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From: seshacha@uokvax.UUCP
Newsgroups: net.math
Subject: Re: Some dandy questions!!
Message-ID: <3300008@uokvax.UUCP>
Date: Fri, 15-Feb-85 15:03:00 EST
Article-I.D.: uokvax.3300008
Posted: Fri Feb 15 15:03:00 1985
Date-Received: Fri, 22-Feb-85 10:04:13 EST
References: <341@aesat.UUCP>
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Nf-From: uokvax!seshacha    Feb 15 14:03:00 1985



 Here is one solution to the problem 1.

 Given 1+11+111+.........+11....11 (n one's)

 Each term could be written as 10^x + 1 with x varying from 0 to n-1

 So the summation would be,

 Summation of (10^x+1) for x varying from 0 to n-1.
 Since there are n terms in the given series, we have,

 Summation of (10^x+1) = summation of(10^x) + n+1, x varying from
1 to n-1.

 Summation of 10^x is a trivial geometric series problem and is
  10(10^(n-1) - 1)/9 giving us the final result as
  ( 10^n + 9n - 1 ) / 9.

The answer to the second problem is trivial when you notice that for
each degree rotation of minute hand the hour hand makes 1/12 degree rotation.
Since the angle between 1 and 4 of the clock is 120 degrees. A 90 degrees
angle would occur approximately at 4.05. The additional fraction that needs
to be added to this is due to the rotation of the hour hand during this period.
I believe it comes out as recurring series of 27 seconds.