Re: <xsl:when test="position()%2=0">

first of all i wish to thank all of you guys (and gals) for providing me with great help. now i have another query

i would like to find the position of a node whether it is odd or even, and tried to use position() mod 2, but i don't know how to do it

can anyone tell me how to find the remainder of a division or else any ideas on how to find whether a node is odd or even

Source:
<?xml version='1.0'?>
<Things>
<Thing>First</Thing>
<Thing>Second</Thing>
<Thing>Third</Thing>
<Thing>Fourth</Thing>
<Thing>Fifth</Thing>
<Thing>Sixth</Thing>
</Things>

<?xml version='1.0'?>
<xsl:stylesheet
 version="2.0"
 xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
 >
<xsl:output method="xml" indent="yes" encoding="UTF-8" />

<xsl:template match="/Things">
<xsl:copy >
<xsl:apply-templates select="Thing" />
</xsl:copy>
</xsl:template>

<?xml version="1.0" encoding="UTF-8"?>
<Things>
   <Remainder>1</Remainder>
   <Thing>First</Thing>
   <Remainder>1</Remainder>
   <Thing>Third</Thing>
   <Remainder>1</Remainder>
   <Thing>Fifth</Thing>
</Things>

If you change the <xsl:when> like this
<xsl:when test="not(position() mod 2)" >