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

$142857\times1=142857$

$142857\times2=285714$

$142857\times3=142857$

$142857\times4=571428$

$142857\times5=714285$

$142857\times6=857142$

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

*Property 2: Product with multiples of 7 have all digits 9. *For $142857$ , it is $7$ . For other cyclic numbers, it is different.

$142857\times7=999999$

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

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: $\underline{14}+\underline{28}+\underline{57}=99$

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

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

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

$857^2-142^2=714285$

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

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Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

Try this link

(Tell me if this link is wrong, please)

Bob Krueger - 7 years, 7 months ago

i like this channel they have good math videos

Felipe Sousa - 7 years, 7 months ago

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

Bruce Wayne - 7 years, 7 months ago

it was awesome.

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

yash gupta - 7 years, 7 months ago

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 - 7 years, 7 months ago

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

To the OP, $142857$ 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:

$0588235294117647*13=7647058823529411$

Jared Low - 7 years, 7 months ago

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

Daniel Chiu - 7 years, 7 months ago

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

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.

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

Shubham Bhargava - 7 years, 7 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$