Pc:p1

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The general result is: If gcd (a,b) = 1, then the period of the repeating block digits in the decimal expansion of a/b is at most (b-1).

First see the mechanism as to how these digits get generated.

We will assume that a<b (if its not replace a, by the remainder when a is divided by b)

Let us take the case when b is not divisible by 2 or 5, so that gcd (10,b) = 1.

Now the mechanism is:

10a = bq+r (q becomes the 1st digit after the decimal)

10r = bq₁ + r₁ (q₁ becomes 2nd digit after decimal)

10r₁ = bq₂+r₂ (q₂ becomes 3rd digit after decimal)

and so on.

Let us focus our attention on the remainders that are generated in each stage as they in turn generate the digits going into the decimal expansion.

A result in Number Theory tells us that the remainders obtained here are a special group: they consist of numbers that are less than b and are prime to it.

There is one more point that we must focus our attention on:

Notice that a remainder at each stage gives rise to a remainder in the next stage. This mapping is unique, a particular remainder gives is mapped only to one remainder in the same set (both could be the same number).

Let us see an example of such a mapping. Take the denominator as 21. First the group of possible remainders are

(1,2,4,5,8,10,11,13,16,17,19,20)

Next the mapping itself. This is a mapping from the set onto itself

1 2 4 5 8 10 11 13 16 17 19 20

10 20 19 8 17 16 5 4 13 2 1 11

Now I am going to write this in a special way. I will start with 1, write its image (10) next, and its image (i.e. the image of 10 which is 16 and so on. Lets start that

(1, 10, 16, 13, 4, 19). The reason I stopped at 19, is that 19's image is 1. What has happened here is that a cycle has been formed. The remaining part of the permutation gets written as (2,20,11,5,8,17). Thus the entire mapping gets written as

(1, 10, 16, 13, 4, 19) (2,20,11,5,8,17). These you notice are disjoint cycles. And a theorem from Group Theory tells us that every such mapping from a set onto itself (a permutation group) consists of a finite number of disjoint cycles.

The termination occurs when you the cycle starting from any particular element terminates

Obviously, the maximum length of the cycle is the number of elements of the set.

And the maximum number of elements in the set occur when b is a prime and that is (b-1).

Just try this process with 7 in the denominator and you will see that there is only one cycle in the case of 7 and thats why you see the period as 6.

The permutation group for 9 consists of 8 singleton cycles, and so the period is always 1. In case of 11 you have 5 cycles of doublets and so always the period is 2.

None of this is JEE syllabus, but I thought that you could absorb the main idea of the proof of the result rather than just mug up the result.

I must add that if b is divisible by 2 or 5, the set of residues is further truncated, and so the result holds in this case too

One mistake in my previous post: In the case of 9, you have 6 singleton sets not 8. But the period is still 1