The Fibonacci Puzzles page has been divided into two. Here is the SECOND part with
puzzles a bit harder than those of the FIRST part which you
are recommended to browse through first!
Harder Fibonacci Puzzles
All these puzzles except one (which??) have the Fibonacci numbers as their answers.
So now you have the puzzle and the answer  so what's left? Just the explanation of
why the Fibonacci numbers are the answer  that's the real puzzle!!
The Fibonacci puzzles are split into two sections:
those with fairly straightforward
and simple explanations as to why the answer is the Fibonacci numbers are on the
Easier Fibonacci Puzzles page.
CONTENTS of this Page
This page contains the second set where it is not so simple to explain why
their answers involve the Fibonacci numbers.
Does a simple explanation exist? If you find a
simple explanation please
email me (click on my name at the foot of the page) and let me know as I'd like to share
the simpler solutions on these pages.
 More coin puzzles
 Games and strategies
 Permutation puzzles
 Electrical resistance puzzles
 Jigsaw and Geometry puzzles
 More Links and References
0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987 ..More..
Pennies for your thoughts  Part 1
Here are two puzzles which are identical  but we count the solutions in two
different ways. Each involves arranging pennies (coins) in rows.
The puzzle here is that only one of these two puzzles involves the Fibonacci
number series!
The other puzzle does not but
just begins with a few of the Fibonacci numbers
and then becomes something different.
One of these puzzles is a fraud, a Fibonacci forgery.
So which is the real Fibonacci puzzle?
Arrange pennies in rows under these two conditions:
 each penny must touch the next in its row
 each penny except ones on the bottom row
touches two pennies on the row below.
There is just 1 pattern with one penny,
and 1 with two pennies
but 2 for three pennies
and 3 with four pennies as shown here:
The first condition means that there are no gaps in any row and the
second means that upper rows are smaller than lower ones.
The following arrangements are not proper combinations for 6 pennies
because the first has a gap in one row
and the second has a penny which is not on the bottom row and is
not touching two beneath it.
If there are P(n) such arrangements for n pennies,
are the P(n) numbers always Fibonacci numbers?
0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987 ..More..
Pennies for your thoughts  Part 2
This puzzle is the same as the previous one and again seems to involve the
Fibonacci numbers  or does it?
The puzzle is exactly the same, but P(n) now counts the number of arrangements
which have n pennies on the bottom row.
Here there is only 1 arrangement with 1 penny on the bottom row so P(1)=1 and
2 arrangements with two on the bottom row, P(2)=2
and 5 patterns with a bottom row of three coins P(3)=5.
What happened to 3? F(4)=3 is missing! You can check that
P(4)=13, so P(n) is clearly not the same as the Fibonacci series
since F(4)=3 and F(6)=8 are missing. This time the question is
Are the P(n) numbers the alternate Fibonacci numbers:
i : 0 1 2 3 4 5 6 7 8
Fib(i): 1 1 2 3 5 8 13 21 34...
P(n): 1 2 5 13 ?
n : 1 2 3 4 5
Which one of these two Pennies puzzles is the forgery (it does not continue
with a pattern of Fibonacci numbers after some point) and which one genuinely
always has Fibonacci numbers of arrangements?
[With thanks to Wendy Hong for brining these two puzzles to my attention.]
References
Richard K Guy, The Second Strong Law of Small Numbers in The Mathematics
Magazine, Vol 63 (1990), pages 321, Examples 45 and 46.
The first Pennies puzzle above was mentioned by F. C. Auluck in On some new types
of partitions associated with Generalized Ferrers graphs in Proceedings of
the Cambridge Philosophical Society, 47 (1951), pages 679686 (examples
45 and 46).
0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987 ..More..
Cities along a river discharge cleanedup water from sewage treatment plants.
It is more efficient to have treatment plants running at maximum capacity and
lessused ones switched off for a period.
So each city has its own treatment plant by the river and also a pipe to
its neighbouring city upstream and a pipe to the next city downstream
along the riverside.
At each city's treatment plant there are three choices:
 either process any water it may receive from one neighbouring city,
together with its
own dirty water, discharging the cleanedup water into the river;
 or send its own dirty water, plus any from its downstream neighbour,
along to the upstream
neighbouring city's treatment plant (provided that city is not already using the pipe
to send it's dirty water downstream);
 or send its own dirty water, plus any from the upstream neighbour, to the
downstream
neighbouring city's plant, if the pipe is not being used.
The choices above ensure that
 every city must have its water treated somewhere and
 at least one city must discharge the cleaned water into the river.
Let's represent a city discharging water into the river as "V" (a downwards flow), passing
water onto its neighbours as ">" (to the next city on its right) or else "<" (to the left).
When we have several cities along the river bank, we assign a symbol to each (V, < or >)
and list the cities symbols in order.
For example, two cities, A and B, can
 each treat their own sewage and each discharges clean water into
the river. So A's action is denoted V as is B's and we write "VV" ;
 or else city A can send its sewage along the pipe (to the right) to B for treatment
and discharge,
denoted ">V" ;
 or else city B can send its sewage to (the left to) A, which
treats it with its own dirty water and discharges (V) the cleaned water
into the river. So A discharges (V) and B passes water to the left (<),
and we denote this situation as "V<".
We could not have "><" since this means A sends its water to B and B sends its own to
A, so both are using the same pipe and this is not allowed. Similarly "<<" is not
possible since A's "<" means it sends its water to a nonexistent city on its
left.
So we have just 3 possible setups that fit the conditions:
Pipe Flow  A   B 
 
RIVER  ~~~~~~ 
Notation  VV 



Now suppose that we have more than two cities along the river bank:
CITY A  
CITY B  
CITY C   ... 
 
 
 
~ ~ ~ ~ ~ ~ R I V E R ~ ~ ~ ~ ~ 
 What are the eight setups possible for 3 cities?
 If S(n) is the number of setups for n cities, then S(1)=1 and we have just shown that
S(2)=3 and S(3)=8. But this does not look like the Fibonacci numbers!
What is S(4)? What is S(5)?
 What is the relationship between the Snumbers here and the Fibonacci series!?
See Fibonacci Numbers and Water Pollution Control R A Deininger in Fibonacci
Quarterly, Vol 10, No 3, 1972, pages 299300 and page 302.
0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987 ..More..
The Fibonacci numbers provide a winning strategy for playing
a game with two piles of matches (or counters or coins etc),
first described by W A Wythoff in 1906.
Players take it in turns to remove some matches (at least one)
from EITHER only one pile OR ELSE an equal number from both piles.
The players can decide how large each heap will be before the game starts
and the winner is the one who takes the LAST match.
A complete heap can be removed as your move if you like.
This is not to be recommended however, since your opponent
can do the same on the next move and so will win by taking the last match!
This leads to the idea of "safe configurations", that is,
ones from which it is possible to force a win, no matter what
your opponent does.
For further details, see
T. H. O'Beirne Puzzles and Paradoxes,
Dover press, 1965, chapter 8.
Ball, W.W.R. and Coxeter, H.S.M.
Mathematical Recreations and Essays,
13th edition, Dover Publications, 1987. A great classic with plenty to
keep you amused and enthused on Maths  definitely worth buying! (You can
order it online via the titlelink.)
0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987 ..More..
Nonneighbour Groups
How often have the list of names in your class been read out in alphabetical order,
or you have been asked to line up in alphabetical order for a firepractice or
when the results of a test are given out?
The trouble with this is that you are always next to the same one or two people
that are on either side of you in the alphabetical order  your
alphabetical neighbours.
You will have got to know them quite well over the course of a year, so
this puzzle is about meeting other people who are not your
alphabetical neighbours.
Suppose that part of the class is needed for a particular task or game. Let's also
say that the group should contain no alphabetical neighbours in it, so it
gives everyone in the group a chance to team up with new people.
In how many ways can you choose such a group from a class of N students?
For instance, if there are 3 people in the class, let's label them according to their
position when in the alphabetical order, so they are 1,
2 and 3.
The puzzle is to select a group from the class
with no pair of successive numbers (positions) in the group.
So if 1 is in the group, then 2
cannot be and 3 may be or not; so we have the groups:
{1} and
{1,3}
If 2 is in the group then, since both
1 and 3 are
2's alphabetical neighbours, then
that group will consist of 2 alone!
{2}
If 3 is in the group then 2
cannot be and 1 may be.
But remember that the group with 3 and
1 in it has already been included above!
So we have the following possible new groups with 3 in:
{3}
All the possible groups of nonneighbours are:
{1,3}
{1} {2}
{3}
{}
Did you notice that the group {} with nobody in it is a nonneighbour
group too? So from a class of 3 people,
there are 5 ways to pick a group consisting solely of nonneighbours.
How many are there in a class of size 4? or 5? or 6? Why?
On The Number of Fibonacci Partitions of a Set Helmut Prodinger
Fibonacci Quarterly Vol 19, 1981, pages 463  465 generlizes this problem.
That's odd!
Aphrodite, Rodrigo and Rodney, who love to play with coloured rods, are now joined by a new friend, Rodderick.
The rods are of different lengths (whole numbers), all the rods of the same length
having the same colour and there are lots of each colour and all the sizes you could want.
In the 2 puzzles of the first Puzzle page, in
No One! all the rods of length 1 were taken by Rodrigo
and in Ones and Twos the only rods Aphrodite had were
of length 1 and 2 as Rodney had the rest.
This time Rodderick has all the rods
of odd lengths: 1, 3,
5, 7, etc.
The puzzle this time is to help Rodderick count the number of different coloured
patterns (arrangements) of rods in a line for a given total length.
E.g. there are 3 ways to make a line of length 4:
4 is 1111,
13 and 31
The last two, 13 and 31
are different since the order of the colours is different.
Steven and Todd
This puzzle is about collections of integers with a restriction on the maximum size.
Steven and Todd are playing a game. They each take turns to name
a number bigger than the previous one, Steven always choosing an
even number (2,4,6,...) and
Todd only an odd (1,3,5,...).
Either of them can end the game at any time but, to prevent it going on for ever,
they agree on a maximum number for each game.
How many different number games are there for a given maximum
when Todd starts?
E.g. if the maximum is 3, we have the following lists:
 1
 1 2
 3
 1 2 3
Don't forget the game when Todd starts but ends it straight away: the list
of numbers played is empty!
So there are 5 lists when Todd starts and they do not
play beyond 3.
If they didn't go beyond 2, then only the top two are allowed,
together with the empty list making 3 lists.
What happens if they increase the maximum number to 4?
All the above lists still apply, but you can add
some more that will include 4.
How many extra lists will there be?
Extend the lists some more by adding those with 5 and so on,
noting how many there are for each choice of maximum.
Now try the same puzzle but with Steven always starting:
How many different number games are there for a given maximum
when Steven starts?
Finally put all the results together:
How many different number games are there for a given maximum
when either of them can start?
Two challenging problems in Combinatorics
section 17 of
Mathematical Chestnuts from Around The World,
R Honsberger, Mathematical Association of America (2001) where this problem is called
Terquem's Problem after Olry Terquem who solved it
in 1839.
0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987 ..More..
Series and Parallel
If we have two electrical resistances of R ohms and S ohms in series:
then the combined resistance is just R+S ohms.
You'll remember that if we have 2 resistances R and S in parallel:
then the combined resistance, T, is given by
1 1 1
 =  + 
T R S
Rungs
Suppose we extend the pattern of parallel resistors into longer and longer ladders,
by putting a 1 ohm resistor between two wires and then keep adding single ohm
resistors in parallel. What is the total resistance?
...
In the diagram above, the 2 resistor ladder has two 1ohm resistors in parallel
so their combined resistance R is given by the equation:
1/R = 1/1 + 1/1 = 2 so R=1/2
For the 3 resistor ladder, we have combined the 2 resistor ladder with
another resistor of 1ohm, in
parallel, so the combined resistance S here is
1/S = 1/(1/2) + 1/1 = 2+1 = 3 so S=1/3
Try computing the overall resistance for yourself for 4 resistors, then with 5 and 6.
What pattern are you getting for the combined resistance?
Can you prove that your pattern always holds?
Rungs and One Side
Now try it with the following pattern of resistances, where one of the legs of the
ladder also has resistance of 1ohm and we alternately add a resistor on a side
leg and then on a rung:
The first ladder has a single resistor so is 1 ohm.
The second ladder has two resistors in series, so the combined resistance is 2.
The third ladder has a 1 ohm resistor in parallel with the second ladder (2 ohms),
so the combined resistance S of 1 ohm and 2 ohms in parallel is
1/S = 1/1 + 1/2 = 3/2 ie S=2/3
Similarly, the next ladder has a 1 ohm resistor in series with the previous ladder,
so its total resistance is 1+2/3=5/3.
What about the next two ladders? What is the general pattern now?
Again, can you prove that your pattern will always hold?
One Side only
Try making a ladder where the only resistances are DOWN ONE SIDE and there is no
resistance on the "rungs". What pattern do you get now?
Have you the Capacity for another puzzle?
Replace the resistors with capacitors in the Rungs and One Side puzzle.
What pattern do you get now?
[Suggested by Bhushit Joshipura.]
References on the Resistance Ladders

The Golden Ratio in an Electrical Network, J Wlodarski
in The Fibonacci Quarterly
Vol 9 (1971) pages 188 and pg 194.


Generalization of Modified MorganVoyce Polynomials,
Fibonacci Quarterly
Vol 38 No 1, 2000, pages 816.
 An advanced mathematical article dealing with resistors, capacitors and inductors.
0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987 ..More..
A Fibonacci Jigsaw puzzle
or How to Prove 64=65
The 8by8 blue square in the diagram here can be cut up into 4 pieces that, when
rearranged, make the red 5by13 rectangle. But the blue square contains
8x8=64 little squares whereas the red rectangle contains
5x13=65. Where has
the extra square come from?
This puzzle can be repeated with other consecutive Fibonacci numbers,
replace 5, 8 and 13 by 8, 13 and 21 or by 3, 5 and 8
If you look at the "8, 13, 21" jigsaw, the square is 13x13=169 but this time
the rectangle is 8x21=168 so we have lost a square this time!
Sometimes there is a square extra, sometimes a square goes missing.
Not convinced? Try this demonstration
Try cutting out the pieces as shown and rearranging them yourself if you are
not sure the puzzle "works".
It works even better as a class demonstration using an overhead projector:
photocopy the square with its grid lines onto an overhead projector transparency,
cut out the shapes and show them as a square on the screen, then
rearrange it into the rectangle, carefully lining up the grid lines to "show I'm
not cheating"!
But what is the explanation?
Hints:
 What is the formula behind these puzzles?
For any three consecutive Fibonacci numbers: F(n1), F(n) and F(n+1),
it relates F(n)^{2} to F(n1)F(n+1);
what is it?
Perhaps you can try to prove it is always true.
 Now look carefully at
one of the jigsaw puzzles. Is it really what it seems? Try taking
a different angle on the problem  perhaps looking at it from a
tangent .
Edward Wakeling in
Rediscovered Lewis Carroll Puzzles
Dover, 1995, says that this
puzzle was found in Lewis Carroll's papers (the original is now kept at Princeton University)
and that this puzzle goes back to Schlomilch, 1868.
Martin Gardner's
Mathematics, Magic and Mystery
a 1956 Dover book, is a book with magic tricks and how the mathematics behind them makes them work.
It has two chapters on such Geometrical Vanishes and has a full explanation
of this and other puzzles. He also traces its origins back to Sam Loyd (senior)
who presented it to the American Chess congress (using an 8by8 chessboard) in 1858,
ten years before Wakeling's reference to Schlomilch in the reference above.
However this also appears not to be the earliest reference...
David Wells in
The Penguin Book of Curious and
Interesting Puzzles (Penguin, 1997) in Puzzle 143 traces its origin back to
William Hooper in Rational Recreations of 1774.
How to Prove 64=63!!
The blue jigsaw of area 64 little squares,
when rearranged into the red positions with 65 little squares,
had seemingly gained a square.
Here is another arrangement. This time the blue puzzle's pieces have been rearranged
as shown here in green and now it loses a square 
there are two 5by6 rectangles + 3 squares in a row joining
them, making a total area of 63!
So what's happened this time???
The second version comes from
Henry E Dudeney's 536 Puzzles and Curious Problems
(which has been edited by Martin Gardner) 1967, Souvenir Press;
Problems 352 and 353 and their answers
Martin Gardner's Mathematics, Magic and Mystery
a 1956 Dover book (mentioned in the first version of this puzzle) says that
Sam Loyd junior (who adopted his father's name and continued his father's puzzle
columns) was the first to discover this new reducedsquare version. This book
has a good explanation of how the two puzzles work and that the Fibonacci numbers
produce other sizes of puzzle with identical variations of an additional and missing
single square. He shows how other generalized Fibonacci sequences
(i.e. starting with two other numbers rather than 0 and 1) can be used to devise variations
where any number of squares can be made to appear and disappear, together with many other
kinds of geometrical dissection puzzles. If you like the puzzles on these two Web pages,
you'll enjoy this book too with number, handkerchief and card puzzles based on mathematics.
Roy Nauw of Kloetinge, the Netherlands found another Fibonacci Puzzle. His
lecturer, Floor van Lamoen, mentioned it on the
Geometry
Puzzles newsgroup (archived at
Math Forum) and it is copied here with Roy's
permission (and my thanks to them both).
It is made up of 4 pieces,
 a smaller green triangle with height 2
and base 5 ;
 a larger red triangle with height
3 and base 8;
 an orange Lshape of height 2 and width 5;
 a blue Lshape of the same width and height but a different shape.
The two Lshaped pieces fit together to make a 3by5 rectangle.
They can all be arranged into a 13by5 triangle as shown here.
Rearranging the 4 pieces shows the triangle has a square missing!
Where does the hole come from?
What's the answer this time and how is it connected with the Fibonacci Numbers?
The puzzle will work with a green triangle height 1 base 3 and
a red triangle height 2 base 5, and two straight
pieces (1by3) that make up a 2by3 rectangle. Rearranging them this time makes the
small rectangle 1 square smaller this time so the two straight pieces have to
overlap.
Similarly, using triangles of height 3 base 8 and height 5 base 13 the rectangle again
loses one square.
 small green triangle 
large red triangle 
rectangle green width
red height
height x base = Area 
rectangle red width
green height
height x base = Area 
Rectangle Area Difference 
 height  base 
height  base 
smaller puzzle 
1  3 
2  5 
2 x 3 = 6 
1 x 5 = 5  1 
puzzle above 
2  5 
3  8 
3 x 5 = 15 
2 x 8 = 16  +1 
larger puzzle 
3  8 
5  13 
5 x 8 = 40 
3 x 13 = 39  1 
larger puzzle 
5  13 
8  21 
8 x 13 = 104 
5 x 21 = 105  +1 
0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987 ..More..
More Links and References
The
Amazing Mathematical Object Factory has an interesting section on
Fibonacci Numbers
which contains explanations for some of the puzzles on this page and the
relationships between them.
© 19962006 Dr Ron Knott
updated 3 September 2005