<div dir="ltr"><div><div><div>Oops. You are correct.<br><br></div>It appears there is no procedure that is to modulo as what<br></div>quotient is to remainder. <br></div>(floor (/ x y)) seems to be the only option if x and y may be negative.<br> <br>Stephan<br></div><div class="gmail_extra"><br><br><div class="gmail_quote">2013/6/11 Laurent <span dir="ltr"><<a href="mailto:laurent.orseau@gmail.com" target="_blank">laurent.orseau@gmail.com</a>></span><br><blockquote class="gmail_quote" style="margin:0 0 0 .8ex;border-left:1px #ccc solid;padding-left:1ex"> <div dir="ltr"><div class="im"><div class="gmail_extra"><br><div class="gmail_quote">On Tue, Jun 11, 2013 at 9:45 AM, Stephan Houben <span dir="ltr"><<a href="mailto:stephanh42@gmail.com" target="_blank">stephanh42@gmail.com</a>></span> wrote:<br> <blockquote class="gmail_quote" style="margin:0px 0px 0px 0.8ex;border-left:1px solid rgb(204,204,204);padding-left:1ex"><div> replace (floor (/ a b)) with (modulo a b)<br> </div></blockquote></div><br><br></div></div><div class="gmail_extra">Actually:<br>> (modulo 10 3)<br>1<br>> (remainder 10 3)<br>1<br>> (quotient 10 3)<br>3<br>> (floor (/ 10 3))<br>3<br><br></div><div class="gmail_extra"> `modulo' is more like `remainder', except for negative numbers.<span class="HOEnZb"><font color="#888888"><br></font></span></div><span class="HOEnZb"><font color="#888888"><div class="gmail_extra"><br></div><div class="gmail_extra"> Laurent<br></div></font></span></div> </blockquote></div><br></div>