tag:blogger.com,1999:blog-15628310.post5311787863446345849..comments2021-04-15T02:41:46.574-04:00Comments on Question of the day: A Whole Range of NumbersAnonymoushttp://www.blogger.com/profile/18153935609499338685noreply@blogger.comBlogger5125tag:blogger.com,1999:blog-15628310.post-66767413039016277382008-07-24T10:19:00.000-04:002008-07-24T10:19:00.000-04:00I'm glad you guys enjoyed the question. It's been...I'm glad you guys enjoyed the question. <BR/><BR/>It's been tough lately trying to come up with new ones.Anonymoushttps://www.blogger.com/profile/18153935609499338685noreply@blogger.comtag:blogger.com,1999:blog-15628310.post-82567502668615721352008-07-24T09:48:00.000-04:002008-07-24T09:48:00.000-04:00Hi Mike, you're right - "From which it be...Hi Mike, you're right - "From which it becomes obvious that if n > 0 no solution can exist" - am blaming a long day at the office for that appalling assumption on my part!!Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-15628310.post-25461346719480732692008-07-23T19:37:00.000-04:002008-07-23T19:37:00.000-04:00Wow, great problem, Mike. Too bad I didn't check i...Wow, great problem, Mike. Too bad I didn't check in yesterday...Abehttps://www.blogger.com/profile/04424868492071587450noreply@blogger.comtag:blogger.com,1999:blog-15628310.post-82309775397139330832008-07-23T09:58:00.000-04:002008-07-23T09:58:00.000-04:00You're on the right track. I may have made a ...You're on the right track. I may have made a mistake somewhere along the way, but 1 is not the only solution.<BR/><BR/>If we let the answer be x and let x = 10*A+ B (as in the A is the 10 digit and B is the 1 digit), then:<BR/>2*10*A+B - 1 = 10*B+A<BR/>Which leads to:<BR/>19*A = 8*B + 1<BR/><BR/>8*B + 1 must be an odd number, therefore 19*A has to be an odd number.<BR/>The largest number 8*B + 1 can be is 73. Therefore 19*A <= 73. A must be 1, 2 or 3. A = 1 doesn't work, but A = 3 can work. If A is 3, then B = 7.<BR/><BR/>Therefore 37 is a possible solution.<BR/><BR/>You can then expand to the third digit and fourth digit and so on to get the possible solutions of:<BR/>37*2 - 1 = 73<BR/>397*2 - 1 = 793<BR/>3997*2 - 1 = 793 <BR/>and so on....<BR/>399...997*2 - 1 = 799...993Anonymoushttps://www.blogger.com/profile/18153935609499338685noreply@blogger.comtag:blogger.com,1999:blog-15628310.post-50016259752937607352008-07-22T13:32:00.000-04:002008-07-22T13:32:00.000-04:00Hint: Consider each number to be a polynomial of t...Hint: Consider each number to be a polynomial of the form:<BR/>a(0)10^n + ... + a(n)10^0<BR/><BR/><BR/><BR/><BR/><BR/><BR/>Answer:<BR/>Now we have well defined functions for what we wish to achieve (multiplication, addition and reversal of ordering).<BR/>Now we search for equations such that:<BR/>2*(a(0)10^n + ... + a(n)10^0 ) - 1 = a(n)10^n + ... + a(0)10^0 <BR/>Rearranging and expanding gives:<BR/>(2a(0) - a(n))10^n + ... + (2a(n) – a(0))10^0 = 1<BR/>From which it becomes obvious that if n > 0 no solution can exist, thus assuming n = 0 we have<BR/>2a(0) – a(0) = 1<BR/>Therefore the polynomial 1 is the only solution to the problem.Anonymousnoreply@blogger.com