Considering a positive whole number X which contains more than one digit, let us define R(X) as the number obtained by reversing the digits of X. Neither X nor R(X) can contain any leading zeroes.

* A Conference commenced on a given day precisely at 10*X/ 143 minutes past P o'clock ( but before P+1 o'clock) and concluded at 10*R(X) / 143 minutes past Q o'clock ( but before Q+1 o'clock) on the same day ; where 11 >=Q > P >=1 with the proviso that P and Q are whole numbers. It was observed that the hour hand and the minute hand had exchanged places during the respective times of commencement and conclusion of the said conference.

* Determine X, R(X), P and Q and,hence or otherwise, find the precise times of commencement and conclusion of the conference.

I enjoy these clock based problems.  Thanks KS.

My best values were X=813 and R(X)=318.  P=4; Q=11.  The conference started at about 4:56:51 and ended at about 11:22:14.

The next best values for X were 243, 615, then 441. followed by the two Charlie found: 486 and 684, and finally 417.  Those are the only six-sigma solutions to the problem.

I didn't use a computer program, so I might have made an error somewhere.  What I did was observe that since the positions of the hour and minute hands were exchanged, given P and Q are whole numbers, the fraction 10*X/143 represents the location of the minute hand at the start of the conference and also indicates how far past the Q o'clock position the hour hand lies.  A similar relationship exists for the fraction 10*R(X)/143, the minute hand at the end of the conference, and how far past the P o'clock position the hour hand lies.

 

Edited on December 9, 2005, 9:17 pm

Edited on December 9, 2005, 9:19 pm

Posted by MindRod on 2005-12-09 17:52:39