Items tagged with puzzle

A string is wound symmetrically around a circular rod. The string goes exactly
4 times around the rod. The circumference of the rod is 4 cm and its length is 12 cm.
Find the length of the string.
Show all your work.

(It was presented at a meeting of the European Mathematical Society in 2001,
"Reference levels in mathematics in Europe at age16").

Can you solve it? You may want to try before seing the solution.
[I sometimes train olympiad students at my university, so I like such problems].

eq:= 2/Pi*cos(t), 2/Pi*sin(t), 3/2/Pi*t; # The equations of the helix, t in 0 .. 8*Pi:
p:=plots[spacecurve]([eq, t=0..8*Pi],scaling=constrained,color=red, thickness=5, axes=none):
plots:-display(plottools:-cylinder([0,0,0], 2/Pi, 12, style=surface, color=yellow),
                         p, scaling=constrained,axes=none);

VectorCalculus:-ArcLength(<eq>, t=0..8*Pi);



Let's look at the first loop around the rod.
If we develop the corresponding 1/4 of the cylinder, it results a rectangle  whose sides are 4 and 12/4 = 3.
The diagonal is 5 (ask Pythagora why), so the length of the string is 4*5 = 20.


The Saturday edition of our local newspaper (Waterloo Region Record) carries, as part of The PUZZLE Corner column, a weekly puzzle "STICKELERS" by Terry Stickels. Back on December 13, 2014, the puzzle was:

What number comes next?

1   4   18   96   600   4320   ?

The solution given was the number 35280, obtained by setting k = 1 in the general term k⋅k!.

On September 5, 2015, the column contained the puzzle:

What number comes next?

2  3  3  5  10  13  39  43  172  177  ?

The proposed solution was the number 885, obtained as a10 from the recursion

where a0 =2.

As a youngster, one of my uncles delighted in teasing me with a similar question for the sequence 36, 9, 50, 55, 62, 71, 79, 18, 20. Ignoring the fact that there is a missing entry between 9 and 50, the next member of the sequence is "Bay Parkway," which is what 22nd Avenue is actually called in the Brooklyn neighborhood of my youth. These are subway stops on what was then called the West End line of the subway that went out to Stillwell Avenue in Coney Island.

Armed with the skepticism inspired by this provincial chestnut, I looked at both of these puzzles with the attitude that the "next number" could be anything I chose it to be. After all, a sequence is a mapping from the (nonnegative) integers to the reals, and unless the mapping is completely specified, the function values are not well defined.

Indeed, for the first puzzle, the polynomial f(x) interpolating the points

(0, 1), (1, 4), (2, 18), (3, 93), (4, 600), (5, 4320)

reproduces the first six members of the given sequence, and gives 18593 (not 35280) for f(7). In other words, the pattern k⋅k! is not a unique representation of the sequence, given just the first six members. The worksheet NextNumber derives the interpolating polynomial f and establishes that f(n) is an integer for every nonzero integer n.

Likewise, for the second puzzle, the polynomial g(x) interpolating the points

(1, 2) ,(2, 3) ,(3, 3) ,(4, 5) ,(5, 10) ,(6, 13) ,(7, 39) ,(8, 43), (9, 172) ,(10, 177)

reproduces the first ten members of the given sequence, and gives -7331(not 885) for g(11). Once again, the pattern proposed as the "solution" is not unique, given that the worksheet NextNumber contains both g(x) and a proof that for integer n, all values of g(n) are integers.

The upshot of these observations is that without some guarantee of uniqueness, questions like "what is the next number" are meaningless. It would be far better to pose such challenges with the words "Find a pattern for the given members of the following sequence" and warn that the function capturing that pattern might not be unique.

I leave it to the interested reader to prove or disprove the following conjecture: Interpolate the first n terms of either sequence. The interpolating polynomial p will reproduce these n terms, but for k>n, p(k) will differ from the corresponding member of the sequence determined by the stated patterns. (Results of limited numerical experiments are consistent with the truth of this conjecture.)


A duck, pursued by a fox, escapes to the center of a perfectly circular pond. The fox cannot swim, and the duck cannot take flight from the water. The fox is four times faster than the duck. Assuming the fox and duck pursue optimum strategies, is it possible for the duck to reach the edge of the pond and fly away without being eaten? If so, how?

there is an animation here

wonder if the equations of motion can be derived usingg maple and an animaton...?

Whassup homies?

tried to solve this using C.Loves program, but didn't quite get their solution...

Vars:= [PN,Name, TV, Dest,Ages,Hair,Lives]:
Name:= [Bob, Keeley, Rachael, Eilish, Amy]:
TV:=[Simpsons, Coronation, Eastenders, Desperate, Neighbours]:
Dest:= [Fra, Aus, Eng, Afr,Ita]:
Ages:= [14, 21, 46, 52, 81]:
Hair:=[afro, long, straight, curly , bald]:
Lives:= [town, city, village, farm, youth]:
Con1:= Desperate=3: Con2:= Bob=1: Con3:= NextTo(Simpsons,youth,PN): Con4:= Succ(Afr,Rachael,PN): Con5:= village=52: Con6:= Aus=straight: Con7:= Afr=Desperate: Con8:= 14=5: Con9:= Amy=Eastenders: Con10:= Ita=long: Con11:= Keeley=village: Con12:= bald=46: Con13:= Eng=4: Con14:= NextTo(Desperate,Neighbours,PN): Con15:= NextTo(Coronation,afro,PN): Con16:= NextTo(Rachael,afro,PN): Con17:= 21=youth: Con18:= Coronation=long: Con19:= 81=farm: Con20:= Fra=town: Con21:= Eilish<>straight:

read "LogicProblem.mpl"; City:= LogicProblem(Vars): with(City);


I'm trying to write a program that solves sudoku's using a Groebner basis. I introduced 81 variables x1 to x81, this is a linearisation of the sudoku board.

The space of valid sudokus is defined by:

for i=1,…,81 : Fi=(xi−1)(xi−2)⋯(xi−9) This represents the fact that all squares have integer values between 1 and 9.

for all xi and xj which...

The June edition of the IBM Ponder This website poses the following puzzle:

Assume that cars have a length of two units and that they are parked along the circumference of a circle whose length is 100 units, which is marked as 100 segments, each one exactly one unit long.

A car can park on any two adjacent free segments (i.e., it does not need any extra maneuvering space).

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