tag:blogger.com,1999:blog-26909122.post114739615720261897..comments2013-07-16T09:10:31.961-07:00Comments on Puzzles, puzzles and puzzles: What's so strange about me?Unknownnoreply@blogger.comBlogger2125tag:blogger.com,1999:blog-26909122.post-1147531299824177902006-05-13T07:41:00.000-07:002006-05-13T07:41:00.000-07:00Bobach, Welcome aboard! Thats a exceelent piece of...Bobach, <BR/>Welcome aboard! Thats a exceelent piece of solution you have there. There is another slight variation of the solution I have come across:<BR/><BR/>The difference is 99(x-z) and we know that x-z lies between 2 and 9. So, the difference is divisible by 99. Such numbers are 198, 297, ....so on. We notice that such numbers have a 9 in the middle digit and the sum of the other 2 digits is equal to 9. So consider the difference number to be abc, this can be represented as 100a+10b+c and its reverse to be 100c+10b+a. Add them and you get 100(a+c)+10(b)+(a+c) = 100*9+10*9+9=1089 and this is true for any 3 digit number that obeys the conditions given. Good solving Bobach. You will go into the list of the people who solved the puzzles.<BR/><BR/>This puzzle is SOLVED!Karthik Nagarajhttps://www.blogger.com/profile/02641166958828638787noreply@blogger.comtag:blogger.com,1999:blog-26909122.post-1147489466387052022006-05-12T20:04:00.000-07:002006-05-12T20:04:00.000-07:00Bobach writes: I enjoy your blog. You guys are go...Bobach writes: I enjoy your blog. You guys are good. Here's my first response: <BR/><BR/>Any 3 digit number can be represented as <BR/> 100x + 10y + z and its reverse would be<BR/> 100z + 10y - x <BR/><BR/>Their difference would be<BR/> 100(x-z) + (z-x) which can be written as 99(x-z).<BR/><BR/>When 1 < (x-z) < 10, 99(x-z) can be written as<BR/> 100(x-z-1) + 90 + (10-(x-z)) whose reverse would be<BR/> 100(10-(x-z)) + 90 + (x-z-1)<BR/><BR/>Adding these last together we find that all the variables cancel each other so we get<BR/> 100(10-1) + 180 + (10-1) which always equals 900+180+9=1089Anonymousnoreply@blogger.com