Relay-Version: version B 2.10 5/3/83; site utzoo.UUCP Posting-Version: version B 2.10.1 6/24/83; site charm.UUCP Path: utzoo!watmath!clyde!burl!ulysses!mhuxr!mhuxb!mhuxn!charm!mam 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