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Conference rusure::math

Title:Mathematics at DEC
Moderator:RUSURE::EDP
Created:Mon Feb 03 1986
Last Modified:Fri Jun 06 1997
Last Successful Update:Fri Jun 06 1997
Number of topics:2083
Total number of notes:14613

644.0. "Potential Future Omni Problems" by BEING::POSTPISCHIL (Always mount a scratch monkey.) Mon Jan 12 1987 00:44

    Exactly one month ago, I received "Trial Test 'A'" from Ronald K.
    Hoeflin, who it seems authored the test in _Omni_ some time ago,
    several problems of which are in previous topics.  The number/spatial
    problems are below. The verbal problems are in JOYOFLEX
    (DIR/TITLE=OMNI).  Although there is no copyright on the test I
    received, I think it would be best not to distribute this further.  One
    month is the recommended time for the entire test.  The answers I have
    are in the first response. 
    
    
    				-- edp
    
                                Number Series
    
    Write the number which best continues each of the following series,
    "best" meaning based on the simplest possible algorithm that
    successfully generates the numbers given in each series.
    
    25.  2 7 23 53 97 151 227
    26.  5 3 5 6 2 9 5 1 4 1
    27.  1 4 17 54 345 368 945
    28.  2 15 1,001 213,441
    29.  7 8 5 3 9 8 1 6 3
    30.  0 6 21 40 5 -504
    
    
                               Spatial Problems
    
    When two or more spatial figures interpenetrate, they may divide one
    another into a number of smaller pieces.  For each of the following
    problems, find the maximum number of pieces that can be produced by the
    particular combination of figures indicated, counting only pieces that
    are not further subdivided.  For interpenetrating two-dimensional
    figures, the pieces are bounded surfaces.  For interpenetrating
    three-dimensional figures, the pieces are bounded volumes.  For
    combinations of two- and three- dimensional figures, you are to
    consider the pieces that are bounded volumes.  Illustrated below [not
    reproduced here] are three interpenetrating circles, yielding seven
    pieces.  You may freely vary the size of the figures to produce the
    maximum number of pieces.
    
    31.  Three circles and two triangles.
    32.  One tetrahedron and one cube.
    33.  One tetrahedron and four spheres.
    34.  Two right circular cones and one torus.
    35.  Two right circular cones and one right circular cylinder.
    36.  One torus and three Mobius strips, each Mobius strip confined
    	 to and encircling the interior of the torus and each having
    	 a 180-degree twist that is evenly distributed along its length.
T.RTitleUserPersonal
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644.1BEING::POSTPISCHILAlways mount a scratch monkey.Mon Jan 12 1987 00:4922
    The answers are below.  The only answer I am confident enough to
    present among the spatial problems is not definite.  I cannot visualize
    worth a damn. 
    
    
    				-- edp
    
    25.  2 7 23 53 97 151 227 311 (1st prime, 4th prime, 9th, . . .)
    26.  5 3 5 6 2 9 5 1 4 1 3 (pi, backwards)
    27.  1 4 17 54 345 368 945
    28.  2 15 1,001 213,441 95,041,567 (2 3*5 7*11*13 . . .)
    29.  7 8 5 3 9 8 1 6 3 3 (pi/4)
    30.  0 6 21 40 5 -504
    31.  Three circles and two triangles.
    32.  One tetrahedron and one cube.
    33.  One tetrahedron and four spheres.
    34.  Two right circular cones and one torus.
    35.  Two right circular cones and one right circular cylinder.
    36.  One torus and three Mobius strips, each Mobius strip confined
    	 to and encircling the interior of the torus and each having
    	 a 180-degree twist that is evenly distributed along its length.
    	 4
644.2BEING::POSTPISCHILAlways mount a scratch monkey.Tue Feb 24 1987 12:0954
    More problems are below.  Some have changed a bit, and I think they are
    more interesting.  The first problem below does not seem to be
    correctly stated, so it is insolvable, but I include it in case I am
    wrong. 
    
    
    				-- edp
    
    Suppose a black box contains ten marbles of unknown colors.  The
    marbles' colors can be determined only by selecting one marble at a
    time at random from the box, but it must be returned to the box and
    mixed thoroughly with the rest before another marble is chosen for
    inspection.  If ten marbles are inspected in this way and all turn out
    to be white marbles, what is the probability at this point that the box
    contains only white marbles?  (Round to the nearest whole percent.)
    
    What is the numerical value of x in the following sequence:  4/10  x/10
    168/1,000  1,229/10,000  9,592/100,000  78,448/1,000,000 . . . ?
    
    What number comes next in each of these sequences?
    4  36  144  400  900  1764  3136  5184  ?
    1  3  8  20  48  112  256  576  1280  ?
    
    Suppose a lump of clay is shaped into a tetrahedron and that it is
    sliced six times with a knife, each knife stroke being perfectly
    straight (i.e., planar), with the pieces formed by the knife strokes
    never being rearranged.  What is the maximum number of tetrahedral
    (i.e., 4-sided) pieces of clay that can thereby be formed, not counting
    pieces that are further subdivided?
    
    Suppose there is a polyhedron all of whose edges are of equal length,
    and suppose there is an ant at each vertex of the polyhedron.  Suppose
    further that each ant randomly selects one of the edges that meet at
    its vertex and crawls along it until it arrives at the next vertex.  If
    all of the ants start out simultaneously, crawl at equal speeds, never
    reversing direction, and arrive at the next vertex simultaneously, what
    is the probability that no two ants will encounter one another?  Solve
    this problem for each of the following polyhedra:
    
    	A tetrahedron.
    	A cube.
    	An octohedron.
    	A dodecahedron.
    	An icosahedron.
    
    [I used a program for those.]
    
    Suppose an ant crawls along the edge of a one-cubic-inch cube at a rate
    of one inch per minute, never reversing direction.  At each corner it
    comes to there is an even chance that it will turn right or left.  If
    the ant starts at a corner, what is the probability that at the end of
    100 minutes it will be back at that same corner?  [I think an
    approximate answer is called for here, say to the nearest percent.] 
          
644.3one answer to .2VINO::JMUNZERTue Mar 03 1987 20:17100
>    Suppose an ant crawls along the edge of a one-cubic-inch cube at a rate
>    of one inch per minute, never reversing direction.  At each corner it
>    comes to there is an even chance that it will turn right or left.  If
>    the ant starts at a corner, what is the probability that at the end of
>    100 minutes it will be back at that same corner?  [I think an
>    approximate answer is called for here, say to the nearest percent.] 


	   G_________________ H		
	   /                /|		Start at A.
	  /                / |		B, C (hidden), D are 1 away from A
	D/________________/  |		E, F, G are 2 away from A
	 |               |F  |		H is 3 away from A
	 |               |   |
	 |               |  / E
	 |               | /
	 |_______________|/
        A                B

Call	A->B, A->C, A->D state 01
	B->A, C->A, D->A state 10
	B->E, B->F, C->E, C->G, D->F, D->G state 12
	E->B, F->B, E->C, G->C, F->D, G->D state 21
	E->H, F->H, G->H state 23
	H->E, H->F, H->G state 32

Note that the ant starts:  state 01, then state 12.

		==	==	==

Use the ant's motion rules to form the following table.  E.g. after B->E (12),
may turn to E->C (21) or to E->H (23).

Transitions  TO:	01	10	12	21	23	32
	FROM:
	01				1
	10		1
	12					1/2	1/2
	21			1/2	1/2
	23							1
	32					1

		==	==	==

Combine the transitions into pairs (e.g. 12-to-23 + 23-to-32 becomes 12-to-32):

Transitions  TO:	10	12	32
	FROM:
	10			1
	12	  	1/4	1/4	1/2
	32		1/2	1/2

		==	==	==

If there were a steady state solution, it could be solved from that table.
P(state) means the probability of being in the state.
    
	P(10) = 1/4 * P(12) + 1/2 * P(32)
	P(12) = P(10) + 1/4 * P(12) + 1/2 * P(32)
	P(32) = 1/2 * P(12)
and
	P(10) + P(12) + P(32) = 1

which yields P(10) = P(32) = 1/4; P(12) = 1/2

		==	==	==

Suppose that you're not at the steady state solution.  Will you get there?

	Let	x = 4 * P(10) - 1
		y = 4 * P(32) - 1

Transitions will make P(10) and P(32) go to 1/4 iff they make x and y go
to zero.

After a pair of transitions, get

	x'	= 4 * P'(10) - 1
		= 4 * [1/4 * P(12) + 1/2 * P(32)] - 1
		= P(12) + 2 * P(32) - 1
		= [1 - P(10) - P(32)] + 2 * P(32) - 1
		= - P(10) + P(32)
		= - (x+1) / 4 + (y+1) / 4
		= - x/4 + y/4
and
	y'	= 4 * P'(32) - 1
		= 4 * 1/2 * P(12) - 1
		= 2 * [1 - P(10) - P(32)] - 1
		= 2 - (x+1) / 2 - (y+1) / 2 - 1
		= - x/2 - y/2

Fortunately,

	|x'| + |y'| <= |x/4| + |y/4| + |x/2| + |y/2| = 3/4 * (|x| + |y|)

so x and y go to zero, and P(10) goes to 1/4.  After 49 pairs of transitions
(that is the original question!) it's about 25% that the ant is going towards
corner A.

John
644.4Re .2VINO::JMUNZERFri Apr 03 1987 13:496
>    What is the numerical value of x in the following sequence:  4/10  x/10
>    168/1,000  1,229/10,000  9,592/100,000  78,448/1,000,000 . . . ?

    x/100 = 25/100.  There are 25 primes less than 100.
    
    John
644.5Re .0VINO::JMUNZERFri Apr 03 1987 13:537
>        30.  0 6 21 40 5 -504

    edp, world:
    
    Has anyone had any ideas about -504?
    
    John
644.6Re .2LATOUR::JMUNZERFri Apr 03 1987 16:388
>    What number comes next in each of these sequences?
>    4  36  144  400  900  1764  3136  5184  ?
>    1  3  8  20  48  112  256  576  1280  ?
    
    4  36  144  400  900  1764  3136  5184  8100  ?
    1  3  8  20  48  112  256  576  1280  2816  ?

    John    
644.7Previous entrants not eligibleSQM::HALLYBAre all the good ones taken?Fri Apr 03 1987 21:327
>    4  36  144  400  900  1764  3136  5184  8100  ?
>    1  3  8  20  48  112  256  576  1280  2816  ?

    4  36  144  400  900  1764  3136  5184  8100  12100  ?
    1  3  8  20  48  112  256  576  1280  2816   6144 ?

      John
644.8CLT::GILBERTeager like a childSat Apr 04 1987 05:092
Re problem 28.
    There seems to be a typo -- the 213,441 should be 215,441.
644.9BEING::POSTPISCHILAlways mount a scratch monkey.Mon Apr 06 1987 15:066
    Re .8:
    
    Right you are.  That's what we call an extra-credit problem.
    
    
    				-- edp
644.10solution to 28CLT::GILBERTBuilderThu Jan 28 1988 12:306

    2 = 2
    15 = 3x5			(product of the next two primes)
    1001 = 7x11x13		(   "    "   "   "   three  "  )
    215441 = 17x19x23x29	(   "    "   "   "   four   "  )
    95041567 = 31x37x41x43x47	(   "    "   "   "   five   "  )
644.11solution to 30RDGCSS::RSMITHFri Jan 29 1988 10:438
(next in series 0, 6, 21, 40, 5, -504, ...)

-4697

Reason follows form feed:

                           3
n th number in series is  n  -  n! 
644.12How not to read notation.ZFC::DERAMOFrom the keyboard of Daniel V. D'EramoFri Jan 29 1988 13:0220
    Re:  reason in .-1
    
    
>>                                  3
>>       n th number in series is  n  -  n!
    
    At first I thought that you meant n^3 - n, with the exclamation
    point because you thought the result was interesting!  I thought,
    "No way those numbers come from n^3 - n ..."  (-:  It took a
    second or two to realize you meant "n cubed minus n factorial"!
    
    Oh so many years ago when I first read something that used the
    notation "n!" -- without a definition, because it was assumed
    the reader knew what it meant -- I mentally pronounced it
    
                            n Wow!
    
    but I couldn't tell from the context why n was so exciting! (-:
    
    Dan