Hint
[Fibonacci numbers Fi are defined for integers
i >= 1 as
Hint
Hint
[This is similar to the random walk problem. That is, at every step you
can make one of two moves:
either p units in the +ve direction or q units in the
-ve direction. The difference is that there is no probability
associated with each move here.]
Hint
Hint
[ Solutions of these problems are nice applications of the pigeon-hole
principle which states that
Hint
Hint
[ Idli is a white fluffy food prepared from rice. Rice grains are first ground
and water is added to it. The pulp is then steamed in a special device to
produce idlis. This food is very popular in the southern part of India.
In case you do not like to eat idlis for some reason or other, eat
whatever you like: cakes or biscuits or fish fries or whatever. Don't
change the numbers in the problem ...
When 22 in the statement of this problem is replaced by 21, a very elegant
solution for the problem exists that uses the pigeon-hole principle. The
solution can be found in Example 4.4, p 128 of
Hint [ L.E. Dickson in 1952 proved that four distinct whole squares can not
be in arithmetic progression (AP). The proof can be found in pages 435-440 of
The proof presented here is a little more involved
than Brown's. Both this proof and Brown's proof reach contradiction by
Fermat's method of infinite descent.
Note that this is not a very easy problem to solve. In my opinion,
this is the most difficult one in this page. Please let me know, if you
find a simpler solution. ]
Hint
Last updated on Nov 12 1998
The solution
2. [The Fibonacci number problem]
Given an integer n > 1, show that there exists a Fibonacci number
divisible by n.
F1 = 1
F2 = 1
Fi = Fi-1 + Fi-2 for all i >= 3
The first few Fibonacci numbers are
1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, ... ]
The solution
3. [The n lamps problem]
Suppose we have n lamps numbered 1 through n. Initially all lamps are off.
We then carry out the n-step process: in the ith step, we toggle the state
of only those lamps whose numbers are multiples of i. (Here toggling the state
of a lamp means if the lamp is on, turn it off, and if the lamp is off,
turn it on.) In other words, we carry out the following algorithm:
lamp L1, L2, ..., Ln;
L1 = L2 = ... = Ln = OFF;
for (i=1; i<=n; i++)
for (j=1; j<=n; j++)
if (i divides j) toggle(Lj);
Which of the lamps will be ON after the execution of the above process?
The solution
4. [The random walk problem]
Let n, p, q be positive integers with n > p + q.
Let x0, x1, ..., xn be integers
satisfying the following conditions:
(a) x0 = xn = 0
(b) for each integer i with 1 <= i <= n,
either xi - xi-1 = p
or xi - xi-1 = -q
Show that there exists a pair (i,j) of indices with i < j and
(i,j) not equal to (0,n) such that xi = xj.
The solution
5. [The partition problem]
Partition the set of numbers {1,2,...,n2} into n pairwise disjoint
subsets each containing n numbers such that the sum of the numbers in
each subset is the same (namely, n(n2+1)/2).
The solution
6. [The (n+1)-subset problem]
Let S be an arbitrary subset of the set {1, 2, ..., 2n}.
If S contains n+1 elements, then show that
If n pigeons are put into m pigeonholes where n > m, then there is a
hole with more than one pigeon.
There are several generalizations of this statement. Though the statement
of the pigeon-hole principle is absolutely obvious and hardly calls for any
proof (Well! If you are interested in a formal proof, go ahead. It's extremely
easy.), it has numerous interesting applications. In addition to the ones
listed in this problem, we have used this principle for solving
the summation problem. You may look at
this website
for a detailed discussion on this principle and for many applications of
it. ]
The solution
7. [All composite problem]
Show that the numbers n4 + 4n are not prime for all
integers n > 1.
The solution
8. [The problem with 132 idli's]
Suppose you want to eat 132 idlis in a total of 77 successive days such that:
Show that whatever be the way you eat the idlis, there will be a set of
successive days during which you must consume a total number of exactly
22 idlis.
C. L. Liu, `Elements of Discrete Mathematics', 2nd Edition, McGraw Hill, 1997
or at the website
http://www.seanet.com/~ksbrown/kmath389.htm.
The proof, however, cannot be straightaway applied to the case of 22 idlis. ]
The solution
9. [Four squares in AP]
There does not exist a solution in positive integers for A, B, C, D and K
satisfying
B2 = A2 + K
C2 = A2 + 2K
D2 = A2 + 3K
Dickson, L. E., History of the Theory of Numbers, Vol. 2:
Diophantine Analysis. New York: Chelsea, 1952.
Kevin S Brown claimed that he and
many other people didn't understand the proof. In particular, Dickson has
assumed a particular form of six expressions. It's apparently not
clear why all these would hold simultaneously. Brown could not give
a justification. So he devised
a proof himself.
Brown's homepage
links to a lot of other problems in number theory and in other branches
of mathematics.
The solution
Page maintained by
Abhijit Das (Barda)
Department of Computer Science and Automation
Indian Institute of Science
Bangalore 560 012
INDIA
e-mail: abhij@csa.iisc.ernet.in
URL: http://geocities.datacellar.net/SiliconValley/Lab/6024/