C# Math.Round(double,Int32) Troubles

I am using Visual C# 2010 Express and am stumped with the Math.Round method.

Two overloads are important to the context of my question:
1- Math.Round(decimal d, int n)
2- Math.Round(double d, int n)

I am trying to round to 5 total base-10 digits. See code snippet below:

int fractional_digits = 4;
double number_to_round = 123.456789;
double rounded_result = Math.Round(number_to_round, fractional_digits);
number_to_round = 11223344.556677;
rounded_result = Math.Round(number_to_round, fractional_digits);
number_to_round = 876.54321E+57;
rounded_result = Math.Round(number_to_round, fractional_digits);

The code is redundant to show the issue three times. The first rounded result gives 123.4568 not 123.46 (or 1.2346E+2) (1 integer + 4 fractional digits).

The second rounded result is 11223344.5567 instead of 11223000 (or 1.1223E+7).
Lastly, the rounded result is 8.7654321E+59 instead of 8.7654E+59.

I need only 5 significant digits. The code example seems to be using method 1 as shown above - not method 2.

I need method 2. (See comment header copied below).

//
// Summary:
//     Rounds a double-precision floating-point value to a specified number of fractional
//     digits.
//
// Parameters:
//   value:
//     A double-precision floating-point number to be rounded.
//
//   digits:
//     The number of fractional digits in the return value.
//
// Returns:
//     The number nearest to value that contains a number of fractional digits equal
//     to digits.
//
// Exceptions:
//   System.ArgumentOutOfRangeException:
//     digits is less than 0 or greater than 15.
public static double Round(double value, int digits);

How do I force the compiler to use method 2? Any other suggestions?

I have done a basic search for help on this site (via Google) but all examples only show a single place-value in the integer position.

Thanks in advance.

-EJ

Tags: C#