+0

Magic Numbers

+3
60
6
+505

I was reading this book and kept noticing that one number kept popping up - 1089. It turns out that when you multiply it, the numbers move sequentially through the columns.

1 - 1089

2 - 2178

3 - 3267

4 - 4365

5 - 5454

6 - 6543

7 - 7632

8 - 8721

9 - 9810

Can anyone else think of some?

helperid1839321  Nov 29, 2017
Sort:

#1
+2

Yes !! 142857

142857 x 1 =142,857

142857 x 2 =285,714 same numbers in different order.

142857 x 3 =428,571  ...................................................

142857 x 4 =571,428.....................................................

142857 x 5 =714,285......................................................

142857 x 6 =857,142......................................................

142857 x 7 =999,999  !!!

Guest Nov 29, 2017
edited by Guest  Nov 29, 2017
#2
+1493
+1

Wow! This pattern that you have displayed actually continues past 7, actually, as I discovered just now. You just have to perform an intermediary manipulation.

$$142857*8=1142856\Rightarrow1+142856=142857$$

$$142857*9=1285713\Rightarrow1+285713=285714$$

$$142857*10=1428570\Rightarrow1+428570=428571$$

It seems to be that any multiple of 7 results in the 999999 result while anything else results in this "cyclic" property.

$$142857*77=10999989\Rightarrow10+999989=999999$$

Here are a few arbitrary attempts I tried to see if the pattern maintained itself. To my relief, it does.

$$142857*43=6142851\Rightarrow6+142851=142857$$

$$142857*6807=972427599\Rightarrow972+427599=428571$$

$$142857*142857=20408122449\Rightarrow20408+122449=142857$$

Amazingly, this pattern maintains itself for incredibly large multipliers. It requires more manipulation, though. This is incredible!

$$142857*758241142857=108320054945122449\Rightarrow108320054945+122449=108320177394\\ 108320177394\Rightarrow108320+177394=285714$$

TheXSquaredFactor  Nov 29, 2017
#3
+505
+2

Wow! Tha's really cool! Do you have any more?

helperid1839321  Nov 30, 2017
#4
+505
+3

Thought of another one: 76923

Multiples:

$$\begin{matrix} 1 && 076923 \\ 10 && 769230 \\ 9 && 692307 \\ 12 && 923076 \\ 3 && 230769 \\ 4 && 307692 \end{matrix} \: \& \: \begin{matrix} 2 && 153846 \\ 7 && 538461 \\ 5 && 384615 \\ 11 && 846153 \\ 6 && 461538 \\ 8 && 615384 \end{matrix}$$

helperid1839321  Nov 30, 2017
#6
+1493
+2

helperid1839321, I decided to delve deeper into this subject. You may like my findings!

First of all, I think I understand why this property occurs. I happen to know that $$142857$$ is the first 6 digits of the decimal expansion of $$\frac{1}{7}$$. At first, I thought this was insignificant, but it turns out that this fact can be used to understand this further. Look at the table below.

 $$\frac{1}{7}=\overline{0.142857142857}$$ Multiply both sides by 10. $$\frac{10}{7}=1.42857\overline{142857}$$ Let me rewrite 10/7 to make things clearer. $$1+\frac{3}{7}=1.42857\overline{142857}$$ Subtract one from both sides. $$\frac{3}{7}=0.42857\overline{142857}$$ WOAH! It cycles! Let's do this again. Multiply both sides by 10 again. $$\frac{30}{7}=4.2857\overline{142857}$$ Rewrite 30/7 again. $$4+\frac{2}{7}=4.2857\overline{142857}$$ Subtract 4 from both sides. $$\frac{2}{7}=0.2857\overline{142857}$$ WOAH! The first 6 digits cycle again. You can continue the pattern, if you wish!

This forced me to wonder if there is any more solutions for $$\frac{1}{p}$$, where p is a whole number that creates this special property. This is because if another number p causes some repetition, then we would have found another number! YAY!

I decided to enlist some help from a computer here. This is what the computer outputted.

7, 17, 19, 23, 29, 47

WHAT! There are more! Yes, these have the same property. Let's check them out, shall we?

 $$p$$ $$\frac{1}{p}$$ 7 .142857... 17 .0588235294117647... 19 .052631578947368421... 23 .0434782608695652173913... 29 .0344827586206896551724137931... 47 .0212765957446808510638297872340425531914893617...

It appears as if 142857 is the only number that does not start with a zero. Let's keep running this simulation! Thankfully, more numbers output! I let it run for some time, too.

59, 61, 97, 109, 113, 131, 149, 167, 179, 181, 193...

In case you are wondering,

$$\frac{1}{193}$$=005181347150259067357512953367875647668393782383419689119170984455958549222797927461139896373056994818652849740932642487046632124352331606217616580310880829015544041450777202072538860103626943...

There are a few patterns that I see here

1) $$p$$ must be a prime number

2) The decimal expansion must have a maximum period decimal expansion of $$p-1$$.

I am also making a conjecture here that I do not know whether or not is true: there are an infinite number for p that create these types of numbers!

TheXSquaredFactor  Dec 2, 2017
#5
+505
+2

Try using 12,345,679!

helperid1839321  Nov 30, 2017

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