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Path: utzoo!watmath!clyde!burl!ulysses!unc!mcnc!decvax!linus!philabs!ttidca!ttidcb!pumphrey
From: pumphrey@ttidcb.UUCP (Larry Pumphrey)
Newsgroups: net.math
Subject: Re: Some dandy questions!!
Message-ID: <278@ttidcb.UUCP>
Date: Tue, 12-Feb-85 14:53:12 EST
Article-I.D.: ttidcb.278
Posted: Tue Feb 12 14:53:12 1985
Date-Received: Fri, 15-Feb-85 05:47:33 EST
References: <341@aesat.UUCP>
Organization: TTI, Santa Monica, CA.
Lines: 93


 from: (Larry Pumphrey @ Citicorp TTI, Santa Monica, CA)

 The poster requested followup via mail; however, we seem to have
 problems using mail from our site, so will post in net.math and
 hope original poster reads this newsgroup.

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 |                                                                         |
 |                                                                         |
 |      Solution to  Sn = 1 + 11 + 111 + ... + 11...11                     |
 |                                             |<--->|                     |
 |                                              n 1's                      |
 |      is easily obtained by re-writing as                                |
 |                                                                         |
 |      (1)  Sn = (10^1 - 1)/9 + (10^2 - 1)/9 + ... + (10^n - 1)/9         |
 |                                                                         |
 |      re-arranging terms gives                                           |
 |                                                                         |
 |      (2)  Sn = 10*(10^n -1)/81  -  n/9                                  |
 |                                                                         |
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 |                                                                         |
 |      Solution to clock problem is as follows:                           |
 |                                                                         |
 |      Let t be the time in hours so that the range of t is 0 to 12 in    |
 |      one complete (12 hour) revolution of the hour hand.  Further       |
 |      assume that t=0 corresponds to 12 o'clock.                         |
 |                                                                         |
 |      The angle A of the minute hand at time t is given by               |
 |                                                                         |
 |      (1)  A = (PI/2 - 2*PI*t)                                           |
 |                                                                         |
 |      and the angle B of the hour hand at time t is given by             |
 |                                                                         |
 |      (2)  B = (PI/2 - PI*t/6)                                           |
 |                                                                         |
 |      For 2 angles (A,B) to differ by PI/2 radians (i.e., to be at       |
 |      right angles to one another), it is both necessary and             |
 |      sufficient that                                                    |
 |                         cos(A-B) = 0                                    |
 |                                                                         |
 |      From (1) and (2) above this produces                               |
 |                                                                         |
 |      (3)  cos(11*PI*t/6) = 0                                            |
 |                                                                         |
 |      But cos(X) = 0 only for X = (2n+1)*PI/2  which implies             |
 |                                                                         |
 |      (4)  t = 3*(2n+1)/11                                               |
 |                                                                         |
 |      The boundary conditions  0 <= t <= 12 give                         |
 |                                                                         |
 |      (5)  0 <= n <= 21   which are all the possible solutions to the    |
 |                          general question "When are the 2 hands of a    |
 |                          clock at right angles to one another?"         |
 |                                                                         |
 |      ---------------------------------------------------------------    |
 |                                                                         |
 |      Note:  The special boundary conditions  4 <= t <= 5  (which was    |
 |             the original problem) is equivalent to  7 <= n <= 8         |
 |             which produce times corresponding to t=45/11 and 51/11      |
 |             hours.  These translate approximately to:                   |
 |                                                                         |
 |                           4:05:27 and 4:38:11                           |
 |                                                                         |
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 |                                                                         |
 |      Now let me pose the following even dandier problem!                |
 |                                                                         |
 |      Given:     A 3 handed clock (hour - minute - second)               |
 |                                                                         |
 |      Question:  Is there ever any time(s) that the 3 hands cut          |
 |                 the clock into 3 identical pie slices?  In              |
 |                 other words is each of the three hands ever             |
 |                 exactly 120 degress from the other two hands?           |
 |                                                                         |
 |      Caveats:   Only "elegant" solutions are acceptable.  For           |
 |                 example, it is easy (although time-consuming)           |
 |                 to determine by brute force --- i.e., simply            |
 |                 calculate the 22 times that the hour hand and           |
 |                 the minute hand are 120 degrees apart (using            |
 |                 an argument similar to that above in solving            |
 |                 the "two-handed case") and then just examine            |
 |                 where the second hand is at each of those 22            |
 |                 occurances.  Such Neanderthal methods are to            |
 |                 be discouraged!  :-)                                    |
 |                                                                         |
 |      Hint:      If you like to solve Diophantine equations or           |
 |                 you are a fan of The Chinese Remainder Theorem,         |
 |                 then you'll love this problem!                          |
 |                                                                         |
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