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From: mam@charm.UUCP (Matthew Marcus)
Newsgroups: net.math
Subject: Some Dandy Questions!! (spoiler warning)
Message-ID: <586@charm.UUCP>
Date: Mon, 11-Feb-85 14:26:58 EST
Article-I.D.: charm.586
Posted: Mon Feb 11 14:26:58 1985
Date-Received: Tue, 12-Feb-85 06:39:17 EST
Organization: Physics Research @ AT&T Bell Labs Murray Hill NJ
Lines: 63

[1+1 sometimes = 2]

A netter asked:
1)	What is the sum  of
1 + 11 + 111 + ... {n ones} ?
2)	At what time between 4:00 and 5:00 do the hour and minute hands of
a clock form a right angle?

My (possible buggy) solutions:
1) This series contains:
n 1's
n-1 10's
n-2 100's
...
1 10^n.

Thus it can be written as

sum from i=1 to n { i 10 sup {n-i+1}}  (using eqn notation)

which can be expressed as

10 sup n+2 sum from i=1 to n {i 10 sup -i-1}

which can be re-written as a derivative:

10 sup n+1 {d over dx } sum from i=0 to n {x sup i}

where the series is evaluated at x=10 sup -1 = .1 .

After a little work we get:

10 sup n+1 {10 sup 1-n times .9n + 1 - 10 sup -n} over .81

which is the answer, I hope.



2)	Let t be the time past 3:00, in hours, so 4:00 -> t=1.
	Let h=position of the hour hand, in units of 90 degrees, clockwise
		from the 12:00 position.
	Let m=position of the minute hand, in the same units.

Then,
	h=1+t/3   (it takes 3 hours for the hour hand to go 90 degrees, and
			at 3:00, it starts at +1 unit).
	m=4t	  (it takes 1/4 hour for the minute hand to go 90 degree,
			and it starts at 0).

We want that the hands be separated by an odd number of right angles:

	h-m = 2n+1   n=integer, not necessarily >=0.

Thus,
	1-11t/3 = 2n+1
	t=-2n*3/11

If n=-2, then t=12/11 = 1:05:27.272727...

Thus the time is 4:05:27.272727..., and the hour hand will be 90 degrees
clockwise of the minute hand.

	{world!BTL}!charm!mam