# 1/7 in decimal form is .142857

1. Jan 12, 2005

### T@P

1/7 in decimal form is .142857
2/7 is a rearangement of those numbers, 3/7 is too etc. anyone know why? (btw 142857 is divible by 999)

2. Jan 12, 2005

### The Bob

$$\frac{1}{7} = 0. \dot1 4285 \dot7$$

$$\frac{2}{7} = 0.\dot2 8571 \dot4$$

$$\frac{3}{7} = 0.\dot4 2857 \dot1$$

$$\frac{4}{7} = 0.\dot5 7142\dot8$$

$$\frac{5}{7} = 0.\dot7 1428\dot5$$

$$\frac{6}{7} = 0.\dot8 5714\dot2$$

The pattern that I can see is that as the nominator increases the first number increases within the numbers there are. 1 then 2 then 4 then 5 etc... Also all the numbers stay in order. Why, though, I do not know.

Last edited: Jan 12, 2005
3. Jan 12, 2005

### dextercioby

Nope.1/7 is:
$$\frac{1}{7}=0.(142857)$$ (1)
,which is a totally different number from the one u've written.Yours is
$$0.142757=\frac{142857}{1000000}$$(2)
and has 6 decimals,while mine is
$$0.(142857)=\frac{142857}{999999}=\frac{1}{7}$$(3)
and has an infinite number of decimals.

What do u mean by rearrangement??

Daniel.
PS.From (3) u get
$$142857\cdot 7=999999=999\cdot 1001=999\cdot 7\cdot 11\cdot 13$$
,from which u get the decomposition of 142857 in prime factors.

4. Jan 13, 2005

### T@P

i meant rearangement of the first 6 digits, since they repeat you can think of it as 142857/999999, and incidentaly it reduces. actually i think that explains why its divisible by 999.

5. Jan 13, 2005

### NateTG

The same holds for any number $n$ which has the property that $\frac{1}{n}$ has $n-1$ repeating digits.
Here's what happens:
Code (Text):

1 = 1 mod 7
10 = 3 mod 7
100 = 2 mod 7
1000 = 6 mod 7
10000 = 4 mod 7
100000 = 5 mod 7

And, calculating the digits after the decimal point is very much like multiplying the denominator by 10, and calculating the last digit and a new remainder, so you cycle through the six possible remainders.

6. Jan 15, 2005

### robert Ihnot

NateTG has it pretty well figured out, but I never did see until now. By the way, they are called cyclic numbers and can be googled.

We have $$1/7 = \overline{.142857}$$. (Where the overline means the period is repeated and repeated)

The first decimal,1, represents how many times 7 goes into 10 whole , 14 represents how many times 7 goes whole into 100, 142 for 1000, etc.

Thus if we look at 100/7 = $$14\overline{.285714}$$

But 100/7 = 14 +2/7, so we end up with 2/7= $$\overline{.285714}$$

Thus as NateTG tells us to get the full array we need to have a cycle of N-1 digits, where 10^(N-1)==1 Mod N. This will only possibly occur when N = p, a prime.

For example, lets look at 1/21 = $$\overline{.047619}$$

The powers of 10^x modulo 21 are 1,10,16,13,4,19,1...for x=0,1,2,3,4,5,6...

i.e. 10^6 == 10^0==1 Modulo 21, and the series has finished.

Thus the cycle will repeat for some numbers, say, 16/21 = $$\overline{.761904}$$

But, of course, since there are only 6 possibilities, some numbers will not cycle that way, for example we have 2/21 = $$\overline{.095238}$$

Last edited: Jan 15, 2005