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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
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6+0 Answers

 #1
avatar
+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
avatar+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
avatar+505 
+2

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

helperid1839321  Nov 30, 2017
 #4
avatar+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
avatar+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
avatar+505 
+2

Try using 12,345,679!

helperid1839321  Nov 30, 2017

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