\[\frac{1}{7} = 0.142857 \ 142857 \ 142857 \ \ldots,\]
with the portion $142857$ recurring indefinitely (it is called the repetend) and having these several striking properties: first,
- $142857 \times 2 = 285714$
- $142857 \times 3 = 428571$
- $142857 \times 4 = 571428$
- $142857 \times 5 = 714285$
- $142857 \times 6 = 857142$
\[\fbox{$142$} \ \fbox{$857$}\]
and add the two "halves", we get
\[142 + 857 = 999,\]
a number wholly composed of $9$'s.
If we try this with the fraction $1/13$, we find something very similar. The repetend is now $076923$, since
\[\frac{1}{13} = 0.076923 \ 076923 \ 076923 \ \ldots,\]
and this number has nearly the same features as the number $142857$; but there are also some curious differences. Thus, its multiples with the numbers $1$, $3$, $4$, $9$, $10$, $12$ yield numbers that have the same circular pattern of digits as $076923$:
- $076923 \times 1 = 076923$
- $076923 \times 3 = 230769$
- $076923 \times 4 = 307692$
- $076923 \times 9 = 692307$
- $076923 \times 10 = 769230$
- $076923 \times 12 = 923076$
- $076923 \times 2 = 153846$
- $076923 \times 5 = 384615$
- $076923 \times 6 = 461538$
- $076923 \times 7 = 538461$
- $076923 \times 8 = 615384$
- $076923 \times 11 = 846153$
\[153846, \quad 384615, \quad 461538, \quad 538461, \quad 615384, \quad 846153\]
all have the same circular arrangement of digits. How very curious!
You may well ask, is there any other property that bonds the numbers $1$, $3$, $4$, $9$, $10$, $12$ together, and similarly bonds the numbers $2$, $5$, $6$, $7$, $8$, $11$ together? Yes, there is! - try to find it.
Finally, note that if we slice the number $076923$ in two halves,
\[\fbox{$076$} \ \fbox{$923$}\]
and add the two halves, we get
\[076 + 923 = 999,\]
exactly like earlier.
Is this a general feature that we find with all recurring decimals? Try it out - go and explore more such decimals! You will almost certainly unearth more patterns of interest.
Here is an associated problem which is of some interest: Find the smallest positive integer with the feature that if the units digit is taken to the "opposite end" of the number, the new number is exactly twice the original number.
Powers sequences strung on a line
Infinite decimals have still more morsels to offer. For example, consider the decimal arising from the fraction $1/99998$:
\[\frac{1}{99998} = 0.00001 \ 00002 \ 00004 \ 00008 \ 00016 \ 00032 \ 00064 \ \ldots.\]
Well, we see the powers of $2$ on display, all strung out on a line!
If we change the fraction to $1/99997$ we get something just as nice:
\[\frac{1}{99997} = 0.00001 \ 00003 \ 00009 \ 00027 \ 00081 \ 00243 \ 00729 \ \ldots.\]
Now we see the powers of $3$. Isn't that a pretty sight?
Once we get the general idea, it isn't at all hard to generalize it. For example, here is $1/9995$:
\[\frac{1}{99995} = 0.00001 \ 00005 \ 00025 \ 00125 \ 00625 \ 03125 \ 15625 \ldots.\]
But I hear you jumping up and saying something urgently to me:
"Hold on, now! These are the decimal expansions of rational numbers, so they ought to yield recurring decimals, right? But where are the recurring sequences in the above decimals? I don't see any!"
Well, I'll leave it to you to figure that one out!
One can even get the Fibonacci numbers strung out on a line:
\[\frac{1000}{998999} = 0.001 \ 001 \ 002 \ 003 \ 005 \ 008 \ 013 \ 021 \ 034 \ 055 \ 089 \ \ldots.\]
Try to figure out which fraction will yield the Lucas numbers.
(The Lucas sequence is less well known than the Fibonacci sequence, but is defined in a very similar way. Its starting numbers are $2$ and $1$, in that order, and following these, each subsequent number is the sum of the two that precede it
\[2, \,\ 1, \,\ 3, \,\ 4, \,\ 7, \,\ 11, \,\ 18, \,\ 29, \,\ 47, \,\ 76, \,\ \ldots.\]
These numbers have extremely close interconnections with the Fibonacci numbers.)
\[\clubsuit \quad \clubsuit \quad \clubsuit \quad \clubsuit \quad \clubsuit \quad \clubsuit \quad \clubsuit \quad \clubsuit \quad \clubsuit \quad \clubsuit\]
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