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Here is a rule: Division by the number 4 must terminate: the

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Senior Manager
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Here is a rule: Division by the number 4 must terminate: the [#permalink] New post 21 Feb 2007, 19:00
Here is a rule: Division by the number 4 must terminate: the remainder when dividing by 4 must be 0, 1, 2 or 3, so the quotient must end with 0, 0.25, 0.5 or 0.75, respectively.

How about if the number is 6 or any other number? Will it be like this?

Division by the number 6 may or may not terminate because the remainder when dividing by 6 must be 0, 1, 2, 3, 4 or 5, so the quotient must end with 0, 1/6 (non-terminating), 2/6, 3/6, 4/6 or 5/6, respectively.

Am I right?

PS: Sorry to them who read another question here just a little while ago. I worked on that and found the result given above. The question is given below.
--------------------------------------------------------------

Am I right?

Any number say m is divided by an odd int or an even int or a prime number. (Given all the numbers are +ve integers here.)

1. m/odd-int = int or terminating-decimal or non-termminating decimal

2. m/even-int = int or terminating decimal or non-terminating decimal

3. m/prime-no = = int or terminating-decimal or non-termminating decimal

Please help!
Senior Manager
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Joined: 12 Mar 2006
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 [#permalink] New post 22 Feb 2007, 20:58
here is how i would approach it

for any number a/b reduce it to the simplest form (i.e. cancel all common factors) now if b contains prime numbers (other than 2 or 5) then the decimal will not terminate...

copy pasting some relevant information from wikipedia

Fractions with prime denominators
A fraction in lowest terms with a prime denominator other than 2 or 5 (i.e. coprime to 10) always produces a recurring decimal. The period of the recurring decimal, 1/ p, where p is prime, is either p − 1 (the first group) or a divisor of p − 1 (the second group).

Examples of fractions of the first group are:

1/7 = 0.142857...; 6 repeating digits
1/17 = 0.0588235294117647...; 16 repeating digits
1/19 = 0.052631578947368421...; 18 repeating digits
1/23 = 0.0434782608695652173913...; 22 repeating digits
1/29 = 0.0344827586206896551724137931...; 28 repeating digits
The list can go on to include the fractions 1/47, 1/59, 1/61, 1/97, 1/109 etc.

This is to say:

106 − 1 = 999,999 (6 digits of 9) is divisible by 7;
1016 − 1 = 9,999,999,999,999,999 (16 digits of 9) is divisible by 17;
1018 − 1 = 999,999,999,999,999,999 (18 digits of 9) is divisible by 19; etc.
These can be deduced from Fermat's little theorem.

It can also be generalised to say that p − 1 digits of 1 (or 2, 3, 4, 5, 6, 7, 8, 9) is divisible by p, which is a prime number other than 2 or 5.

The following multiplications exhibit an interesting property:

2/7 = 2 × 0.142857... = 0.285714...
3/7 = 3 × 0.142857... = 0.428571...
4/7 = 4 × 0.142857... = 0.571428...
5/7 = 5 × 0.142857... = 0.714285...
6/7 = 6 × 0.142857... = 0.857142...
That is, these multiples can be obtained from rotating the digits of the original decimal of 1/7. The reason for the rotating behaviour of the digits is apparent from an arithmetics exercise of finding the decimal of 1/7.

Of course 142857 × 7 = 999999, and 142 + 857 = 999.

Decimals of other prime fractions such as 1/17, 1/19, 1/23, 1/29, 1/47, 1/59, 1/61, 1/97, 1/109 all exhibit the same property.

Fractions of the second group are:

1/3 = 0.333.... which has 1 repeating digit.
1/11 = 0.090909... which has 2 repeating digits.
1/13 = 0.076923... which has 6 repeating digits.
Note that the following multiples of 1/13 exhibit the discussed property of rotating digits:

1/13 = 0.076923...
3/13 = 0.230769...
4/13 = 0.307692...
9/13 = 0.692307...
10/13 = 0.769230...
12/13 = 0.923076...
And similarly these multiples:

2/13 = 0.153846...
5/13 = 0.384615...
6/13 = 0.461538...
7/13 = 0.538461...
8/13 = 0.615384...
11/13 = 0.846153...
Again, 076923 × 13 = 999999, and 076 + 923 = 999.


[edit] Why all repeating or terminating decimals must be rational numbers
Given a repeating decimal, it is possible to calculate the fraction which produced it. For example:

x = 0.333333...
10x = 3.33333... (multiplying each side of the above line by 10)
9x = 3 (subtracting the 1st line from the 2nd)
x = 3/9 = 1/3 (simplifying)
Another example:

x = 0.18181818...
100x = 18.181818...
99x = 18
x = 18/99 = 2/11
From this kind of argument, we can see that the period of the repeating decimal of a fraction n/d will be (at most) the smallest number k such that 10k − 1 is divisible by d.

For example, the fraction 2/7 has d = 7, and the smallest k that makes 10k − 1 divisible by 7 is k = 6, because 999999 = 7 × 142857. The period of the fraction 2/7 is therefore 6.


for more ....
http://en.wikipedia.org/wiki/Recurring_ ... nominators
  [#permalink] 22 Feb 2007, 20:58
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Here is a rule: Division by the number 4 must terminate: the

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