College Mathematics I.S.I. and C.M.I. Entrance Math Olympiad Problem Solving Marathon USA Math Olympiad

A Math Conversation – I

Inspired by the book of Precalculus written in a dialogue format by L.V.Tarasov, I also wanted to express myself in a similar fashion when I found that the process of teaching and sharing knowledge in an easy way is nothing but the output of a lucid conversation between a student and a teacher inside the person only.

Instead of presenting in paragraphs, I will express that discussion in the raw format of conversation to you.

Student: Do you think problem – solving is so important in learning math?

Teacher: Yes, surely without problem-solving, you will not at all enjoy the process of learning of mathematics. If you learn math without problem-solving is like charging the phone without the battery.

S: But why is it so important?

T: Math is nothing but layers of answers to a large number of Whys, Hows, Can We, etc. All the thoughts, questions and answers are refined into something called theorems and lemmas to make them look concise. Theory develops in this way only. Often someone gives a new way of looking into things or connecting two different fields of study, that becomes the definitions, but that was actually done to solve a new problem or question that occurred in their minds or make the things precise. Problems and Theory are the two sides of the same coin.

S: Interesting! So, a student how can we build our own new theory?

T: That’s a good question. As I told you, it is about not developing a new theory, it is about solving a new problem always. So find your own problems or existing ones and start spending time with them.

S: What if I couldn’t solve a problem?

T: That is the best part when you solve problems. You try to understand what doesn’t work. You have to understand why it doesn’t work. Then only you can engineer an alternate pathway, and that is eventually called a new theory if it is too influential. But whatever you have thought you can call it your own theory. Even if you cannot develop a full alternated pathway, something new you have thought will help you solve another problem you were pondering upon for sometime. This actually happenned with me.

S: Wow! This problem solving path seems to be quite adventurous. I never thought it this way.

T: Yes, it is. Infact human beings think and behave in this way only, when their rational and logical part of the brain work. Only they have to conscious of the steps they are taking and they have to take the opportunity to learn at every failure.

S: So, are there any problem solving tips for me?

T: I really don’t believe in tricks while learning, tricks are important for quick problem solving that is only required in the exam. But those are temporary. Once you develop your own strategies, first of all we will get to know something new from you and also you will never forget those strategies. But yes, from my experience I can help you in approaching problems.

S: Surely, what are they?

T: First of all, you need to understand the working principle of problem-solving. It works like whenever you think or see some problem, you are either blank or you have some ideas to approach based on your previous experiences and knowledge.

S: Yes, then?

T: Well, if you have some ideas, you approach along that path till you can’t proceed further. You have to find everytime where is the problem occurring and how to bypass it. This is why different people have different solutions to a problem because they create different bypasses.

S: What if I am blank?

T: You have to learn how to bring in ideas into you.

S: But how?

T: First of all, without knowledge or past experience ideas will rarely come, because that is how humans think using past data. Do you that our visual memory is the maximum memory among all the senses?

S: I felt that was true, but I knew that. But how is it related to math and problem-solving?

T: well, try to draw a picture in your mind or your copy. Try to draw a mindmap, some flowcharts to get some visual aspect of the theory or the problems. Now your visual memory comes into play automatically. That’s why, the important step when you learn something new or try to solve something, ALWAYS TRY TO DRAW DIAGRAMS AND PICTURES WHENEVER IT IS POSSIBLE.

S: That’s seems quite satisfying. I will start doing it from today only. What else?

T: Actually this is the first and important step. Also, intuition is also gained by the method of HIT AND TRIAL. When you have a problem, you play with the numbers in the problem or the expressions and often it happened to me that something interesting came out. So, also whenever you are blank, you can follow this HIT AND TRIAL methodology to start your idea engines and this helps you in recognizing patterns.

S: Hit and Trial seems exciting and adventurous too. What did you mean by pattern recognition?

T: Pattern recognition is the most important part of math. You have to observe a pattern and justify it with logic in a new problem. The first step to problem-solving is to find the pattern. Then comes the next part of logically justifying the pattern. To help you see the pattern and proceeding through you are guided by DRAWING PICTURES and HIT AND TRIAL methodologies.

S: What about logically justifying a problem?

T: This is the most difficult part and rewarding part of mathematics. Often people are not being properly trained in this domain that’s why they find it uninteresting. The method of proofs depends on the pattern you see and observe and that varies from person to person. But proving that the pattern actually holds requires some serious training. Isn’t it so beautiful, when you show that what you observe actually works? It is like painting your patterns by the brush of logic.

S: How do I develop that?

T: That is solely practice and you must enjoy that process. That process requires some knowledge and how people approached on similar ideas and patterns. Once you develop that habit, you will find it super rewarding. It is also a continuously learning path.

S: How do we learn like that that interesting way?

T: Yes, that is a really nice question and those who can impart this knowledge into you very fluidly and interestingly are good teachers. If you want to be a good teacher, you have to start practicing that too. You have to understand and question why every step works and are there any other ways to do it? Then only you will learn out of a problem, or a solution you have learned. Even if you are learning, you have to question WHY THIS WAY AND NOT THAT WAY. Develop the habit of questioning at every step of your learning process, then only you will enjoy.

S: Nice! I learnt a lot of new things today. Let me put that into action. See you later. Have a nice day.

T: Yes, you must put them into action. Yes see ya later. Have a nice day. Bye.

S: Bye.

I.S.I. and C.M.I. Entrance Informatics Olympiad Math Olympiad Problem Solving Marathon USA Math Olympiad

The 3n+1 Problem | Learn Collatz Conjecture

The 3n+1 Problem is known as Collatz Conjecture.

Consider the following operation on an arbitrary positive integer:

  • If the number is even, divide it by two.
  • If the number is odd, triple it and add one.

The conjecture is that no matter what value of the starting number, the sequence will always reach 1.

Observe that once it reaches 1, it will do like the following oscillation

1 -> 4 -> 2 -> 1-.> 4 ->2 ….

So, we see when it converges to 1, or does it?

We will investigate the problem in certain details as much as possible with occasional exercises and some computer problems ( python ).

Let’s start!

Let’s start by playing with some numbers and observe what is happening actually.

We will need this frequently. So let’s make a little piece of code to do this.

n = 12
while n > 1:
    if n % 2 == 0 :
        n = n//2
        print (n)
        n = 3*n+1
        print (n)

This will give the output :

12 6 3 10 5 16 8 4 2 1

Now depending on the input of “n” you can get different sequences.

Exercise: Please copy this code and changing the input value of “n”, play around with the sequences here. (Warning: Change it to python before implementing the code.) Eg: Do it for 7.

3n+1 Problem - sequence

If you observe that different numbers requires different time to converge and that is the unpredictable which makes the problem so hard.

There are in total two possibilities:

  • It may occur a particular value occurs repeatedly i.e. after some time the sequence starts repeating. We define the last non-repeating digit to be the Terminating Number. The total number of numbers in the sequence until the terminating number is called the Length of the Sequence. For eg: For 5 the Terminating Number is 1 and the Length of the sequence is 6.
  • We may never end the sequence. In that case, the sequence never terminates.

Exercise: Find out the length of the sequence for all the numbers from 1 to 100 by a computer program. I will provide you the code. Your job is to find a pattern among the numbers.

c = 1
for n in range(2,101):
    i = n 
    while i > 1:
        if (i % 2 == 0):
            i = i//2
            c = c + 1
            if i == 1 :
                print( "The length of the sequence of {} is {}".format(n, c) )
                c = 1
            i = 3*i+1
            c = c + 1 

Exercise: Consider the length of a sequence corresponding to a starting number “n’ as L(n). Consider numbers of the form n = (8k+4) and n+1 = (8k+5), then find the relationship between L(n) and L(n+1). Try to observe the sequence of lengths for the numbers till 100 and predict the conjecture..

Exercise: Prove the conjecture that you discovered.


8k+4 -> 2k+1 -> 6k+4 -> 3k+2

8k+5 -> 24k+ 16 -> 3k+2

Exercise: Prove that if n = 128k + 28, then L(n) = L(n+1) =L(n+2)


128k+28 -> 48k+11 -> 81k+20

128k+29 -> 48k+11 -> 81k+20

128k+30 -> 81k+20

Exercise: Try to generalize the result for general k consecutive elements. Maybe you can take help of the computer to observe the pattern. Modify the code as per your choice.

You can try to do an easier problem and get the intuition as an exercise.

Exercise: Try to understand the behavior and the terminating number of the sequence if the rule is the following:

Consider the following operation on an arbitrary positive integer:

1. If the number is even, divide it by two.
2. If the number is odd, add 1.

Exercise: Try to understand the behavior and the terminating number of the sequence if the rule is the following:
Consider the following operation on an arbitrary positive integer:

1. If the number is even, divide it by two.
2. If the number is odd, add any 3.

Exercise: Try to understand the behavior and the terminating number of the sequence if the rule is the following:
Consider the following operation on an arbitrary positive integer:

1. If the number is even, divide it by two.
2. If the number is odd, add ANY ODD NUMBER.

Eager to hear your approaches and ideas. Please mention them in the comments.

Other useful links:-

College Mathematics I.S.I. and C.M.I. Entrance Math Olympiad Problem Solving Marathon Themes in Mathematics USA Math Olympiad

The Dhaba Problem | ISI and CMI Entrance

Suppose on a highway, there is a Dhaba. Name it by Dhaba A.

You are also planning to set up a new Dhaba. Where will you set up your Dhaba?

Model this as a Mathematical Problem. This is an interesting and creative part of the BusinessoMath-man in you.

You have to assume something for Mathematical simplicity to model a real life phenomenon via math.


  • The length of the highway is 1 unit. Assume the highway is denoted by the interval [0,1].
  • The number of Persons in a road is proportional to the road length. P = c. R, where P is the Persons and R is the road length, c is an arbitrary constant.
  • The profit is modeled as the Total Number of Persons arriving in the Dhaba.
  • A person will visit a dhaba which is nearer over another which is further away.
  • When a customer is at the same distance from two shops then the customer choose randomly.
  • Your aim is to maximize your profits.

Observe that the assumptions are valid and are actually followed in real life. Just sit and think for some time placing yourself in that position.

The Explicit Mathematical Problem

distance of the dhaba
distance of the dhaba
dhaba problem distance

Profit Calculation of your Dhaba B

Now based on the lengths of the roads above and the assumptions made, we have to calculate the profit of Dhaba B.

solution of the dhaba problem

So the profit made by B = c.R , where R is the length of the road on which B has monopoly. Here, as shown R = \c.( x + \frac{|d-x|}{2}\).

We have assumed in this case that \(0 \leq x \leq d\) by the diagram.

So, the profit for the Dhaba B is c.\(\frac{d+x}{2}\) .

Hence as \(0 \leq x \leq d\), the profit is maximized if x = d.

Exercise: Show that the profit of Dhaba B as a function of x is c.(\( 1 – \frac{d+x}{2}\)) if \(d \leq x \leq 1\) .

Also, observe that in this case the profit is maximized at x = d.

Exercise: Calculate the maximum profit in both cases. Do you observe something fishy? What steps and what arguments will you give to understand the fishiness?

Please share your views in the comments section.

Eager to listen to your beautiful ideas.

Other useful links:-

College Mathematics I.S.I. and C.M.I. Entrance Math Olympiad USA Math Olympiad

The Organic Math of Origami

Did you know that there exists a whole set of seven axioms of Origami Geometry just like that of the Euclidean Geometry?

Related image
The Origami in building the solar panel maximizing the input of Solar Power and minimizing the Volume of the satellite

Instead of being very mathematically strict, today we will go through a very elegant result that arises organically from Origami.

Before that, let us travel through some basic terminologies. Be patient for a few more minutes and wait for the gem to arrive.

In case you have forgotten what Origami is, the following pictures will remove the dust from your memories.

Origami Dinosaurs
Origami Papers

Origami (from the Japanese oru, “to fold,” and kami, “paper”) is a traditional Japanese art of folding a sheet of paper, usually square, into a representation of an object such as a bird or flower.

Flat origami refers to configurations that can be pressed flat, say between the pages of a book, without adding any new folds or creases.

Non- Flat Origami
Image result for flat origami
Flat Origami

When an origami object is unfolded, the resulting diagram of folds or creases on the paper square is called a crease pattern.

We denote mountain folds by unbroken lines and valley folds by dashed
lines. A vertex of a crease pattern is a point where two or more folds intersect, and a flat vertex fold is a crease pattern with just one vertex.

The Origami and the Crease Pattern

In a crease pattern, we see two types of folds, called mountain folds and valley folds.

Related image

Now, if you get to play with your hands, you will get to discover a beautiful pattern.

Related image
The blue lines denotes the mountain folds and the red lines indicate the valley folds.

The positive difference between the dotted lines and the full lines is always 2 at a given vertex. The dotted line denotes the mountain fold and the full line denotes the valley fold.

This is encoded in the following theorem.

Maekawa’s Theorem: The difference between the number of mountain
folds and the number of valley folds in a flat vertex fold is two.

Isn’t it strange ?

Those who are familiar graph theory may think it is related to the Euler Number.

We will do the proof step by step but you will weave together the steps to understand it yourself. The proof is very easy.

Step 1:

Let n denote the number of folds that meet at the vertex, m of which
are mountain folds and v that are valley folds, so that n = m + v. (m for mountain folds and v for valley folds.)

Step 2:

Consider the cross section of a flat vertex.

Cross Section of a Flat Vertex

Step 3:

Consider the creases as shown and fold it accordingly depending on the type of fold – mountain or valley.

Folding the Cross Section of a Flat Vertex along the creases

Step 4:

Now observe that we get the following cross section. Observe that the number of sides of the formed polygon is n = m + v.

Step 5:

We will also count the angle sum of the n sided polygon in the following way. Consider the polygon formed.

We get a four sided polygon here with internal angles 0 degree and 360 degrees.
An upper view of the four sided triangle formed.

Observe that the vertex 2 is the Valley Fold and other vertices are Mountain Fold. Also the angle subtended the vertex due to valley fold is 360 degrees and that of due to the mountain fold is 0 degrees.

Therefore, the sum of the internal angles is v.360 degrees.

Step 6:

Now we also know that the sum of internal angles formed by n vertices is 180.(n-2), which is = v.360.

Hence we get by replacing n by m+v, that m – v = 2.


So simple and yet so beautiful and magical right?

But it is just the beginning!

There are lot more to discover …

Do you observe any pattern or any different symmetry about the creases or even in other geometry while playing with just paper and folding them?

We will love to hear it from you in the comments.

Do you know Cheenta is bringing out their third issue of the magazine “Reason, Debate and Story” this summer?

Do you want to write an article for us?

Email us at

College Mathematics I.S.I. and C.M.I. Entrance Math Olympiad USA Math Olympiad

Natural Geometry of Natural Numbers

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How does this sound?

The numbers 18 and 30 together looks like a chair. 

The Natural Geometry of Natural Numbers is something that is never advertised, rarely talked about. Just feel how they feel!

Let’s revise some ideas and concepts to understand the natural numbers more deeply.

We know by Unique Prime Factorization Theorem that every natural number can be uniquely represented by the product of primes.

So, a natural number is entirely known by the primes and their powers dividing it.

Also if you think carefully the entire information of a natural number is also entirely contained in the set of all of its divisors as every natural number has a unique set of divisors apart from itself.

We will discover the geometry of a natural number by adding lines between these divisors to form some shape and we call that the natural geometry corresponding to the number.

Let’s start discovering by playing a game.

Take a natural number n and all its divisors including itself.

Consider two divisors a < b of n. Now draw a line segment between a and b based on the following rules:

  • a divides b.
  • There is no divisor of n, such that a < c < b and a divides c and c divides b.

Also write the number \(\frac{b}{a}\) over the line segment joining a and b.

Let’s draw for number 6.

Number 6 has the shape like a square

Now, whatever shape we get, we call it the natural geometry of that particular number. Here we call that 6 has a natural geometry of a square or a rectangle. I prefer to call it a square because we all love symmetry.

What about all the numbers? Isn’t interesting to know the geometry of all the natural numbers?

Let’s draw for some other number say 30.

The Natural Geometry of 30 – A Cube

Observe this carefully, 30 has a complicated structure if seen in two dimensions but its natural geometrical structure is actually like a cube right?

The red numbers denote the divisors and the black numbers denote the numbers to be written on the line segment.

Beautiful right!

Have you observed something interesting?

  • The numbers on the line segments are always primes. 

Actually, it shows that to build this shape the requirement of the line segments is as important as the prime numbers to build the number.

Exercise: Prove from the rules of the game that the numbers on the line segment always correspond to prime numbers.

Did you observe this?

  • In the pictures above, the parallel lines have the same prime number on it.

Exercise: Prove that the numbers corresponding to the parallel lines always have the same prime number on it.

Actually each prime number corresponds to a different direction. If you draw it perpendicularly we get the natural geometry of the number.

Let’s observe the geometry of other numbers.

Try to draw the geometry of the number 210. It will look like the following:

Image result for Is four dimensional hyper cube graph

The natural geometry of the number 210 – Tesseract
Image result for tesseract cube
This is the three dimensional projection of the structure of the Tesseract.

Obviously, this is not the natural geometry as shown. But neither we can visualize it. The number 210 lies in four dimensions. If you try to discover this structure, you will find that it has four different directions corresponding to four different primes dividing it. Also, you will see that it is actually a four-dimensional cube, which is called a tesseract. What you see above is a two dimensional projection of the tesseract, we call it a graph.

A person acquainted with graph theory can understand that the graph of a number is always k- regular where k is the number of primes dividing the number.

Now it’s time for you to discover more about the geometry of all the numbers.

I leave some exercises to help you along the way.

Exercise: Show that the natural geometry of \(p^k\) is a long straight line consisting of k small straight lines, where p is a prime number and k is a natural number.

Exercise: Show that all the numbers of the form \(p.q\) where p and q are two distinct prime numbers always have the natural geometry of a square.

Exercise: Show that all the numbers of the form \(p.q.r\) where p, q and r are three distinct prime numbers always have the natural geometry of a cube.

Research Exercise: Find the natural geometry of the numbers of the form \(p^2.q\) where p and q are two distinct prime numbers. Also, try to generalize and predict the geometry of \(p^k.q\) where k is any natural number.

Research Exercise: Find the natural geometry of \(p^a.q^b.r^c\) where p,
q, and r are three distinct prime numbers and a,b and c are natural numbers.

Let’s end with the discussion with the geometry of {18, 30}. First let us define what I mean by it.

We define the natural geometry of two natural numbers quite naturally as a natural extension from that of a single number.

Take two natural numbers a and b. Consider the divisors of both a and b and follow the rules of the game on the set of divisors of both a and b. The shape that we get is called the natural geometry of {a, b}.

You can try it yourself and find out that the natural geometry of {18, 30} looks like the following:

Looks like a chair indeed!

Sit on this chair, grab a cup of coffee and set off to discover.

The numbers are eagerly waiting for your comments. 🙂

Please mention your observations and ideas and the proofs of the exercises in the comments section. Also think about what type of different shapes can we get from the numbers.

Also visit: Thousand Flowers Program

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Math Olympiad Themes in Mathematics USA Math Olympiad

Thousand Flowers Program: Paradigm shift in Olympiad Training

The central theme of the thousand flowers program is: connected ideas and connected problems. We will illustrate the idea using some examples.

But before we do so, let’s point out the theoretical motivation behind such a program. It is greatly borrowed from the pedagogical experiments of Rabindranath Thakur. (Reference:বিশ্বভারতী). One of his major criticisms of existing pedagogical methods is this (I won’t try to translate this):

এই শিক্ষাপ্রণালীর সকলের চেয়ে সাংঘাতিক দোষ এই যে, এতে গোড়া থেকে ধরে নেওয়া হয়েছে যে আমরা নিঃস্ব। যা-কিছু সমস্তই আমাদের বাইরে থেকে নিতে হবে—আমাদের নিজের ঘরে শিক্ষার পৈতৃক মূলধন যেন কানাকড়ি নেই। এতে কেবল যে শিক্ষা অসম্পূর্ণ থাকে তা নয়, আমাদের মনে একটা নিঃস্ব-ভাব জন্মায়। আত্মাভিমানের তাড়নায় যদি-বা মাঝে মাঝে সেই ভাবটাকে ঝেড়ে ফেলতে চেষ্টা করি তা হলেও সেটাও কেমনতরো বেসুরো রকম আস্ফালনে আত্মপ্রকাশ করে। আজকালকার দিনে এই আস্ফালনে আমাদের আন্তরিক দীনতা ঘোচে নি, কেবল সেই দীনতাটাকে হাস্যকর ও বিরক্তিকর করে তুলেছি।

Opposing the piggy bank method of education (where there is a teacher who ‘knows’ and a student who ‘does not know’) we want to seek the students’ input to solve problems with their own creativity. We are assuming that the student ‘knows’ and is ‘creative’; that he/she can do stuff. While he/she explores that inner strength, we catalyze the process with some inputs (skills, interesting problems) from time to time.

(Note that this program is run by Cheenta for young students, usually of age 7/8 to 10/11. It is designed as a launching pad for advanced Olympiad programs. Our central goal is to expose the students to rigorous creative problem-solving. This cannot be achieved by simple formula-learning. We must allow young minds to be creative. It takes years of hard work).


(each of the themes presented below may span over 6 to 8 classes (of 90 minutes). They should be punctuated by exercises and software simulations):

Primes and Algorithm of the Sieve of Eratosthenes (Connecting Mathematics and Computer Science)

  • We begin with a description of prime numbers and how they are useful to ‘build’ other numbers. Next, we try to find methods of checking if a number is prime. An elementary number theoretic investigation reveals that to check n is prime, it is sufficient to perform \( \sqrt n \) divisions.
  • A natural follow up question would be: how many primes are there between 1 and n? This leads to Sieve of Eratosthenes. It is unnatural to compute the sieve by hand. Here comes the introduction to algorithms. A simple implementation using Python does the trick.This theme is usually spread over 4 to 5 sessions. It is a fantastic introduction to elementary number theory and computer programming cum algorithms.
  • Exercises: Divisibility 1 from Mathematical Circles by Fomin // Simple algorithms

Area, Irrationals and Algebraic Identities (Connecting Algebra and Geometry)

  • We begin with simple algebraic identities such as \( (a+b)^2 = a^2 + 2ab + b^2 \). We show the geometric implementation of these identities. This immediately brings us to the discussion of ‘area’. We define the area of a unit square as 1 and follow up with a development of area formula as a product of length and width (students realize that we are actually counting the number of unit squares.
  • We immediately define rational numbers (as ratios of integers) and show that geometrically we can chop off the unit square into smaller pieces. Next, we use the area to draw pictures of more intricate identities. Finally, we show that some numbers (like \( \sqrt 2 \) cannot be expressed as ratios of integers. We prove that using parity argument and also present a geometric construction of such numbers (compass-straight edge construction)
  • Exercises: Compass -straight edge construction of rational and irrational numbers starting with integers, Geometric proof of intricate algebraic identities, parity -argument proof of irrationality.

Vectors, Angle and Motion (Connecting Mathematics and Physics)

  • We begin with a simple description of points on the plane (Cartesian).  We clearly describe that points can be visualized as static objects or as a representation of motion. For example (1, 2) can be regarded as just a point or some physical phenomena with a magnitude of \( \sqrt {1^2 + 2^2} \) and direction \( \tan^{-1} 2 \).This is a great point to introduce the notion of ‘angle’. We describe it as a measure of rotation (an isometry of the plane). We specify that angle is just a ratio of arc over radius (and geometrically why that ratio is important). Next, we go over some physical phenomena that can be described using points.
  • We draw pictures of position-time graphs. We define velocity and acceleration draw graphs of several combinations of those quantities. We solve problems of kinematics and describe the physical and mathematical aspects of it. Simulations in Geogebra or other software may aid the process.
  • Exercises: Prove that arc over radius is invariant in the rotation. Plot points and vectors and differentiate them. Plot position -time, velocity-time graphs, Kinematics problems from Irodov.

As students, we mostly want to be inspired. A closely followed (connected) second would be to get intellectually challenged. A holistic problem-oriented approach to mathematical science (mathematics + computer science + physics and part of natural sciences) may serve both purposes.

Note for teachers: It is useless to lecture for a large span of time. In fact, it is extremely important to throw clever problems every now and then, that puts all the beads of ideas together.

Note for students/parents: The world is not separated by subjects and classrooms. It is important to approach a problem in a holistic manner. This approach, in fact, provides room for creativity and experiments.

We want to create a futuristic program for our children who will grow in a world of artificial intelligence and advanced technologies. If children of today are not allowed to be creative, then they won’t be able to respond to a world where most mundane tasks will be done by machines anyways.

I will add more ideas and themes here. Let me know your opinion in the comments section or at We are still a development stage for this program.

Link to our YouTube Channel:

Math Olympiad USA Math Olympiad

AMC 10A 2016

  1. What is the value of \( \dfrac{11!-10!}{9!}\)?
    (A) 99
    (B) 100
    (C) 110
    (D) 121
    (E) 132
  2. For what value of \( x \) does \( 10^x \cdot 100^{2x} = 1000^5 \)?
    (A) 1
    (B) 2
    (C) 3
    (D) 4
    (E) 5
  3. For every dollar Ben spent on bagels, David spent 25 cents less. Ben paid $12.50 more than David. How much did they spend in the bagel store together?
    (A) $37.50
    (B) $50.00
    (C) $87.50
    (D) $90.00
    (E) $ 92.50
  4. The remainder function can be defined for all real numbers \( x \) and \( y \) with \( y \neq 0 \) by $$ rem (x,y) – x – y \left \lfloor \frac{x}{y} \right \rfloor $$
    where \( \left \lfloor \frac{x}{y} \right \rfloor \) denotes the greatest integer less than or equal to \( \frac{x}{y} \). What is the value of \( rem(\frac{3}{8}, – \frac{2}{5}) \)
    (A) \( -\frac{3}{8} \)
    (B) \( -\frac{1}{40} \)
    (C) 0
    (D) \( \frac{3}{8} \)
    (E) \( \frac{31}{40} \)
  5. A rectangular box has integer side lengths in the ratio 1:3:4. What is the volume of the box?
    (A) 48
    (B) 56
    (C) 64
    (D) 96
    (E) 144
  6. Ximena lists the whole numbers 1 through 30 once. Emilio copies Ximena’s numbers, replacing each occurrence of the digit 2 by digit 1. Ximena adds her numbers and Emilio adds his numbers. How much larger is Ximena’s sum than Emilio’s?
    (A) 13
    (B) 26
    (C) 102
    (D) 103
    (E) 110
  7. The mean, median, and mode of the 7 data values 60, 100, x, 40, 50, 200, 90 are all equal to \( x \). What is the value of \( x \) ?
    (A) 50
    (B) 60
    (C) 75
    (D) 90
    (E) 100
  8. Trickster Rabbit agrees with Foolish Fox to double Fox’s money every time Fox crosses the bridge by Rabbit’s house, as long as Fox pays 40 coins in toll to Rabbit after each crossing. The payment is made after the doubling. Fox is excited about his good fortune until he discovers that all his money is gone after crossing the bridge three times. How many coins did Fox have at the beginning?
    (A) 20
    (B) 30
    (C) 35
    (D) 40
    (E) 45
  9. A triangular array of 2016 coins has 1 coin in the first row, 2 coins in the second row, 3 coins in the third row, and so on upto \( N \) coins in the \( N \)th row. What is the sum of the digits of \( N \) ?
    (A) 6
    (B) 7
    (C) 8
    (D) 9
    (E) 10
  10. A rug is made with three different colors as shown. the areas of the three differently colored regions from an arithmetic progression. The inner rectangle is one foot wide, and each 0f two shaded region is1 foot wide on all four sides. What is the length in feet of the inner rectangle?
    (A) 1
    (B) 2
    (C) 4
    (D) 6
    (E) 8
  11. What is the area of the shaded region of the given 8 X 5 rectangle?
    (A) \( 4 \dfrac{3}{4} \)
    (B) 5
    (C) \( 5 \dfrac{1}{4} \)
    (D) \( 6 \dfrac{1}{4} \)
    (E) 8
  12. Three distinct integers are selected at random between 1 and 2016, inclusive. What should be the correct statement about the probability \( p \) that the product of the three integers is odd?
    (A)  \( p > \frac{1}{8} \)
    (B)  \( p = \frac{1}{8} \)
    (C) \( \frac{1}{8} < p < \frac{1}{3} \)
    (D) \( p = \frac{1}{3} \)
    (E) \( p < \frac{1}{3} \)
  13. Five friends sat in a movie theatre in a row containing 5 seats, numbered 1 to 5 from left to right. (The direction “left” and “right” are from the point of view of the people as they sit in the seats.) During the movie Ada went to the lobby to get some popcorn. When she returned. she found that Bea had moved two seats to the right, Ceci had moved one seat to the left, and Dee and Edie had switched seats, leaving an end seat for Ada. In which seat had Ada been sitting before she got up?
    (A) 1
    (B) 2
    (C) 3
    (D) 4
    (E) 5
  14. How many ways are there to write 2016 as the some of twos and threes, ignoring order? (For example,\( 1008 \cdot 2 + 0 \cdot 3 \) and \( 402 \cdot 2 + 404 \cdot 3 \) are two such ways.)
    (A) 236
    (B) 336
    (C) 337
    (D) 403
    (E) 672
  15. Seven cookies of radius 1 inch are cut from a circle of cookie dough, as shown. Neighboring cookies are tangent, and all except the centre cookie are tangent to the edge of the dough. The leftover scrap is reshaped to form another cookie of the same thickness. What is the radius in inches of the scrap cookie?
    amc pic 1
    (A) \( \sqrt{2}\)
    (B) 1.5
    (C) \( \sqrt{\pi}\)
    (D) \( \sqrt{2\pi}\)
    (E) \( \pi \)
  16. A triangle with vertices \( A(0,2)\), \( B(-3,2)\), and \( C(-3,0)\) is reflected about the x axis; then the image \( \triangle A’B’C’ \) is rotated counterclockwise around the origin by \( 90^{\circ} \) to produce \( \triangle A”B”C” \). What is the transformation will return \( \triangle A”B”C” \) to \( \triangle ABC \) ?
    (A) counterclockwise rotation around the origin by \( 90^{\circ} \)
    (B) clockwise rotation around the origin by \( 90^{\circ} \)
    (C) reflection about the x-axis
    (D) reflection about the line y-x
    (E) reflection about the y-axis
  17. Let \( N \) be a positive multiple of 5. One red ball and \( N \) green balls are arranged in a line in random order. Let \( P(N) \) be the probability that at least \( \frac{3}{5} \) of the green balls are on the same side of the red ball. Observe that \( P(5) \)=1 and that  \( P(N) \) approaches  \( \frac{4}{5} \) as \( N \) grows large. What is the sum of the digits of the least value of \( N \) such that \( P(N) < \frac{321}{400} \) ?
    (A) 12
    (B) 14
    (C) 16
    (D) 18
    (E) 20
  18. Each vertex of a cube is to be labeled with an integer from 1 through 8, with each integer being used once, in such a way that the sum of the four numbers on the vertices of a face is the same for each face. Arrangements that can be obtained from each other through rotations of the cube are considered to be the same. How many different arrangements are possible?
    (A) 1
    (B) 3
    (C) 6
    (D) 12
    (E) 24
  19. In rectangle ABCD, AB=6 and BC=3. Point E between B and C, and point F between E and C are such that BE=EF=FC. segment \( \bar{AE} \) and \( \bar{AF} \) intersect \( \bar{BD} \) at P and Q respectively. The ratio BP:PQ:QD can be written as r:s:t, where the greatest common factor of r,s, and t is 1. what is \( r+s+t \) ?
    (A) 7
    (B) 9
    (C) 12
    (D) 15
    (E) 20
  20. For some particular value of \( N \), when \( (a+b+c+d+1)^N \) is expanded and like terms are combined, the resulting expression contains exactly 1001 terms that include all four variables, a,b,c and d, each to some positive power. What is \( N \) ?
    (A) 9
    (B) 14
    (C) 16
    (D) 17
    (E) 19
  21. Circles with centres P,Q, and R, having radii 1,2, and 3, respectively, lie on the same side of line l and are tangent to l at P’,Q’, and R’, respectively, with Q’ between P’ and R’. The circle with center Q is ex tangent to each of the othe other two circles. What is the area of \( \triangle PQR \) ?
    (A) 0
    (B) \( \sqrt{\frac{2}{3}} \)
    (C) 1
    (D) \( \sqrt{6} – \sqrt{2}\)
    (E) \( \sqrt{\frac{3}{2}} \)
  22. For some positive integer \( n \), the number \( 110x^2 \) has 110 positive integer divisors, including 1 and the number \( 110x^2 \). How many positive integer divisors does the \( 81x^2 \) have?
    (A) 110
    (B) 191
    (C) 261
    (D) 325
    (E) 425
  23. A binary operation \( \diamondsuit \) has the properties that \(  a\diamondsuit(b\diamondsuit c)-(a\diamondsuit b) \cdot c \) and that \( a\diamondsuit a=1 \) for all nonzero real numbers a,b, and c. (Here the dot. represents the usual multiplication operation.) the solution to the equation \( 2016 \diamondsuit (6 \diamondsuit x) – 100 \) can be written as \( \frac{p}{q} \), where \( p \) and \( q \) are relatively prime positive integers. What is \(  p+q \)?
    (A) 109
    (B) 201
    (C) 301
    (D) 3049
    (E) 33601
  24. A quadrilateral is inscribed in a circle of radius \( 200\sqrt{2} \). Three of the sides of this quadrilateral have length 200. What is the length of the fourth side?
    (A)  200
    (B)  \( 200\sqrt{2} \)
    (C)   \( 200\sqrt{3} \)
    (D) \( 300\sqrt{2} \)
    (E) 500
  25. How many ordered triples \( (x,y,z) \) of positive integers satisfy lcm(x,y)=72, lcm(x,z)=600, and lcm(y,z)=900?
    (A) 15
    (B) 16
    (C) 24
    (D)  27
    (E) 64
AIME I Math Olympiad USA Math Olympiad

Geometry Problems in AIME; problems and discussions.

Let’s have a problem discussion of Geometry problems in AIME. (American Invitational Mathematics Competitions). Give them a try.

  1. In \Delta ABC, AB = 3, BC = 4 , and CA = 5. Circle \omega intersects\overline{AB} at E and B, \overline{BC} at B and D, and \overline{AC} at F and G. Given that EF=DF and \displaystyle \dfrac{DG}{EG} = \frac{3}{4} , length \displaystyle DE=\dfrac{a\sqrt{b}}{c} , where a and c are relatively prime positive integers, and b is a positive integer not divisible by the square of any prime. Find a+b+c. (2014 AIME I Problems/Problem 15)
  2. Circle C with radius 2 has diameter \overline{AB}. Circle D is internally tangent to circle C at A. Circle E is internally tangent to circle C, externally tangent to circle D, and tangent to \overline{AB}. The radius of circle D is three times the radius of circle E, and can be written in the form \sqrt{m}-n, where m and n are positive integers. Find m+n. (2014 AIME II Problems/Problem 8)
  3. A rectangle has sides of length a and 36. A hinge is installed at each vertex of the rectangle, and at the midpoint of each side of length 36. The sides of length a can be pressed toward each other keeping those two sides parallel so the rectangle becomes a convex hexagon as shown. When the figure is a hexagon with the sides of length a parallel and separated by a distance of 24, the hexagon has the same area as the original rectangle. Find a^2. (2014 AIME II Problems/Problem 3)
  4. In \triangle{ABC}, AB=10, \angle{A}=30^\circ , and \angle{C=45^\circ}. Let H, D, and M be points on the line BC such that AH\perp{BC}, \angle{BAD}=\angle{CAD}, and BM=CM. Point N is the midpoint of the segment HM, and point P is on ray AD such that PN\perp{BC}.  Then AP^2=\dfrac{m}{n}, where m and n are relatively prime positive integers. Find m+n. (2014 AIME II Problems/Problem 14)
  5. In triangle RED, measured \angle DRE=75^{\circ} and measured \angle RED=45^{\circ}. abs{RD}=1. Let M be the midpoint of segment \overline{RD}. Point C lies on side \overline{ED} such that \overline{RC} \perp \overline{EM}. Extend segment \overline{DE} through E to point A such that CA=AR. Then AE=\frac{a-\sqrt{b}}{c}, where a and c are relatively prime positive integers, and b is a positive integer. Find a+b+c. (2014 AIME II Problems/Problem 11)
  6. In triangle ABC, AB= \frac{20}{11} AC. The angle bisector of angle A intersects BC at point D, and point M is the midpoint of AD. Let P be the point of the intersection of AC and BM. The ratio of CP to PA can be expressed in the form \dfrac{m}{n}, where m and n are relatively prime positive integers. Find m+n. (2011 AIME II Problems/Problem 4)
  7. The degree measures of the angles in a convex 18-sided polygon form an increasing arithmetic sequence with integer values. Find the degree measure of the smallest angle. (2011 AIME II Problems/Problem 3)
  8. On square ABCD, point E lies on side AD and point F lies on side BC, so that BE=EF=FD=30. Find the area of the square ABCD. (2011 AIME II Problems/Problem 2)
  9. Point P lies on the diagonal AC of square ABCD with AP > CP. Let O_{1} and O_{2} be the circumcenters of triangles ABP and CDP respectively. Given that AB = 12 and {\angle O_{1}PO_{2}} = 120^{\circ}, then AP = \sqrt{a} + \sqrt{b}, where a and b are positive integers. Find a + b. (2011 AIME II Problems/Problem 13)
  10. Gary purchased a large beverage, but only drank m/n of it, where m and n are relatively prime positive integers. If he had purchased half as much and drunk twice as much, he would have wasted only 2/9 as much beverage. Find m+n. (2011 AIME II Problems/Problem 1)
  11. Let ABCDEF be a regular hexagon. Let G, H, I, J, K, and L be the midpoints of sides AB, BC, CD, DE, EF, and AF, respectively. The segments \overbar{AH} , \overbar{BI}, \overbar{CJ}, \overbar{DK}, \overbar{EL}, and \overbar{FG} bound a smaller regular hexagon. Let the ratio of the area of the smaller hexagon to the area of ABCDEF be expressed as a fraction \frac {m}{n} where m and n are relatively prime positive integers. Find m + n. (2010 AIME II Problems/Problem 9)
  12. Triangle ABC with right angle at C, \angle BAC < 45^\circ and AB = 4. Point P on \overbar{AB} is chosen such that \angle APC = 2\angle ACP and CP = 1. The ratio \frac{AP}{BP} can be represented in the form p + q\sqrt{r}, where p, q, r are positive integers and r is not divisible by the square of any prime. Find p+q+r. (2010 AIME II Problems/Problem 14)
  13. Two noncongruent integer-sided isosceles triangles have the same perimeter and the same area. The ratio of the lengths of the bases of the two triangles is 8: 7. Find the minimum possible value of their common perimeter. (2010 AIME II Problems/Problem 12)
  14. In triangle{ABC} with AB = 12, BC = 13, and AC = 15, let M be a point on \overline{AC} such that the incircles of triangle{ABM} and triangle{BCM} have equal radii. Let p and q be positive relatively prime integers such that \frac {AM}{CM} = \frac {p}{q}. Find p + q. (2010 AIME I Problems/Problem 15)
  15. Rectangle ABCD and a semicircle with diameter AB are coplanar and have nonoverlapping interiors. Let \mathcal{R} denote the region enclosed by the semicircle and the rectangle. Line ell meets the semicircle, segment AB, and segment CD at distinct points N, U, and T, respectively. Line ell divides region \mathcal{R} into two regions with areas in the ratio 1: 2. Suppose that AU = 84, AN = 126, and UB = 168. Then DA can be represented as m\sqrt {n}, where m and n are positive integers and n is not divisible by the square of any prime. Find m + n. (2010 AIME I Problems/Problem 13)
  16. Let \mathcal{R} be the region consisting of the set of points in the coordinate plane that satisfy both |8 - x| + y \le 10 and 3y - x \ge 15. When \mathcal{R} is revolved around the line whose equation is 3y – x = 15, the volume of the resulting solid is \frac {m\pi}{n\sqrt {p}}, where m, n, and p are positive integers, m and n are relatively prime, and p is not divisible by the square of any prime. Find m + n + p. (2010 AIME I Problems/Problem 11)
Math Olympiad Themes in Mathematics USA Math Olympiad

Number Theory in Math Olympiad – Beginner’s Toolbox

This article is aimed at entry level Math Olympiad (AMC and AIME in U.S. , SMO Junior in Singapore, RMO in India). We have complied some of the most useful results and tricks in elementary number theory that helps in problem solving at this level. Note that only with a lot of practice and conceptual discussions, you may make practical use of these ideas.


A proper guess or a hunch is sometimes instrumental in the solution of a problem.

  • Say x, y and n are positive integers such that xy = n^2 . It is sometimes useful to show that GCD (x, y) = 1 because in that case x and y are individually perfect squares.
  • GCD of two consecutive numbers is 1 i.e. GCD(n, n+1) =1
  • GCD (a, b) = GCD (a, a-b)
  • The possible candidates for GCD of two numbers a and b are always less than (at best equal to) a-b.
  • Only numbers that have odd number of divisors are perfect squares.
  • Sum of perfect squares is zero implies each one is 0.
  • Well Ordering Principle; Every set of natural numbers (positive integers) has a least element.
  • To check if a number is a perfect square (or to disprove it) show that the number is divisible by a particular prime but not by it’s square. Generally the easiest case is to show that the number is divisible by 3 (that is sum of the digits is divisible by 3) but not by 9.
  • Non Linear Diophantine Equations: Simplest situations are like this x^2 - y^2 = 31 . In cases like this factorize both sides and consider all possibilities (keeping in mind that x and y are integers).


  • Fermat’s Theorem: a^p \equiv a  mod p if a is any positive number and p is a prime.
  • Euler’s Totient function: \phi (n) = n \times (1 - \frac{1}{p_1})(1 - \frac{1}{p_2}) ... (1 - \frac{1}{p_k}) where n = \prod _{i=1}^k {p_i}^{r_i} is the prime factorization of n. This gives the number of numbers smaller than and co prime to n.
  • Pythagorean Triplet: If a^2 + b^2 = c^2 be a pythagorean equation (a, b and c positive integers). Then there exists positive integers u and  v such that a = u^2 - v^2 , b = 2 u v , c = u^2 + v^2 provided GCD(a, b, c) =1.
  • Number Theoretic Functions: If n = \prod _{i=1}^k {p_i}^{r_i} is the prime factorization of n then
    • Number of Divisors of n = (r_1 +1 ) (r_2 + 1) + ... + (r_k +1) = d(n)
    • Sum of the Divisors of n = \prod_{i=1}^{k} \frac{ {p_i}^{{r_i} + 1} -1 }{{p_i} -1 }
    • Product of Divisors of n = n^{d(n) / 2}
  • Highest power of a prime in n! = \sum _{k=1}^{\infty} [ {\frac{n}{p^k}} ] where [x] = greatest integer smaller than x.
  • Bezout’s Theorem: Suppose a and b be two positive integers and x, y be arbitrary integers (positive, negative or zero). Then the set ax + by is precisely the set of multiples of the gcd(a, b). More importantly there exists integers x, y (not unique) such that ax + by = d where d = gcd (a, b).
  • Congruence Notation:
    • a \equiv b mod m if m divides a – b.
    • You may raise both sides of a congruency to same power, multiply, add or subtract constants.
    • However note that ac \equiv bc mod m does not necessarily imply a \equiv b mod m. If m does not divides c then the above follows.
  • Wilson’s Theorem: (p-1)! \equiv -1 latex mod p if p is a prime. The converse also holds. That is if (n-1)! \equiv -1 mod n then n is a prime.
  • Sophie Germain Identity: a^4 + 4b^4 is factorizable
  • x^3 + y^3 + z^3 - 3xyz = (x+y+z)(x^2 +y^2 + z^2 - xy - yz - zx ) \frac {1}{2} (x + y + z) ( (x-y)^2 + ( y-z)^2  + (z-x)^2 ) $