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."
