# Elementary problems in Ramsey number theory for RMO

Question 1:

Show that in any group of 6 people there will always be a subgroup of 3 people who are pairwise acquainted or a subgroup of 3 people who are pairwise strangers.

Solution 1:

Let $\{ A, B, C, D, E, F\}$ be a group of 6 people. Suppose that the people known to A are seated in room Y and the people NOT known to A are seated in room Z; A is not in either room. Then, there are necessarily at least 3 people in either room Y or in room Z; (a) Suppose B, C, D to be in room Y. Either these 3 people are mutual strangers (and so the given theorem is true), or, at least two of them (say, B and C) know each other. In the latter case, A, B and C form a group of 3 mutual acquaintances — and again, the theorem is true. (b) In (a), replace room Y by Z and interchange the notion of ‘”acquaintances” and “strangers”‘.

Question 2:

Show that in any group of 10 people there is always (a) a subgroup of 3 mutual strangers or a subgroup of 4 mutual acquaintances, and (b) a subgroup of 3 mutual acquaintances or a subgroup of 4 mutual strangers.

Solution 2:

(a) Let A be one of the ten people; the remaining 9 people can be assigned to two rooms: those who are known to A are in room Y and those who are not known to A are in room Z. Either room Y has at least 6 people or room Z has at least 6 people. For, (i) suppose room Y has at least 6 people. Then, by previous problem number 1, there is either a subgroup of 3 mutual acquaintances or a subgroup of 3 mutual strangers (thus, the theorem is true) in this room. In the former case, A and these 3 people constitute 4 mutual acquaintances (ii) Suppose room Z has at least 4 people. Either these 4 people know one another or at least 2 of them, say B and C, do not know each other. In the former case, we have a subgroup of 4 mutual acquaintances. In the latter case A, B and C constitute 3 mutual strangers.

(b) In the previous scenario, let people who are strangers become acquaintances, and let people who are acquaintances pretend they are strangers. The situation is symmetric.

Question 3:

Show that in any subgroup of 20 people there will always be either a subgroup of 4 mutual acquaintances or a subgroup of 4 mutual strangers.

Solution 3:

Suppose A is one of these 20 people. People known to A are in room Y and people not known to A are room Z. Either room Y has at least 10 people or room Z has at least 10 people. (i) If Y has at least 10 people, then by part B of problem number 2 here, there is either a subgroup of 3 mutual acquaintances or a subgroup of 4 mutual strangers — as asserted — in this room. In the former case, A and these mutual acquaintances will form a subgroup of 4 mutual acquaintances. (ii) Switch ‘”acquaintances” and “strangers”‘ in (i).

Question 4:

Let p and q be 2 positive integers. A positive integer r is said to have the (p,q) – Ramsey property, if in any group of r people either there is a subgroup of p people known to one another or there is a subgroup of q people not known to one another. {By Ramsey’s theorem, all sufficiently large integers r have the (p,q)-Ramsey property.} The smallest r with the (p,q)-Ramsey property is called the Ramsey number R(p,q). Show that (a) R(p,q) = R (q,p). (b) $R(p,1)=1$, and (c) R(p,2)=p.

Solution 4:

(a) By parts (b) of the previous three questions, we have proved part a of the proof here.

(b) This is obvious.

(c) In any group of p people, if all of them are not known to one another, there will be at least 2 people who do not know each other.

Question 5:

Prove that $R(3,3)=6$.

Solution 5:

Question 1 and its proof in this blog article imply that $R(3,3) \leq 6$.

To prove that $R(3,3)>5$, it is sufficient to consider a seating arrangement of 5 people about a round table in which each person knows only the 2 people on either side. In such a situation, there is no set of 3 mutual acquaintances and no set of 3 people not known to one another.

Question 6:

Show that if m and n are integers both greater than 2, then

$R(m,n) \leq R(m-1,n) + R(m,n-1)$.

(this recursive inequality gives a non-sharp upper bound for R(m,n)).

Solution 6:

Let $p \equiv R(m-1,n)$, $q=R(m,n-1)$ and $r \equiv p + q$. Consider a group $\{ 1,2, 3, \ldots, r\}$ of r people. Let L be the set of people known to person 1 and M be the set of people NOT known to person 1. The two sets together have $r-1$ people, so either L has at least p people or M has at least q people. (a) If L has $p \equiv R(m-1,n)$ people, then, by definition, it contains a subset of $(m-1)$ people known to one another or it contains a subset of n people unknown to one another. In the former case, the $(m-1)$ people and person 1 constitute m people known to one another.

Thus, in their case, a group of $R(m-1,n) + R(m,n-1)$ people necessarily includes m mutual acquaintances or n mutual strangers. That is, $R(m,n) \leq R(m-1,n) + R(m,n-1)$.

(b) By the usual symmetry argument, the same conclusion follows when M contains q people.

Question 7:

(Remark: A pretty property of Ramsey numbers related to combinatorics).

Show that if m and n are integers greater than 1, then $R(m,n) \leq { {m+n-2} \choose {m-1}}$ — a non-recursive upper bound.

Solution 7:

When $m=2$, or $n=2$, (i) holds with equality (see problem 4 in this blog article). The proof is by induction on $k=m+n$. As we have just seen, the result is true when $k=4$. Assume the result true for $k-1$. Then,

$R(m-1,n) \leq {{m+n-3} \choose {m-2}}$  and $R(m,n-1) \leq {{m+n-3} \choose {m-1}}$

Now, Pascal’s identity gives:

${{m+n-3} \choose {m-2}} + {{m+n-3} \choose {m-1}} = {{m+n-2}} \choose {m-1}$ so that $R(m-1,n) + R(m,n-1) \leq {{m+n-2}} \choose {m-1}$

But, from the previous question and its solution, we get $R(m,n) \leq R(m-1,n) + R(m, n-1)$

PS: As Richard Feynman, used to say, you will have to “piddle” with smallish problems as particular cases of these questions in order to get a grip over theory or formal language of this introduction.

PS: Additionally, you can refer to any basic Combinatorics text like Brualdi, or Alan Tucker or even Schaum Series outline ( V K Balakrishnan).

# How to find the number of proper divisors of an integer and other cute related questions

Question 1:

Find the number of proper divisors of 441000. (A proper divisor of a positive integer n is any divisor other than 1 and n):

Solution 1:

Any integer can be uniquely expressed as the product of powers of prime numbers (Fundamental theorem of arithmetic); thus, $441000 = (2^{3})(3^{2})(5^{3})(7^{2})$. Any divisor, proper or improper, of the given number must be of the form $(2^{a})(3^{b})(5^{c})(7^{d})$ where $0 \leq a \leq 3$, $0 \leq b \leq 2$, $0 \leq c \leq 3$, and $0 \leq d \leq 2$. In this paradigm, the exponent a can be chosen in 4 ways, b in 3 ways, c in 4 ways, d in 3 ways. So, by the product rule, the total number of proper divisors will be $(4)(3)(4)(3)-2=142$.

Question 2:

Count the proper divisors of an integer N whose prime factorization is: $N=p_{1}^{\alpha_{1}} p_{2}^{\alpha_{2}} p_{3}^{\alpha_{3}}\ldots p_{k}^{\alpha_{k}}$

Solution 2:

By using the same reasoning as in previous question, the number of proper divisors of N is $(\alpha_{1}+1)(\alpha_{2}+1)(\alpha_{3}+1)\ldots (\alpha_{k}+1)-2$, where we deduct 2 because choosing all the factors means selecting the given number itself, and choosing none of the factors means selecting the trivial divisor 1.

Question 3:

Find the number of ways of factoring 441000 into 2 factors, m and n, such that $m>1, n>1$, and the GCD of m and n is 1.

Solution 3:

Consider the set $A = {2^{3}, 3^{2}, 5^{3}, 7^{2}}$ associated with the prime factorization of 441000. It is clear that each element of A must appear in the prime factorization of m or in the prime factorization of n, but not in both. Moreover, the 2 prime factorizations must be composed exclusively of elements of A. It follows that the number of relatively prime pairs m, n is equal to the number of ways of partitioning A into 2 unordered nonempty, subsets (unordered as mn and nm mean the same factorization; recall the fundamental theorem of arithmetic).

The possible unordered partitions are the following:

$A = \{ 2^{3}\} + \{ 3^{2}, 5^{3}, 7^{2}\} = \{3^{2}\}+\{ 2^{3}, 5^{3}, 7^{2}\} = \{ 5^{3}\} + \{ 2^{3}, 3^{2}, 7^{2}\} = \{ 7^{2}\}+\{ 2^{3}, 3^{2}, 5^{3}\}$,

and $A = \{ 2^{3}, 3^{2}\} + \{ 5^{3}, 7^{2}\}=\{ 2^{3}, 5^{3}\} + \{3^{2}, 7^{2} \} = \{ 2^{3}, 7^{2}\} + \{ 3^{2}, 5^{3}\}$

Hence, the required answer is $4+3=2^{4-1}+1=7$.

Question 4:

Generalize the above problem by showing that any integer has $2^{k-1}-1$ factorizations into relatively prime pairs m, n ($m>1, n>1$).

Solution 4:

Proof by mathematical induction on k:

For $k=1$, the result holds trivially.

For $k=2$, we must prove that a set of k distinct elements, $Z = \{ a_{1}, a_{2}, a_{3}, \ldots, a_{k-1}, a_{k}\}$ has $2^{k-1}-1$ sets. Now, one partition of Z is

$Z = \{ a_{k}\} \bigcup \{ a_{1}, a_{2}, a_{3}, \ldots, a_{k-1}\} \equiv \{ a_{k}\} \bigcup W$

All the remaining partitions may be obtained by first partitioning W into two parts — which, by the induction hypothesis, can be done in $2^{k-2}-1$ ways — and then, including $a_{k}$ in one part or other — which can be done in 2 ways. By the product rule, the number of partitions of Z is therefore

$1 + (2^{k-2})(2)=2^{k-1}-1$. QED.

Remarks: Question 1 can be done by simply enumerating or breaking it into cases. But, the last generalized problem is a bit difficult without the refined concepts of set theory, as illustrated; and of course, the judicious use of mathematical induction is required in the generalized case.

Cheers,

Nalin Pithwa.

# Some number theory (and miscellaneous) coaching for RMO and INMO: tutorial (problem set) III

Continuing this series of slightly vexing questions, we present below:

1. Prove the inequality $\frac{A+a+B+b}{A+a+B+b+c+r} + \frac{B+b+C+c}{B+b+C+c+a+r} > \frac{C+c+A+a}{C+c+A+a++b+r}$, where all the variables are positive numbers.

2. A sequence of numbers: Find a sequence of numbers $x_{0}$, $x_{1}$, $x_{2}, \ldots$ whose elements are positive and such that $a_{0}=1$ and $a_{n} - a_{n+1}=a_{n+2}$ for $n=0, 1, 2, \ldots$. Show that there is only one such sequence.

3. Points in a plane: Consider several points lying in a plane. We connect each point to the nearest point by a straight line. Since we assume all distances to be different, there is no doubt as to which point is the nearest one. Prove that the resulting figure does not containing any closed polygons or intersecting segments.

4. Examination of an angle: Let $x_{1}$, $x_{2}, \ldots, x_{n}$ be positive numbers. We choose in a plane a ray OX, and we lay off it on a segment $OP_{1}=x_{1}$. Then, we draw a segment $P_{1}P_{2}=x_{2}$ perpendicular to $OP_{1}$ and next a segment $P_{2}P_{3}=x_{3}$ perpendicular to $OP_{2}$. We continue in this way up to $P_{n-1}P_{n}=x_{n}$. The right angles are directed in such a way that their left arms pass through O. We can consider the ray OX to rotate around O from the initial point through points $P_{1}$, $P_{2}, \ldots, P_{n}$ (the final position being $P_{n}$). In doing so, it sweeps out a certain angle. Prove that for given numbers $x_{i}$, this angle is smallest when the numbers $x_{i}$, that is, $x_{1} \geq x_{2} \geq \ldots x_{n}$ decrease; and the angle is largest when these numbers increase.

5. Area of a triangle: Prove, without the help of trigonometry, that in a triangle with one angle $A = 60 \deg$ the area S of the triangle is given by the formula $S = \frac{\sqrt{3}}{4}(a^{2}-(b-c)^{2})$ and if $A = 120\deg$, then $S = \frac{\sqrt{3}}{12}(a^{2}-(b-c)^{2})$.

More later, cheers, hope you all enjoy. Partial attempts also deserve some credit.

Nalin Pithwa.

# Some number theory problems: tutorial set II: RMO and INMO

1. A simplified form of Fermat’s theorem: If x, y, z, n are natural numbers, and $n \geq z$, prove that the relation $x^{n} + y^{n} = z^{n}$ does not hold.

2. Distribution of numbers: Find ten numbers $x_{1}, x_{2}, \ldots, x_{10}$ such that (a) the number $x_{1}$ is contained in the closed interval $[0,1]$ (b) the numbers $x_{1}$ and $x_{2}$ lie in different halves of the closed interval $[0,1]$ (c) the numbers $x_{1}$, $x_{2}$, $x_{3}$ lie in different thirds of the closed interval $[0,1]$ (d) the numbers $x_{1}$, $x_{2}$, $x_{3}$ and $x_{4}$ lie in different quarters of the closed interval $[0,1]$,  etc., and finally, (e) the numbers $x_{1}$, $x_{2}$, $x_{3}, \ldots, x_{10}$ lie in different tenths of the closed interval $[0,1]$

3. Is generalization of the above possible?

4. Proportions: Consider numbers A, B, C, p, q, r such that: $A:B =p$, $B:C=q$, $C:A=r$, write the proportion $A:B:C = \Box : \Box : \Box$ in such a way that in the empty squares, there will appear expressions containing p, q, r only; these expressions being obtained by cyclic permutation of one another expressions.

5. Give an elementary proof of the fact that the positive root of $x^{5} + x = 10$ is irrational.

I will give you sufficient time to try these. Later, I will post the solutions.

Cheers,

Nalin Pithwa.

# Pure plane geometry problems for RMO training or practice

Similar Figures:

Definition:

Polygons which are equiangular and have their corresponding sides proportional are called similar.

If, in addition, their corresponding sides are parallel, they are said to be similarly situated or homothetic.

Theorem 1:

If O is any fixed point and ABCD…X any polygon, and if points $A^{'}, B^{'}, C^{'}, \ldots X^{'}$ are taken on OA, OB, OC, …OX (or those lines produced either way) such that $\frac{OA^{'}}{OA} = \frac{OB^{'}}{OB} = \ldots = \frac{OX^{'}}{OX}$, then the polygons $ABCD...X$ and $A^{'}B^{'}C^{'} \ldots X^{'}$ are homothetic.

Before we state the next theorem, some background is necessary.

If O is a fixed point and P is a variable point on a fixed curve S, and if $P^{'}$ is a point on OP such that $\frac{OP^{'}}{OP} = k$, a constant, then the locus of $P^{'}$ is a curve $S^{'}$, which is said to be homothetic to S; and $P, P^{'}$ are corresponding points.

O is called the centre of similitude of the two figures.

If $P$ and $P^{'}$ lie on the same side of O, the figures are said to be directly homothetic w.r.t.O, and O is called the external centre of similitude.

If $P$ and $P^{'}$ lie on the opposite sides of O, the figures are said to be inversely homothetic w.r.t. O, and O is called the internal centre of similitude.

If we join A and B in the first case, we say that the parallel lines $AB, A^{'}B^{'}$ are drawn in the same sense, and in the second case, in opposite senses.

Theorem 2:

Let A, B be the centres of any two circles of radii a, b; AB is divided externally at O and internally at $O_{1}$ in the ratio of the radii, that is, $\frac{AO}{BO} = \frac{AO_{1}}{O_{1}B} = \frac{a}{b}$ then the circles are directly homothetic w.r.t. O and inversely homothetic w.r.t. $O_{2}$, and corresponding points lie on the extremities of parallel radii.

Now, prove the above two hard core basic geometry facts. ! 🙂 🙂 🙂

Cheers,

Nalin Pithwa.

# Some number theory training questions: RMO and INMO

Question 1:

Let us write an arbitrary natural number (for example, 2583), and then add the squares of its digits. ($2^{2}+5^{2}+8^{2}+3^{2}=102$). Next, we do the same thing to the number obtained. Namely, $1^{2}+0^{2}+2^{2}=5$. Now proceed further in the same way:

$5^{2}=25$, $2^{2}+5^{2}=29$, $2^{2}+9^{2}=85, \ldots$.

Prove that unless this procedure leads to number 1 (in which case, the number 1 will, of course, recur indefinitely), it must lead to the number 145, and the following cycle will repeat again and again:

145, 42, 20, 4, 16, 37, 58, 89.

Question 2:

Prove that the number $5^{5k+1} + 4^{5k+2} + 3^{5k}$ is divisible by 11 for every natural k.

Question 3:

The number $3^{105} + 4^{105}$ is divisible by 13, 49, 181 and 379, and is not divisible by either 5 or by 11. How can this result be confirmed?

Cheers,

Nalin Pithwa.