Processing math: 100%
 
+0  
 
+3
1511
6
avatar+633 

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?

 Nov 29, 2017
 #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  !!!

 Nov 29, 2017
edited by Guest  Nov 29, 2017
 #2
avatar+2446 
+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.

 

1428578=11428561+142856=142857

1428579=12857131+285713=285714

14285710=14285701+428570=428571

 

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

 

14285777=1099998910+999989=999999

 

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

 

14285743=61428516+142851=142857

1428576807=972427599972+427599=428571

142857142857=2040812244920408+122449=142857

 

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

142857758241142857=108320054945122449108320054945+122449=108320177394108320177394108320+177394=285714

TheXSquaredFactor  Nov 29, 2017
 #3
avatar+633 
+2

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

helperid1839321  Nov 30, 2017
 #4
avatar+633 
+3

Thought of another one: 76923

Multiples: 

10769231076923096923071292307632307694307692&2153846753846153846151184615364615388615384

helperid1839321  Nov 30, 2017
 #6
avatar+2446 
+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 17. 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.

 

17=¯0.142857142857 Multiply both sides by 10.
107=1.42857¯142857 Let me rewrite 10/7 to make things clearer.
1+37=1.42857¯142857 Subtract one from both sides.
37=0.42857¯142857 WOAH! It cycles! Let's do this again. Multiply both sides by 10 again.
307=4.2857¯142857 Rewrite 30/7 again.
4+27=4.2857¯142857 Subtract 4 from both sides.
27=0.2857¯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 1p, 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 1p    
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,

 1193=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 p1.

 

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+633 
+2

Try using 12,345,679!

 Nov 30, 2017

2 Online Users

avatar