Knot theory is the study of mathematical objects called knots that behave like 3D or 2D closed circular paths. Knot theory has many fascinating problems that can be easy to state, but that can become relatively complex to solve. Try looking at this example: it is a knot on a necklace that I found a few days ago.
Can you turn this knot into an “unknot”, or a circle, without cutting or puncturing the knot in any way?
The Tower of Hanoi is a popular puzzle invented by E. Lucas in 1883. The idea is that there are n disks arranged from largest to smallest on one rod together with two empty rods. What is the smallest number of moves that must be made in order to transfer the disks completely from one rod to an adjacent one if:
1) Only one disk may be moved at a time.
2) A large disk may not rest on top of a smaller one.
Hint: Use induction. First, try n = 2, then n = 3, etc and determine the pattern.
Challenge: What is the most efficient algorithm for moving these disks?
Given a regular n-gon P1P2…Pn. Let point O be the center of this polygon. If we take arbitrary point P in the plane, prove that |PP1|2+|PP2|2+|PP3|2+…+|PPn|2 = n|PO|2+n.
Hint: Complex numbers make this problem easier!
Here is a really interesting, “messy” math puzzle problem:
Say you have 16 two-sided puzzle pieces that fit together into a circle. Each puzzle piece has either two “in” knobs or two “out” knobs. Eight of these 16 pieces have two “in” knobs and the remaining pieces have two “out” knobs. See picture.
In how many ways can this puzzle be made?
This beautiful math challenge is by Justin Solonynka.
A little late for this year’s AMC, but definitely a problem worth looking at! This problem looks difficult at first but inductive thinking makes a very messy problem slightly less intimidating.
The product 
, where the second factor has 
digits, is an integer whose digits have a sum of 
. What is ?
I haven’t put up a contest problem in a while but this one is just fantastic. This is one of the prettiest problems I have seen lately; although it looks very messy at first, it has a clean solution. This comes from 1986 NYSML power round.
Let F(r,c) be defined for all nonnegative integers r,c>=0, according to the following rules:
1) F(r,0) = 1 for r >=0
2) F(r,c) = 0 when c>r
3) If F(r,c) = F(r,c+1), then F(r+1,c+1) = 0
4) If F(r,c) not equal F(r, c+1), then F(r+1,c+1) = 1
I a)Complete the entries on the printed table for 0<= r,c <=10.
b)Compute the value of each of the following: F(1986,1), F(1986,2), F(1986,1986), F(1986,993)
II a) Prove: If 0 <= c <= r, then F(r,c) = F(r,r-c).
b) Prove: If r = 2m+1 is odd then F(r,0) + F(r,2) +…+F(r,2m) = F(r,1) + F(r,3) +…+F(r,2m+1).
c) Let S_r be the sum of the entries in row r. Form and prove a conjecture about the value of S_r.
III a) Compute the sum of the entries in row r, where r = 2^100+1
b) Compute all values of r where 7<=r<=63, for which S_r = 8
c) Compute S_1934.
Solutions to Proofs:
Prove that between any two real numbers there exists a rational number. Let a,b ∈ R such that a < b. Subtracting a from both sides gives 0 < (b−a). By the Archimedian Principle, we know that for any real numbers x and y there exists a natural number n such that n*x > y. Rearranging this, we know that x/y>1/n for any x and y. Let b−a=x/y. Thus we know that there exists some positive natural number n such that 0 < 1/n< b − a. Rearranging n this inequality, we get a < b − 1/n.
Furthermore, let m be a natural number such that m/n<b and b <(m+1)/n. Therefore a<b−1/n<m/n<b. Therefore,there exists m/n such that a<m/n<b. Thus, there exists a rational number between any two real numbers.
Proof 2: As an extreme case let us assume that we have countably many disjoint sets that are each countable.
We list our countable sets as A_1, A_2, A_3…, A_n,… for all n in natural numbers. The elements can be denoted as a_{1,1}, a_{1,2}, a_{1,3}…., a_{1,n},…, a_{2,1} , a_{2,2}…, a_{2,n}…, a_{j, 1}, …a_{j,n}…. For every A_j there exists f_j : A_j –>N. Thus this forms a matrix that is in 1:1 correspondence with NxN. This 1:1 correspondence is f(a_{i,j}) = (i,j). To show that the union of countably many countable sets is countable, we must show that there is a 1:1 correspondence between NxN and N. It is sufficient to show that there exists a 1:1 function from NxN–>N to show that NxN is countable. Let this 1:1 function be g((i,j))=2^i *3^j. This function is clearly 1:1. Thus we have established that NxN is countable. Therefore, the union of countably many countable sets is also countable.
Here are two great items to prove about our number systems (they may seem intuitive at first, but rigorously proving them can be challenging). See if you can answer them.
1) Prove that between any two real numbers there exists a rational number. Similarly, prove that between any two real numbers there exists an irrational number
Hint: Use Archimedean property.
2) Prove that the union of a finite number of countable sets is countable.
Hey guys,
Check out the new Android app that I wrote, “Codester: A Brainteaser Puzzle”. It’s lots of fun and it teaches programming concepts!
Find it here: https://play.google.com/store/apps/details?id=com.codesterapp&hl=en or just type in “Codester” in the Play Store search.
It works on any Android device, tablet, or smartphone. Let me know what you think, thanks.
Another interesting problem!
Suppose you have a quadrilateral ABCD where AB is parallel to CD and such that the length of DA = AB = BC and BD = DC.
What are the angles of this quadrilateral?
Hint: Obviously, a trigonometry solution is possible with some work. But there’s another way, without writing down a single number! Keep in mind your shapes and rules about certain, specific edges being equal to one another as in this problem.
Mess or Math? 
Written
on April 1, 2014