Go back
Curious number

Curious number

Posers and Puzzles

F

Joined
11 Nov 05
Moves
43938
Clock
09 Apr 08
1 edit
Vote Up
Vote Down

142857 is a curious number:

1 * 142857 = 142857
2 * 142857 = 285714
3 * 142857 = 428571
4 * 142857 = 571428
5 * 142857 = 714285
6 * 142857 = 857142
Note that the numbers are the same and in the same order circularly

7 * 142857 however = 999999.

Explanation, anyone? Or is it just a coincidence?

m

Joined
07 Sep 05
Moves
35068
Clock
09 Apr 08
Vote Up
Vote Down

Not an explanation, but I did immediately notice that these are the first six digits of 1/7.

D

Joined
12 Sep 07
Moves
2668
Clock
09 Apr 08
Vote Up
Vote Down

http://en.wikipedia.org/wiki/Cyclic_number

E
Seeker

Going where needed.

Joined
16 May 07
Moves
3366
Clock
09 Apr 08
Vote Up
Vote Down

Originally posted by FabianFnas
142857 is a curious number:

1 * 142857 = 142857
2 * 142857 = 285714
3 * 142857 = 428571
4 * 142857 = 571428
5 * 142857 = 714285
6 * 142857 = 857142
Note that the numbers are the same and in the same order circularly

7 * 142857 however = 999999.

Explanation, anyone? Or is it just a coincidence?
Then, if you continue the series,

142857 * 8 = 1142856 (the 7 on the end splits into a 1 and a 6, the 1 going to the beginning of the number and the six remaining at the end).

142857 * 9 = 1258713 (the 4 on the end splits into a 1 and a 3).

Yes, that number is very interesting.

g

Joined
15 Feb 07
Moves
667
Clock
09 Apr 08
1 edit
Vote Up
Vote Down

I have sometimes wondered if Cyclical numbers exist in other bases, Base 12 for instance..

I suspect so, and I suspect 1/7 in Base 12 would share similar properties, because 12 % 7 is 5, which is a prime modulo root, a term I didn't know before but an aspect of number theory I had noticed before.

(Composite numbers don't have prime modulo roots unless they are the power of odd prime numbers. The main property of a prime modulo root is as follows...

We define the following:
N is the base number we want to find a prime modulo root for.
P is the candidate for prime modulo root.
f(x) = (P ^ x) mod N

If we start with x=1 and examine the value of f(x) until we find a value of x where f(x)=1, we find length of the cyclical pattern for f(x).

If the length of this pattern is equal to the number of relatively prime numbers below N, then P is a prime modulo root of N.

3

Joined
06 Apr 08
Moves
1871
Clock
09 Apr 08
Vote Up
Vote Down

Composite numbers don't have prime modulo roots unless they are the power of odd prime numbers.
Actually, that is not true.

If n is a positive integer the congruence classes coprime to n (that is a set of all integerts 0 < m < n-1 such that gcd(m, n) = 1) form a group with multiplication modulo n). This group, usually denoted az Z/nZ, is cyclic if and only if n is 1, 2, 4, p^k, or 2p^k where p^k is a power of an odd prime number. If Z/nZ is cyclic, the generators are primitive root modulo n.

For example, for n = 14, Z/14Z is { 1, 3, 5, 9, 11, 13 } which is isomorphic to C6 (cyclic group with 6 elements), and the generators are 3 and 5. Hence 3 and 5 are primitive roots modulo 14.

more on this truly interesting topic is here:
http://en.wikipedia.org/wiki/Primitive_root_modulo_n
http://en.wikipedia.org/wiki/Multiplicative_group_of_integers_modulo_n

G

Joined
13 Dec 06
Moves
792
Clock
09 Apr 08
Vote Up
Vote Down

1/7 = .142857142857142857142857142857142857142857142857...
2/7 = .285714285714285714285714285714285714285714285714...
3/7 = .428571428571428571428571428571428571428571428571...
4/7 = .571428571428571428571428571428571428571428571428...
5/7 = .714285714285714285714285714285714285714285714285...
6/7 = .857142857142857142857142857142857142857142857142...
7/7 = .999999999999999999999999999999999999999999999999... = 1

You get this cycle because when you do 1 divided by 7 there are 6 possible remainders, each of which gets used once. Then you hit a remainder that has already happened before and the cycle repeats (try it).

There are many other numbers like this; Wikipedia has some information here: http://en.wikipedia.org/wiki/Recurring_decimal

g

Joined
15 Feb 07
Moves
667
Clock
09 Apr 08
Vote Up
Vote Down

Originally posted by 3v1l5w1n
Actually, that is not true.

If n is a positive integer the congruence classes coprime to n (that is a set of all integerts 0 < m < n-1 such that gcd(m, n) = 1) form a group with multiplication modulo n). This group, usually denoted az Z/nZ, is cyclic if and only if n is 1, 2, 4, p^k, or 2p^k where p^k is a power of an odd prime number. If Z/nZ is cyclic ...[text shortened]... /Primitive_root_modulo_n
http://en.wikipedia.org/wiki/Multiplicative_group_of_integers_modulo_n
I missed the 2p^k term... My apologies.

I was mainly thinking about numbers like 15, 21, 35, 33... Composite numbers with at least 2 odd prime divisors.

Cookies help us deliver our Services. By using our Services or clicking I agree, you agree to our use of cookies. Learn More.