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    <p>I don't think there are any standard numeric format strings which will always omit trailing insignificant zeroes, I'm afraid.</p> <p>You could try to write your own decimal normalization method, but it could be quite tricky. With the <code>BigInteger</code> class from .NET 4 it would be reasonably feasible, but without that (or something similar) it would be very hard indeed.</p> <p>EDIT: Okay, I think this is what you want:</p> <pre><code>using System; using System.Numerics; public static class DecimalExtensions { // Avoiding implicit conversions just for clarity private static readonly BigInteger Ten = new BigInteger(10); private static readonly BigInteger UInt32Mask = new BigInteger(0xffffffffU); public static decimal Normalize(this decimal input) { unchecked { int[] bits = decimal.GetBits(input); BigInteger mantissa = new BigInteger((uint) bits[0]) + (new BigInteger((uint) bits[1]) &lt;&lt; 32) + (new BigInteger((uint) bits[2]) &lt;&lt; 64); int sign = bits[3] &amp; int.MinValue; int exponent = (bits[3] &amp; 0xff0000) &gt;&gt; 16; // The loop condition here is ugly, because we want // to do both the DivRem part and the exponent check :( while (exponent &gt; 0) { BigInteger remainder; BigInteger divided = BigInteger.DivRem(mantissa, Ten, out remainder); if (remainder != BigInteger.Zero) { break; } exponent--; mantissa = divided; } // Okay, now put it all back together again... bits[3] = (exponent &lt;&lt; 16) | sign; // For each 32 bits, convert the bottom 32 bits into a uint (which won't // overflow) and then cast to int (which will respect the bits, which // is what we want) bits[0] = (int) (uint) (mantissa &amp; UInt32Mask); mantissa &gt;&gt;= 32; bits[1] = (int) (uint) (mantissa &amp; UInt32Mask); mantissa &gt;&gt;= 32; bits[2] = (int) (uint) (mantissa &amp; UInt32Mask); return new decimal(bits); } } class Program { static void Main(string[] args) { Check(6.000m); Check(6000m); Check(6m); Check(60.00m); Check(12345.00100m); Check(-100.00m); } static void Check(decimal d) { Console.WriteLine("Before: {0} - after: {1}", d, d.Normalize()); } } } </code></pre>
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    1. COJust cast to a string and recursively remove right hand zeros before returning, no?
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    2. CO@drachenstern: Well you could... but you'd also have to be careful with, say, "6000".
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    3. CO@JonSkeet ~ Very true, I should add that to my answer :p
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