<html xmlns:v="urn:schemas-microsoft-com:vml" xmlns:o="urn:schemas-microsoft-com:office:office" xmlns:w="urn:schemas-microsoft-com:office:word" xmlns:m="http://schemas.microsoft.com/office/2004/12/omml" xmlns="http://www.w3.org/TR/REC-html40"><head><meta http-equiv=Content-Type content="text/html; charset=utf-8"><meta name=Generator content="Microsoft Word 14 (filtered medium)"><style></style></head><body lang=EN-US link=blue vlink=purple><div class=WordSection1>This day almost slipped by without due notice:<o:p></o:p><o:p> </o:p>Today, November 02, 2011, or 11-02-2011, or 11022011, is super palindrome day! (For those parts of the world where the date is expressed as day-month-year their most recent palindrome day, 1-10-2011 or 1102011 has already passed by sadly unmarked.)<o:p></o:p><o:p> </o:p>But Wait! There’s More! The following is from The Beacon, the student paper of the University of Portland (Oregon):<o:p></o:p><o:p> </o:p>This year is special because it contains two palindrome calendar dates: Jan. 10, 2011 expressed as 1-10-2011 (or simply 1102011) and Nov. 2, 2011 written as 11-02-2011 (11022011). The first one has already occurred and the second one, 11022011, is coming up. (Note that in most of the world where day-month-year date format is used, this year also has two palindrome dates. Date 11022011 representing 11 February 2011 already passed and 1102011 corresponding to 1 October 2011 is about to occur.)<o:p></o:p>After 2011, there will be one more year in this (21st) century containing two palindrome dates. That will be 2021, with palindrome dates Jan. 20, 2021 (1202021) and Dec. 2, 2021 (12022021). (In the rest of the world, after 2011, there will also be one more year in this century containing two palindrome dates, but that year will be 2012 instead of 2021.)<o:p></o:p><o:p> </o:p>Nov. 2, 2011 represented as 11022011 is a one-of-a-kind palindrome date with respect to all palindrome dates contained in all four-digit years. Why?<o:p></o:p>First, number 11022011 equals 7 x 7 x 11 x 11 x 11 x 13 x 13, that is, the product of seven square, eleven cube and thirteen square where numbers seven, eleven and thirteen are three consecutive prime numbers! So, number 11022011 is divisible by the product of the squares of three consecutive prime numbers! Furthermore, it's also divisible by the cube of the middle prime of the three consecutive primes! In fact, 11022011 = 72 x 113 x 132 where, interestingly enough, the three superscripts side-by-side constitute 232 which is also a palindrome! Fascinating, isn't it? No other such palindrome date exists in all four-digit years.<o:p></o:p><o:p> </o:p>In addition, since 7 x 11 x 13 yields 1001, another palindrome number, one could also express date 11022011 as 1001 x 11 x 1001 where the left and right sides of this expression divided in the middle are almost mirror images of one another! Isn't that something?<o:p></o:p>Also, if date number 11022011 is split into four two-digit numbers as 11, 02, 20 and 11, the first two numbers add up to 13 and the sum of the last two is 31, and 13 and 31 put side-by-side yield 1331 = 11 x 11 x 11! Now, by introducing two zeros in-between the digits of the first and last elevens, this expression will change from 11 x 11 x 11 to 1001 x 11 x 1001 = 11022011! Wow!<o:p></o:p><o:p> </o:p>(by Aziz Inan, professor of electrical engineering at UP. He can be contacted at ainan@up.edu)<o:p></o:p><o:p> </o:p>ddf<o:p></o:p><o:p> </o:p><div>Delwin D Fandrich<o:p></o:p>Piano Design & Fabrication<o:p></o:p>6939 Foothill Court SW, Olympia, Washington 98512 USA<o:p></o:p>Phone  360.515.0119 — Cell 360.388.6525<o:p></o:p><a href="mailto:del@fandrichpiano.com">del@fandrichpiano.com </a>— <a href="mailto:ddfandrich@gmail.com">ddfandrich@gmail.com</a><o:p></o:p></div><o:p> </o:p></div></body></html>