Cyclic Numbers

I came across an awesome set of numbers called as CYCLIC NUMBERS. The smallest such number is 142857. Some properties of 142857 are:

Property 1: Multiples have same digits

142857×1=142857142857\times1=142857

142857×2=285714142857\times2=285714

142857×3=142857142857\times3=142857

142857×4=571428142857\times4=571428

142857×5=714285142857\times5=714285

142857×6=857142142857\times6=857142

142857×54=7714278142857\times54=7714278. Take the last 6 digits and add to it, the number formed using the remaining digits. 714278+7=714285714278+7=714285

*Property 2: Product with multiples of 7 have all digits 9. *For 142857142857, it is 77. For other cyclic numbers, it is different.

142857×7=999999142857\times7=999999

142857×112=15999984142857\times112=15999984. Take the last 6 digits and add to it, the number formed using the remaining digits. 999984+15=999999999984+15=999999

Property 3: Sum of numbers formed using the digits of the cyclic numbers equals numbers having all digits 9

Using two digits at a time: 14+28+57=99\underline{14}+\underline{28}+\underline{57}=99

Using three digits at a time: 142+857=999\underline{142}+\underline{857}=999

Using four digits at a time (Each digit must be used equal number of times): 1428+5714+2857=9999\underline{1428}+\underline{5714}+\underline{2857}=9999

Property 4: Difference of squares equals the number having the same digits as the cyclic number

85721422=714285857^2-142^2=714285

Try to prove the above results. Also, try to arrive at the formula for finding cyclic numbers.

For hints, solutions (and of course, the source), do check out Cyclic numbers on numberphile

Note by Bruce Wayne
5 years, 11 months ago

No vote yet
17 votes

  Easy Math Editor

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

  • Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
  • Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
  • Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.
  • Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events.

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list

  • bulleted
  • list

1. numbered
2. list

  1. numbered
  2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1

paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
    # 4 spaces, and now they show
    # up as a code block.

    print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.
2 \times 3 2×3 2 \times 3
2^{34} 234 2^{34}
a_{i-1} ai1 a_{i-1}
\frac{2}{3} 23 \frac{2}{3}
\sqrt{2} 2 \sqrt{2}
\sum_{i=1}^3 i=13 \sum_{i=1}^3
\sin \theta sinθ \sin \theta
\boxed{123} 123 \boxed{123}

Comments

Sort by:

Top Newest

Try this link

(Tell me if this link is wrong, please)

Bob Krueger - 5 years, 11 months ago

Log in to reply

i like this channel they have good math videos

Felipe Sousa - 5 years, 11 months ago

Log in to reply

It is an awesome channel. They show some very interesting stuff on numbers.

Bruce Wayne - 5 years, 11 months ago

Log in to reply

it was awesome.

thanks!!!!!!!!!!!!!!

yash gupta - 5 years, 11 months ago

Log in to reply

A cyclic number can be found out by the formula - (10^N)/N Where n is a prime. Only certain primes work. Regards, Keshav.

Keshav Gupta - 5 years, 11 months ago

Log in to reply

You mean 10n1n\frac{10^n-1}{n}, where nn is a prime.

To the OP, 142857142857 is also the only cyclic number that does not have a leading zero. The zeroes in front are important for the cyclic nature of the larger numbers such as:

058823529411764713=76470588235294110588235294117647*13=7647058823529411

Jared Low - 5 years, 11 months ago

Log in to reply

Actually, it should be 10n11n\dfrac{10^{n-1}-1}{n} where nn is prime.

Daniel Chiu - 5 years, 11 months ago

Log in to reply

It is true that cyclic numbers need to have the leading digit=0. 142857 is an exception.

Bruce Wayne - 5 years, 11 months ago

Log in to reply

Ok, Totally forgot to mention the leading zero thing. And im sorry for writing out the wrong formula. missed the -1 part! Thanks for correctig me! Regards, Keshav.

Keshav Gupta - 5 years, 11 months ago

Log in to reply

I learnt about this from numberphile. It is a pretty good channel.

Shubham Bhargava - 5 years, 11 months ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...