Miscellaneous questions: part II: tutorial practice for preRMO and RMO

Problem 1:

Let a_{1}, a_{2}, \ldots, a_{10} be ten real numbers such that each is greater than 1 and less than 55. Prove that there are three among the given numbers which form the lengths of the sides of a triangle.

Problem 2:

In a collection of 1234 persons, any two persons are mutual friends or enemies. Each person has at most 3 enemies. Prove that it is possible to divide this collection into two parts such that each person has at most 1 enemy in his subcollection.

Problem 3:

A barrel contains 2n balls numbered 1 to 2n. Choose three balls at random, one after the other, and with the balls replaced after each draw. What is the probability that the three element sequence obtained has the properties that the smallest element is odd and that only the smallest element, if any, is repeated?

That’s all, folks !!

You will need to churn a lot…!! In other words, learn to brood now…learn to think for a long time on a single hard problem …

Nalin Pithwa

Miscellaneous questions: Part I: tutorial practice for preRMO and RMO

Problem 1:

The sixty four squares of a chess board are filled with positive integers one on each in such a way that each integer is the average of the of the integers on the neighbouring squares. (Two squares are neighbours if they share a common edge or vertex. Thus, a square can have 8,5 or 3 neighbours depending on its position). Show that all sixty four entries are in fact equal.

Problem 2:

Let T be the set of all triples (a,b,c) of integers such that 1 \leq a < b < c \leq 6. For each triple (a,b,c) in T, take the product abc. Add all these products corresponding to all triples in I. Prove that the sum is divisible by 7.

Problem 3:

In a class of 25 students, there are 17 cyclists, 13 swimmers, and 8 weight lifters and no one in all the three. In a certain mathematics examination, 6 students got grades D or E. If the cyclists, swimmers and weight lifters all got grade B or C, determine the number of students who got grade A. Also, find the number of cyclists, who are swimmers.

Problem 4:

Five men A, B, C, D, E are wearing caps of black or white colour without each knowing the colour of his cap. It is known that a man wearing a black cap always speaks the truth while a man wearing a white cap always lies. If they make the following statements, find the colour of the cap worn by each of them:

A: I see three black and one white cap.
B: I see four white caps.
C: I see one black and three white caps.
D: I see four black caps.

Problem 5:

Let f be a bijective (one-one and onto) function from the set A=\{ 1,2,3,\ldots,n\} to itself. Show that there is a positive integer M>1 such that f^{M}(i)=f(i) for each i \in A. Note that f^{M} denotes the composite function f \circ f \circ f \ldots \circ f repeated M times.

Problem 6:

Show that there exists a convex hexagon in the plane such that:
a) all its interior angles are equal
b) its sides are 1,2,3,4,5,6 in some order.

Problem 7:

There are ten objects with total weights 20, each of the weights being a positive integer. Given that none of the weights exceed 10, prove that the ten objects can be divided into two groups that balance each other when placed on the pans of a balance.

Problem 8:

In each of the eight corners of a cube, write +1 or -1 arbitrarily. Then, on each of the six faces of the cube write the product of the numbers written at the four corners of that face. Add all the fourteen numbers so writtein down. Is it possible to arrange the numbers +1 and -1 at the corners initially so that this final sum is zero?

Problem 9:

Given the seven element set A = \{ a,b,c,d,e,f,g\} find a collection T of 3-element subsets of A such that each pair of elements from A occurs exactly in one of the subsets of T.

Try these !!

Nalin Pithwa

Towards Baby Analysis: Part I: INMO, IMO and CMI Entrance

\bf{Reference: \hspace{0.1in}Introductory \hspace{0.1in} Real Analysis: \hspace{0.1in} Kolmogorov \hspace{0.1in} and \hspace{0.1in} Fomin; \hspace{0.1in}Dover \hspace{0.1in }Publications}

\bf{Equivalence \hspace{0.1in} of \hspace{0.1in} Sets \hspace{0.1in} The \hspace{0.1in}Power \hspace{0.1in }of \hspace{0.1in }a \hspace{0.1in}Set}

\bf{Section 1}:

\bf{Finite \hspace{0.1in} and \hspace{0.1in} infinite \hspace{0.1in} sets}

The set of all vertices of a given polyhedron, the set of all prime numbers less than a given number, and the set of all residents of NYC (at a given time) have a certain property in common, namely, each set has a definite number of elements which can be found in principle, if not in practice. Accordingly, these sets are all said to be \it{finite}.\it{Clearly \hspace{0.1in} we \hspace{0.1in}can \hspace{0.1in} be \hspace{0.1in} sure \hspace{0.1in} that \hspace{0.1in} a \hspace{0.1in} set \hspace{0.1in}is \hspace{0.1in}finite \hspace{0.1in} without \hspace{0.1in} knowing \hspace{0.1in} the \hspace{0.1in} number \hspace{0.1in} of elements \hspace{0.1in}in \hspace{0.1in}it.}

On the other hand, the set of all positive integers, the set of all points on the line, the set of all circles in the plane, and the set of all polynomials with rational coefficients have a different property in common, namely, \it{if \hspace{0.1in } we \hspace{0.1in}remove \hspace{0.1in} one \hspace{0.1in} element \hspace{0.1in}from \hspace{0.1in}each \hspace{0.1in}set, \hspace{0.1in}then \hspace{0.1in}remove \hspace{0.1in}two \hspace{0.1in}elements, \hspace{0.1in}three \hspace{0.1in}elements, \hspace{0.1in}and \hspace{0.1in}so \hspace{0.1in}on, \hspace{0.1in}there \hspace{0.1in}will \hspace{0.1in}still \hspace{0.1in}be \hspace{0.1in}elements \hspace{0.1in}left \hspace{0.1in}in \hspace{0.1in}the \hspace{0.1in}set \hspace{0.1in}in \hspace{0.1in}each \hspace{0.1in}stage}. Accordingly, sets of these kind are called \it{infinite} sets.

Given two finite sets, we can always decide whether or not they have the same number of elements, and if not, we can always determine which set has more elements than the other. It is natural to ask whether the same is true of infinite sets. In other words, does it make sense to ask, for example, whether there are more circles in the plane than rational points on the line, or more functions defined in the interval [0,1] than lines in space? As will soon be apparent, questions of this kind can indeed be answered.

To compare two finite sets A and B, we can count the number of elements in each set and then compare the two numbers, but alternatively, we can try to establish a \it{one-\hspace{0.1in}to-\hspace{0.1in}one \hspace{0.1in}correspondence} between the elements of set A and set B, that is, a correspondence such that each element in A corresponds to one and only element in B, and vice-versa. It is clear that a one-to-one correspondence between two finite sets can be set up if and only if the two sets have the same number of elements. For example, to ascertain if or not the number of students in an assembly is the same as the number of seats in the auditorium, there is no need to count the number of students and the number of seats. We need merely observe whether or not there are empty seats or students with no place to sit down. If the students can all be seated with no empty seats left, that is, if there is a one-to-one correspondence between the set of students and the set of seats, then these two sets obviously have the same number of elements. The important point here is that the first method(counting elements) works only for finite sets, while the second method(setting up a one-to-one correspondence) works for infinite sets as well as for finite sets.

\bf{Section 2}:

\bf{Countable \hspace{0.1in} Sets}.

The simplest infinite set is the set \mathscr{Z^{+}} of all positive integers. An infinite set is called \bf{countable} if its elements can be put into one-to-one correspondence with those of \mathscr{Z^{+}}. In other words, a countable set is a set whose elements can be numbered a_{1}, a_{2}, a_{3}, \ldots a_{n}, \ldots. By an \bf{uncountable} set we mean, of course, an infinite set which is not countable.

We now give some examples of countable sets:

\bf{Example 1}:

The set \mathscr{Z} of all integers, positive, negative, or zero is countable. In fact, we can set up the following one-to-one correspondence between \mathscr{Z} and \mathscr{Z^{+}} of all positive integers: (0,1), (-1,2), (1,3), (-2,4), (2,5), and so on. More explicitly, we associate the non-negative integer $n \geq 0$ with the odd number 2n+1, and the negative integer n<0 with the even number 2|n|, that is,

n \leftrightarrow (2n+1), if n \geq 0, and n \in \mathscr{Z}
n \leftrightarrow 2|n|, if n<0, and n \in \mathscr{Z}

\bf{Example 2}:

The set of all positive even numbers is countable, as shown by the obvious correspondence n \leftrightarrow 2n.

\bf{Example 3}:

The set 2,4,8,\ldots 2^{n} is countable as shown by the obvious correspondence n \leftrightarrow 2^{n}.

\bf{Example 4}:    The set latex \mathscr{Q}$ of rational numbers is countable. To see this, we first note that every rational number \alpha can be written as a fraction \frac{p}{q}, with q>0 with a positive denominator. (Of course, p and q are integers). Call the sum |p|+q as the “height” of the rational number \alpha. For example,

\frac{0}{1}=0 is the only rational number of height zero,

\frac{-1}{1}, \frac{1}{1} are the only rational numbers of height 2,

\frac{-2}{1}, \frac{-1}{2}, \frac{1}{2}, \frac{2}{1} are the only rational numbers of height 3, and so on.

We can now arrange all rational numbers in order of increasing “height” (with the numerators increasing in each set of rational numbers of the same height). In other words, we first count the rational numbers of height 1, then those of height 2 (suitably arranged), then those of height 3(suitably arranged), and so on. In this way, we assign every rational number a unique positive integer, that is, we set up a one-to-one correspondence between the set Q of all rational numbers and the set \mathscr{Z^{+}} of all positive integers.

\it{Next \hspace{0.1in}we \hspace{0.1in} prove \hspace{0.1in}some \hspace{0.1in}elementary \hspace{0.1in}theorems \hspace{0.1in}involving \hspace{0.1in}countable \hspace{0.1in}sets}


\bf{Every \hspace{0.1in} subset \hspace{0.1in}of \hspace{0.1in}a \hspace{0.1in}countable \hspace{0.1in}set \hspace{0.1in}is \hspace{0.1in}countable}.


Let set A be countable, with elements a_{1}, a_{2}, a_{3}, \ldots, and let set B be a subset of A. Among the elements a_{1}, a_{2}, a_{3}, \ldots, let a_{n_{1}}, a_{n_{2}}, a_{n_{3}}, \ldots be those in the set B. If the set of numbers n_{1}, n_{2}, n_{3}, \ldots has a largest number, then B is finite. Otherwise, B is countable (consider the one-to-one correspondence i \leftrightarrow a_{n_{i}}). \bf{QED.}


\bf{The \hspace{0.1in}union \hspace{0.1in}of \hspace{0.1in}a \hspace{0.1in}finite \hspace{0.1in}or \hspace{0.1in}countable \hspace{0.1in}number \hspace{0.1in}of \hspace{0.1in}countable \hspace{0.1in}sets \hspace{0.1in}A_{1}, A_{2}, A_{3}, \ldots \hspace{0.1in}is \hspace{0.1in}itself \hspace{0.1in}countable.}


We can assume that no two of the sets A_{1}, A_{2}, A_{3}, \ldots have any elements in common, since otherwise we could consider the sets A_{1}, A_{2}-A_{1}, A_{3}-(A_{1}\bigcup A_{2}), \ldots, instead, which are countable by Theorem 1, and have the same union as the original sets. Suppose we write the elements of A_{1}, A_{2}, A_{3}, \ldots in the form of an infinite table

\begin{array}{ccccc} a_{11} & a_{12} & a_{13} & a_{14} &\ldots \\ a_{21} &a_{22} & a_{23} & a_{24} & \ldots \\ a_{31} & a_{32} & a_{33} & a_{34} & \ldots \\ a_{41} & a_{42} & a_{43} & a_{44} & \ldots \\ \ldots & \ldots & \ldots & \ldots & \ldots \end{array}

where the elements of the set A_{1} appear in the first row, the elements of the set A_{2} appear in the second row, and so on. We now count all the elements in the above array “diagonally”; that is, first we choose a_{11}, then a_{12}, then move downwards, diagonally to “left”, picking a_{21}, then move down vertically picking up a_{31}, then move across towards right picking up a_{22}, next pick up a_{13} and so on (a_{14}, a_{23}, a_{32}, a_{41})as per the pattern shown:

\begin{array}{cccccccc}   a_{11} & \rightarrow & a_{12} &\hspace{0.1in} & a_{13} & \rightarrow a_{14} & \ldots \\     \hspace{0.1in} & \swarrow & \hspace{0.1in} & \nearrow & \hspace{0.01in} & \swarrow & \hspace{0.1in} & \hspace{0.1in}\\     a_{21} & \hspace{0.1in} & a_{22} & \hspace{0.1in} & a_{23} \hspace{0.1in} & a_{24} & \ldots \\     \downarrow & \nearrow & \hspace{0.1in} & \swarrow & \hspace{0.1in} & \hspace{0.1in} & \hspace{0.1in} & \hspace{0.1in}\\     a_{31} & \hspace{0.1in} & a_{32} & \hspace{0.1in} & a_{33} & \hspace{0.1in} & a_{34} & \ldots \\     \hspace{0.1in} & \swarrow & \hspace{0.1in} & \hspace{0.1in} & \hspace{0.1in} & \hspace{0.1in} & \hspace{0.1in} & \hspace{0.1in}\\    a_{41} & \hspace{0.1in} & a_{42} &\hspace{0.1in} & a_{43} &\hspace{0.1in} &a_{44} &\ldots\\     \ldots & \hspace{0.1in} & \ldots & \hspace{0.1in} & \ldots & \hspace{0.1in} & \ldots & \hspace{0.1in} \end{array}

It is clear that this procedure associates a unique number to each element in each of the sets A_{1}, A_{2}, \ldots thereby establishing a one-to-one correspondence between the union of the sets A_{1}, A_{2}, \ldots and the set \mathscr{Z^{+}} of all positive integers. \bf{QED.}


\bf{Every \hspace{0.1in}infinite \hspace{0.1in}subset \hspace{0.1in}has \hspace{0.1in}a \hspace{0.1in}countable \hspace{0.1in}subset.}


Let M be an infinite set and a_{1} any element of M. Being infinite, M contains an element a_{2} distinct from a_{1}, an element a_{3} distinct from both a_{2} and a_{1}, and so on. Continuing this process, (which can never terminate due to “shortage” of elements, since M is infinite), we get a countable subset A= \{ a_{1}, a_{2}, a_{3}, \ldots, a_{n}, \ldots\} of the set M. \bf{QED.}


Theorem 3 shows that countable sets are the “smallest” infinite sets. The question of whether there exist uncountable (infinite) sets will be considered below.


\bf{Equivalence \hspace{0.1in} of \hspace{0.1in} sets}

We arrived at the notion of a countable set M by considering one-to-one correspondences between set M and the set \mathscr{Z^{+}} of all positive integers. More generally, we can consider one-to-one correspondences between any two sets M and N.


Two sets M and N are said to be \bf{equivalent} (written M \sim N) if there is a one-to-one correspondence between the elements of M and the elements of N.

The concept of equivalence is applicable both to finite and infinite sets. Two finite sets are equivalent if and only if they have the same number of elements. We can now define a countable set as a set equivalent to the set \mathscr{Z^{+}} of all positive integers. It is clear that two sets are equivalent to a third set are equivalent to each other, and in particular that any two countable sets are equivalent.


The sets of points in any two closed intervals $[a,b]$ and $[c,d]$ are equivalent; you can “see’ a one-to-one correspondence by drawing the following diagram: Step 1: draw cd as a base of a triangle. Let the third vertex of the triangle be O. Draw a line segment “ab” above the base of the triangle; where “a” lies on one side of the triangle and “b” lies on the third side of the third triangle. Note that two points p and q correspond to each other if and only if they lie on the same ray emanating from the point O in which the extensions of the line segments ac and bd intersect.


The set of all points z in the complex plane is equivalent to the set of all points z on a sphere. In fact, a one-to-one correspondence z \leftrightarrow \alpha can be established by using stereographic projection. The origin is the North Pole of the sphere.


The set of all points x in the open unit interval (0,1) is equivalent to the set of all points y on the whole real line. For example, the formula y=\frac{1}{\pi}\arctan{x}+\frac{1}{2} establishes a one-to-one correspondence between these two sets. \bf{QED}.

The last example and the examples in Section 2 show that an infinite set is sometimes equivalent to one of its proper subsets. For example, there are “as many” positive integers as integers of arbitrary sign, there are “as many” points in the interval (0,1) as on the whole real line, and so on. This fact is characteristic of all infinite sets (and can be used to define such sets) as shown by:


\bf{Every \hspace{0.1in} infinite \hspace{0.1in} set \hspace{0.1in}is \hspace{0.1in} equivalent \hspace{0.1in} to \hspace{0.1in}one \hspace{0.1in}of \hspace{0.1in}its \hspace{0.1in}proper \hspace{0.1in}subsets.}


According to Theorem 3, every infinite set M contains a countable subset. Let this subset be A=\{a_{1}, a_{2}, a_{3}, \ldots, a_{n}, \ldots \} and partition A into two countable subsets A_{1}=\{a_{1}, a_{3}, a_{5}, \ldots \} and A_{2}=\{a_{2}, a_{4}, a_{6}, \ldots \}.

Obviously, we can establish a one-to-one correspondence between the countable subsets A and A_{1} (merely let a_{n} \leftrightarrow a_{2n-1}). This correspondence can be extended to a one-to-one correspondence between the sets A \bigcup (M-A)=M and A_{1} \bigcup (M-A)=M-A_{2} by simply assigning x itself to each element x \in M-A. But M-A_{2} is a proper subset of M. \bf{QED}.

More later, to be continued,

Nalin Pithwa

Find a flaw in this proof: RMO and PRMO tutorial

What ails the following proof that all the elements of a finite set are equal?

The following is the “proof”;

All elements of a set with no elements are equal, so make the induction assumption that any set with n elements has all its elements equal. In a set with n elements, the first and the last n are equal by induction assumption. They overlap at n, so all are equal, completing the induction.

End of “proof:


Nalin Pithwa

Problem Solving approach: based on George Polya’s opinion: Useful for RMO/INMO, IITJEE maths preparation

I have prepared the following write-up based on George Polya’s classic reference mentioned below:


First. “You have to understand the problem.”

What is the unknown ? What are the data? What is the condition?

Is it possible to satisfy the condition? Is the condition sufficient to determine the unknown? Or is it insufficient? Or redundant ? Or contradictory?

Draw a figure/diagram. Introduce a suitable notation. Separate the various parts of the condition. Can you write them down?



Find the connection between the data and the unknown. You may be obliged to consider auxiliary problems if an immediate connection cannot be found. You should eventually obtain a plan for the solution.”

Have you seen it before? Or have you seen the problem in a slightly different form? Do you know a related problem? Do you know a theorem that could be useful? Look at the unknown! And try to think of a familiar problem having the same or a similar unknown. Here is a problem related to yours and solved before. Could you use it? Could you use its result? Could you use its method? Should you restate it differently? Go back to definitions.

If you cannot solve the proposed problem, try to solve some related problem. Could you imagine a more accessible related problem? A more general problem? A more special problem? An analogous problem? Could you solve a part of the problem? Keep only a part of the condition, drop the other part, how far is the unknown then determined, how can it vary? Could you derive something useful from the data? Could you think of other data appropriate to determine the unknown? Could you change the unknown of the data, or both, if necessary, so that the new unknown and the new data are nearer to each other? Did you use all the data? Did you use the whole condition? Have you taken into account all essential notions involved in the problem?


“Carry out your plan.”

Carrying out your plan of the solution, check each step. Can you clearly see that the step is correct? Can you prove that it is correct?


Examine the solution.

Can you check the result? Can you check the argument? Can you derive the result differently? Can you see it at a glance? Can you see the result, or the method, for some other problem?



How to Solve It: A New Aspect of Mathematical Method — George Polya.

Amazon India Link:


The above simple “plan” can be useful even to crack problems from a famous classic, Problem Solving Strategies, by Arthur Engel, a widely-used text for training for RMO, INMO and IITJEE Advanced Math also, perhaps. 

Reference: Problem-Solving Strategies by Arthur Engel; available on Amazon India


Concept of order in math and real world

  1. Rise and Shine algorithm: This is crazy-sounding, but quite a perfect example of the need for “order” in the real-world: when we get up in the morning, we first clean our teeth, finish all other ablutions, then go to the bathroom and first we have to remove our pyjamas/pajamas and then the shirt, and then enter the shower; we do not first enter the shower and then remove the pyjamas/shirt !! 🙂
  2. On the number line, as we go from left to right: a<x<b, that is any real number to the left of another real number is always “less than” the number to the right. (note that whereas the real numbers form an “ordered field”, the complex numbers are only “partially ordered”…we will continue this further discussion later) .
  3. Dictionary order
  4. Alphabetical order (the letters A \hspace{0.1in} B \ldots Z in English.
  5. Telephone directory order
  6. So a service like JustDial certainly uses “order” quite intensely: let us say that you want to find the telephone clinic landline number of Dr Mrs Prasad in Jayanagar 4th Block, Bengaluru : We first narrow JustDial to “Location” (Jayanagar 4th Block, Bengaluru), then narrow to “doctors/surgeons” as the case may be, and then check in alphabetic order, the name of Dr Mrs Prasad. So, we clearly see that the “concept” and “actual implementation” of order (in databases) actually speeds up so much the time to find the exact information we want.
  7. So also, in math, we have the concept of ordered pair; in Cartesian geometry, (a,b) means that the first component a \in X-axis and b \in Y-axis. This order is generalized to complex numbers in the complex plane or Argand’s diagram.
  8. There is “order” in human “relations” also: let us (x,y) represents x (as father) and y (as son). Clearly, the father is “first” and the son is “second”.
  9. So, also any “tree” has a “natural order”: seed first, then roots, then branches.


Nalin Pithwa.

Why do we need proofs? In other words, difference between a mathematician, physicist and a layman

Yes, I think it is a very nice question, which kids ask me. Why do we need proofs? Well, here is a detailed explanation (I am not mentioning the reference I use here lest it may intimidate my young enthusiastic, hard working students or readers. In other words, the explanation is not my own; I do not claim credit for this…In other words, I am just sharing what I am reading in the book…)

Here it goes:

What exactly is the difference between a mathematician, a physicist, and a layman? Let us suppose that they all start measuring the angles of hundreds of triangles of various shapes, find the sum in each case and keep a record. Suppose the layman finds that with one or two exceptions, the sum in each case comes out to be 180 degrees. He will ignore the exceptions and say “the sum of the three angles in a triangle  is 180 degrees.” A physicist will be more cautious in dealing with the exceptional cases. He will examine them more carefully. If he finds that the sum in them is somewhere between 179 degrees to 180 degrees, say, then he will attribute the deviation to experimental errors. He will then state a law: The sum of three angles of any triangle is 180 degrees. He will then watch happily as the rest of the world puts his law to test and finds that it holds good in thousands of different cases, until somebody comes up with a triangle in which the law fails miserably. The physicist now has to withdraw his law altogether or else to replace it by some other law which holds good in all cases tried. Even this new law may have to be modified at a later date. And, this will continue without end.

A mathematician will be the fussiest of all. If there is even a single exception he will refrain from saying anything. Even when millions of triangles are tried without a single exception, he will not state it as a theorem that the sum of the three angles in ANY triangle is 180 degrees. The reason is that there are infinitely many different types of triangles. To generalize from a million to infinity is as baseless to a mathematician as to generalize from one to a million. He will at the most make a conjecture and say that there is a strong evidence suggesting that the conjecture is true. But that is not the same thing as a proving a theorem. The only proof acceptable to a mathematician is the one which follows from earlier theorems by sheer logical implications (that is, statements of the form : If P, then Q). For example, such a proof follows easily from the theorem that an external angle of a triangle is the sum of the other two internal angles.

The approach taken by the layman or the physicist is known as the inductive approach whereas the mathematician’s approach is called the deductive approach. In the former, we make a few observations and generalize. In the latter, we deduce from something which is already proven. Of course, a question can be raised as to on what basis this supporting theorem is proved. The answer will be some other theorem. But then the same question can be asked about the other theorem. Eventually, a stage is reached where a certain statement cannot be proved from any other earlier proved statement(s) and must, therefore, be taken for granted to be true. Such a statement is known as an axiom or a postulate. Each branch of math has its own axioms or postulates. For examples, one of the axioms of geometry is that through two distinct points, there passes exactly one line. The whole beautiful structure of geometry is based on 5 or 6 axioms such as this one. Every theorem in plane geometry or Euclid’s Geometry can be ultimately deduced from these axioms.

PS: One of the most famous American presidents, Abraham Lincoln had read, understood and solved all of Euclid’s books (The Elements) by burning mid-night oil, night after night, to “sharpen his mental faculties”. And, of course, there is another famous story (true story) of how Albert Einstein as a very young boy got completely “addicted” to math by reading Euclid’s proof of why three medians of a triangle are concurrent…(you can Google up, of course).


Nalin Pithwa