Question

I already know when a fraction is repeating decimals. Here is the function.

public bool IsRepeatingDecimal
{
    get
    {
        if (Numerator % Denominator == 0)
            return false;

        var primes = MathAlgorithms.Primes(Denominator);

        foreach (int n in primes)
        {
            if (n != 2 && n != 5)
                return true;
        }

        return false;
    }
}

Now, I’m trying to get the repeated number. I’m checking this web site: http://en.wikipedia.org/wiki/Repeating_decimal

public decimal RepeatingDecimal()
{
    if (!IsRepeatingDecimal) throw new InvalidOperationException("The fraction is not producing repeating decimals");

    int digitsToTake;
    switch (Denominator)
    {
        case 3:
        case 9: digitsToTake = 1; break;
        case 11: digitsToTake = 2; break;
        case 13: digitsToTake = 6; break;
        default: digitsToTake = Denominator - 1; break;
    }

    return MathExtensions.TruncateAt((decimal)Numerator / Denominator, digitsToTake);
}

But I really realized, that some numbers has a partial decimal finite and later infinite. For example: 1/28

Do you know a better way to do this? Or an Algorithm?

Answer 1

A very simple algorithm is this: implement long division. Record every intermediate division you do. As soon as you see a division identical to the one you’ve done before, you have what’s being repeated.

Example: 7/13.

1. 13 goes into   7 0 times with remainder  7; bring down a 0.
2. 13 goes into  70 5 times with remainder  5; bring down a 0.
3. 13 goes into  50 3 times with remainder 11; bring down a 0.
4. 13 goes into 110 8 times with remainder  6; bring down a 0.
5. 13 goes into  60 4 times with remainder  8; bring down a 0.
6. 13 goes into  80 6 times with remainder  2; bring down a 0.
7. 13 goes into  20 1 time  with remainder  7; bring down a 0.
8. We have already seen 13/70 on line 2; so lines 2-7 have the repeating part

The algorithm gives us 538461 as the repeating part. My calculator says 7/13 is 0.538461538. Looks right to me! All that remains are implementation details, or to find a better algorithm!