hal@aesat.UUCP (Hal Patnaik) (02/08/85)
I've got a few nice problems here, and could use one or two suggestions:
1) What is the value of the sum:
1 + 11 + 111 + 1111 + 11111 + ... + 11...1 ?
n 1's
2) At what times between 4:00 and 5:00 do the hands of a clock
form a right angle?
Mail me back with you suggesstions, as I don't normally read this
newsgroup.
Thanks a lot!!
Hal Patnaik
allegra \
ihnp4 \
----- ! utzoo ! aesat ! hal
linus /
decvax /
AES DATA, Inc. The opinions expressed above are strictly my
1900 Minnesota Crt. own and do not represent those of my employer.
Mississauga, Ont.
CANADAmam@charm.UUCP (Matthew Marcus) (02/11/85)
[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!mampumphrey@ttidcb.UUCP (Larry Pumphrey) (02/12/85)
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. =========================================================================== | | | | | 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 | | | =========================================================================== | | | 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 | | | =========================================================================== | | | 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! | | | ===========================================================================
seshacha@uokvax.UUCP (02/15/85)
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.
stchakra@uokvax.UUCP (02/20/85)
Please contact me. I am an Oriya.