 0]$ (slash separates logical groups - we just do "minus one" from each number in the first [3,6,9] group, then we do "minus two", until "three minus three" gives zero).

If I count in 4's like $(4, 8, 12, 16, 20, 24, 28, ...)$ - then the last digit forms a repeating sequence $[4, 8, 2, 6, 0]$. I don't see any (easy mnemonic) rule here - like in the above "count in 3's" case.

Question: if I count in 2's, 3's, 4's, 6's, 7's, 8's, 9's - is there a general rule for n (n=2, n=3, ... n=9) how to produce that repeating sequence of last digits?

For 3's there is a simple (though not general) rule - take $3, 6, 9$ and subtract one until $3-3=0$ (see the first paragraph).

Maybe there is a similar simple rule for others (4's, 6's, 7's, 8's, 9's) or even a general rule (one simple trick for them all).

P.S. Originaly I wanted to provide a nice life-hack for my child learning to count in 3's, 4's etc - this hack is highly wanted, but besides I got interested myself. Sorry, I'm not a professional mathematician. Maybe modular arithmetics or some modern algebra / number theory findings (ring of residues modulo N?) does that (mnemonic / generalizing) trick?

Code Complete
  • 993
  • 1
    It's not clear what you mean by "do minus one / two" hence it is not clear what sort of general rule you seek. Please elaborate. – Bill Dubuque Apr 26 '20 at 16:32
  • @Gone "If I count in 3's like (3,6,9,12,15,...) - then the last digit forms a repeating sequence [3,6,9/2,5,8/1,4,7/0]" - now in details: 3,6,9 -> 2,5,8 is done by subtracting 1: 3-1=2, 6-1=5, 9-1=8. Again, 2,5,8 -> 1,4,7 is also done by subtracting 1: 2-1=1, 5-1=4, 8-1=7. General Rule, or any Partial Rule (non-General) for counting by 4's, 7's, etc I seek can be anything - I don't have any special requirements for it (well, I expect it to be rather simple and not very complicated - so that it is relatively easy/mnemonic). – Code Complete Apr 26 '20 at 17:40
  • 2
    The pattern you see for multiples of $3$ exists because $3\cdot 3 = 9\equiv -1\pmod{!10}.,$ The same will hold true for any factor of the radix minus one. Since $3$ is the only nontrivial factor of $10-1$ there are no further such patterns (except the trivial factors $1$ and $9$ which add or subtract one from the trivial periods of length one). But in hex (radix $16$) we have two nontrivial examples since $,15=3\cdot 5,$ so e.g. for $,5,$ we have $,\color{#c00}5,\color{#0a0}{10},15;, \color{#c00}4,\color{#0a0}9,14;,\ldots,$ and similarly for the factor $3\ \ $ – Bill Dubuque Apr 26 '20 at 18:15

2 Answers2

2

If you want to explain this property to a child, just write the digits $0$ to $9$ on a circle and view it as an Harry Potter's style imaginary clock with only $10$ hours. If you count by $3$ around this clock, you will find successively $0, 3, 6, 9, 2, 5, 8, 1, 4, 7, 0$ and if you count by $4$, you will find $0, 4, 8, 2, 6, 0$.

You can experiment with a clock with $12$, $7$ or $15$ hours to see what happens.

Now, the mathematical explanation. If you have a clock with $n$ hours and you count in $c$ ($O, c, 2c, 3c$, etc.), the length of the cycle will be $n/d$ where $d$ is the greatest common divisor of $n$ and $c$. For instance for $n =10$ and $c = 3$, one has $d = 1$ and thus the length of the cycle is $10$. If $c = 4$, one has $d =2$ and thus the length of the cycle is $10/2 = 5$.

J.-E. Pin
  • 42,871
  • 1
    @Code You can impart visual intuition about such cyclic (sub)groups using toys that concretely implement periodic models such as star polygons and/or roulette curves, e.g. the Spirograph toy. See here for more on such. – Bill Dubuque May 03 '20 at 17:33
1

Your rule for $4$ does have a somewhat similar simple pattern, $[4,8/2,6/0]$, where $2$ is subtracted from the first group of two to get the second group of two.

The rule for $7$ is the reverse of the rule for $3$, $[7,4,1/8,5,2/9,6,3/0]$, while the rule for $6$ is the reverse of the rule for $4$, $[6,2/8,4/0]$.

The rules for $1$ and $2$ are fairly simple, while the rules for $9$ and $8$ are the reverses of these.

The answer of J.-E. Pin and the comments of Gone explain some of what is going on here. It's a good exercise to try this in other number bases to see how things differ, and to understand why. The groups of three that you see for $3$ and $7$ in base $10$ are related to the fact that $10-1$ is divisible by $3$. Since $12-1=11$ is prime, you aren't going to have such nice patterns in base $12$. For example, for $5$ you have $[5,A,3,8,1,6,B,4,9,2,7,0]$. You can think up different ways of grouping these, but they won't be as nice as what you get for $3$ in base $10$.

Will Orrick
  • 19,218