rhm (05/27/82)
Find a non-zero number x with the property that when you write the number in decimal notation, then remove the right-hand digit and add it to the left end of the number, the result is 2*x. i.e. 214 -> 421 is close, but no cigar.
jpj (05/28/82)
Note that the problem is *not* a math problem at all, it is a *string* problem! Given this, you can crank out answers by noticing that if you choose a value for the right-most digit (say 2) and then multiply by 2 mod 10 to get the next digit (keeping track of carries as necessary), when you get to the left-most digit, multiplying it by 2 and comparing it with the digit you started with will tell you if you have a solution. What I have found indicates that there are *no* solutions for less than 18 digits, that the solution at 18 is a string that can be permuted 8 different ways and that you thus find new solutions at every string of length n*18 - each time 8 solutions! These strings were generated on a VAX in under 2 seconds... n now equals 18 105263157894736842 157894736842105263 210526315789473684 263157894736842105 315789473684210526 368421052631578947 421052631578947368 473684210526315789 . . . n now equals 90 105263157894736842105263157894736842105263157894736842105263157894736842105263157894736842 157894736842105263157894736842105263157894736842105263157894736842105263157894736842105263 210526315789473684210526315789473684210526315789473684210526315789473684210526315789473684 263157894736842105263157894736842105263157894736842105263157894736842105263157894736842105 315789473684210526315789473684210526315789473684210526315789473684210526315789473684210526 368421052631578947368421052631578947368421052631578947368421052631578947368421052631578947 421052631578947368421052631578947368421052631578947368421052631578947368421052631578947368 473684210526315789473684210526315789473684210526315789473684210526315789473684210526315789 Cheers... Jim Jenal BTL/CB
rhm (05/29/82)
The continued fraction representation of a number x is written as x = (n1, n2, n3, ... ) where n1, etc. are positive integers, and x = n1 + 1/(n2 + 1/(n3 + ... )) A continued fraction that terminates (zero from some point on) represents a rational number. 1. Does the continued fraction rep. of every rational number terminate? If not, does it at least repeat? 2. What are the numbers whose continued fraction rep's repeat? 3. The continued fraction (2, 1, 2, 1, 1, 4, 1, 1, 6, 1, 1, 8, 1, 1, 10, ... ) seems to equal something like 2.718... Is it possible that this is a simple representation of "e", the base of natural logarithhms, which is a transcendental number? Is it in fact equal to e?
G:shallit (05/30/82)
To answer your questions, it is not hard to prove that the continued fraction for x terminates if and only if x is rational. One way to see this is to associate CF's with the Euclidean algorithm for GCD; the two algorithms are sort of "mirrors" of each other, since in one you are interested in remainders; in the other, the quotients. Second, if the CF for x is eventually periodic, then x is the root of a quadratic equation. The converse is also true. Third, it is true that e = [ 2, 1, 2, 1, 1, 4, 1, 1, 6, ... ]. This representation is due to Euler and Hurwitz. There are similar expansions of interest for exp (1/m) where m is a positive integer and related transcendental quantities. Two books that have interesting things about CF's are "Continued Fractions by C. D. Olds, and Knuth, Art of Computer Programming, V. II. /Jeff Shallit, Department of Mathematics, University of California, Berkeley.
rhm (06/02/82)
I was informed recently by a swami of irreproachable accuracy that the world would end the next time the first day of a century falls on a Sunday. Can anyone help me find out when this will be? [The rules for the calendar currently in use are that century years are not leap years unless they are divisible by 400. I.e. 2000 is a leap year; 1900 was not. The first day of the 20th century was Tuesday, Jan 1, 1901 NOT Jan 1, 1900.]
rhm (06/07/82)
Someone once told me that there has to be at least one place on Earth where there is no wind. Does anyone know how to prove this without bringing in a lot of jazz about topology?
rhm (06/08/82)
Someone once told me that the probability of Christmas falling on a Sunday is not precisely 1/7. Anyone know what it is?
eagle@ihuxs.UUCP (John Blumenstein) (08/02/84)
From: Michael Rios (at AT&T Bell Labs, I think)
Re: A personal headache.
When I was in high school, I was a very active member of the Math Team (much
to the other members' annoyance, most of the time). We would always (to keep
in practice) generate problems to give to each other and set a time limit
(typically a week) to solve them. The only problem ever to defeat me was one
that I made up myself:
A regular pentagon of side 10 has a line drawn inside it, parallel
to one side, which divides the pentagon into two sections of equal
area. What is the length of this line (no decimal approximations,
please)?
Could someone please help me with this?
("Is this the right room for an arguement?" "I've told you once.")
Michael Rios
Chicago, Il.
Earth
("No you haven't.")
--
John T. Blumenstein
ihuxr!eaglericharde@tektronix.UUCP (Richard Elliott) (08/07/84)
<<A regular pentagon of side 10 has a line drawn inside it, parallel
<<to one side, which divides the pentagon into two sections of equal
<<area. What is the length of this line?
The easiest way (in my mind) to solve this problem is to express the
dividing line segment as the base of a trapezoid with an area equal
to half that of the pentagon. The other base of the trapezoid will be
ten units, the length of a side of the pentagon, while the angles to
the sides from the small base (unknown segment) and the large base can
be easily found to be 108(deg) (the inside angle of a pentagon) and
72(deg) (the angle supplementary to 108) respectively. Thus the only
unknowns are the height and large base of the trap. The next step is
to establish a relationship between these two unknowns and substitute
this equation in terms of B(large base) for h(height) into the equation
for the area of a trap. This final equation can be solved for B as
it is the only unknown. I find the line segment to be 14.55 units.
Equations and formulas used:
Area of a regular polygon (n=# of sides, l=length of side)
(1) A = (n*l(l*tan(180-360/n)/2)/2)/2
which in the case of a pentagon with l=10 proves to be 344 sq units.
Area of a trapezoid (large base=B, small base=b, height=h)
(2) A = h(B+b)/2
equation relating h and B (with b=10)
(3) h = tan72(deg)(B-10)/2
equation substituting (3) into (2)
(4) 2A(trap) = (B*B - 100)tan72(deg)
or
B = sqt((2A+100tan72(deg))/tan72(deg))
= 14.55 approx.
Richard Elliott
richarde@tektronix