Here is an example function i found on msdn,that will only produce even nearest numbers
seems to fit your case well ,
using System;
class Example
{
public static void Main()
{
// Define a set of Decimal values.
decimal[] values = { 1.45m, 1.55m, 123.456789m, 123.456789m,
123.456789m, -123.456m,
new Decimal(1230000000, 0, 0, true, 7 ),
new Decimal(1230000000, 0, 0, true, 7 ),
-9999999999.9999999999m,
-9999999999.9999999999m };
// Define a set of integers to for decimals argument.
int[] decimals = { 1, 1, 4, 6, 8, 0, 3, 11, 9, 10};
Console.WriteLine("{0,26}{1,8}{2,26}",
"Argument", "Digits", "Result" );
Console.WriteLine("{0,26}{1,8}{2,26}",
"--------", "------", "------" );
for (int ctr = 0; ctr < values.Length; ctr++)
Console.WriteLine("{0,26}{1,8}{2,26}",
values[ctr], decimals[ctr],
Decimal.Round(values[ctr], decimals[ctr]));
}
}
// The example displays the following output:
// Argument Digits Result
// -------- ------ ------
// 1.45 1 1.4
// 1.55 1 1.6
// 123.456789 4 123.4568
// 123.456789 6 123.456789
// 123.456789 8 123.456789
// -123.456 0 -123
// -123.0000000 3 -123.000
// -123.0000000 11 -123.0000000
// -9999999999.9999999999 9 -10000000000.000000000
// -9999999999.9999999999 10 -9999999999.9999999999
"When rounding midpoint values, the rounding algorithm performs an equality test. Because of problems of binary representation and precision in the floating-point format, the value returned by the method can be unexpected."