# miscellaneous

# Number theory: let’s learn it the Nash way !

Reference: A Beautiful Mind by Sylvia Nasar.

*Comment: This is approach is quite similar to what Prof. Joseph Silverman explains in his text, “A Friendly Introduction to Number Theory.”*

Peter Sarnak, a brash thirty-five-year-old number theorist whose primary interest is the Riemann Hypothesis, joined the Princeton faculty in the fall of 1990. He had just given a seminar. The tall, thin, white-haired man who had been sitting in the back asked for a copy of Sarnak’s paper after the crowd had dispersed.

Sarnak, who had been a student of Paul Cohen’s at Stanford, knew Nash by reputation as well as by sight, naturally. Having been told many times Nash was completely mad, he wanted to be kind. He promised to send Nash the paper. A few days later, at tea-time, Nash approached him again. He had a few questions, he said, avoiding looking Sarnak in the face. At first, Sarnak just listened politely. But within a few minutes, Sarnak found himself having to concentrate quite hard. Later, as he turned the conversation over in his mind, he felt rather astonished. Nash had spotted a real problem in one of Sarnak’s arguments. What’s more, he also suggested a way around it. “The way he views things is very different from other people,” Sarnak said later. ‘He comes up with instant insights I don’t know I would ever get to. Very, very outstanding insights. Very unusual insights.”

They talked from time to time. After each conversation, Nash would disappear for a few days and then return with a sheaf of computer printouts. Nash was obviously very, very good with the computer. He would think up some miniature problem, usually very ingeniously, and then play with it. If something worked on a small scale, in his head, Sarnak realized, Nash would go to the computer to try to find out if it was “also true the next few hundred thousand times.”

{What really bowled Sarnak over, though, was that Nash seemed perfectly rational, a far cry from the supposedly demented man he had heard other mathematicians describe. Sarnak was more than a little outraged. Here was this giant and he had been all but forgotten by the mathematics profession. And the justification for the neglect was obviously no longer valid, if it had ever been.}

Cheers,

Nalin Pithwa

PS: For RMO and INMO (of Homi Bhabha Science Foundation/TIFR), it helps a lot to use the following: (it can be used with the above mentioned text of Joseph Silverman also): TI nSpire CAS CX graphing calculator.

# Solutions to “next number in sequence”: preRMO, pRMO and RMO

What is the next number in sequence?

A) 15, 20, 20, 6, 6, 19, 19, 5, 14, 20, 5, ?

Solution to A:

Ans is 20. The sequence is the position in the letter of the alphabet of the first letter in the numbers 1 to 12, when given in full. e.g. ONE: O=15.

B) 1, 8, 11, 18, 80, ?

Ans is 81. The sequence comprises whole numbers beginning with a vowel.

C) 1, 2, 4, 14, 21, 22, 24, 31, ?

Ans is 32, The sequence comprises whole numbers containing the letter O.

D) 4, 1, 3, 1, 2, 4, 3, ?

Ans. is 2. The sequence is as follows: there is one number between the two I’s, two numbers between the two 2’s, three numbers between the two 3’s and four numbers between the two 4’s.

E) 1, 2, 4, 7, 28, 33, 198, ?

Ans is 205.

F) 17, 8, 16, 23, 28, 38, 49, 62, ?

Answer is 70. Sum of digits in all previous numbers in the sequence.

G) 27, 216, 279, 300, ?

Ans is 307. Difference divided by 3 and added to the last number.

H) 9,7,17,79,545, ?

Answer is 4895. Each number is multiplied by its rank in the sequence, and the next number is subtracted.

I) 2,3,10,12,13, 20,?

Answer is 21. They all begin with the letter T.

J) 34, 58, 56, 60, 42, ?

Answer is 52. The numbers are the totals of the letters in the words ONE, TWO, THREE, FOUR, FIVE, SIX when A=1, B=2, C=3, etc.

Regards,

Nalin Pithwa.

# Age related questions: pRMO, preRMO training: and their solutions

**Age old questions:**

Dave is younger than Fred and older than George.

Alan is younger than Ian and older than Colin.

Ian is younger than George and older than John.

John is younger than Colin and older than Edward.

Fred is younger than Barry and older than Harry.

Harry is older than Dave.

**Who is the youngest?**

Answer: You have to try to create an “ascending” or “a descending” sequence and try to “fill in the gaps” :— answer is Edward.

Regards,

Nalin Pithwa.

# Next number in sequence: PreRMO, pRMO, RMO

What is the next number in the sequence?

a) 15, 20, 20, 6, 6, 19, 19, 5, 14, 20, 5, ?

b) 1,8,11, 18, 80, ?

c) 1,2,4,14,21,22,24,31,?

d) 4,1,3,1,2,4,3,?

e) 1,2,4,7,28,33,198,?

f) 17,8,16,23, 28, 38, 49, 62, ?

g) 27, 216, 279, 300, ?

h) 9,7,17,79,545,?

i) 2,3,10,12,13,20,?

j) 34, 58, 56, 60, 42, ?

Regards,

Nalin Pithwa.

# Miscellaneous Questions: part I: solution to chess problem by my student RI

Some blogs away I had posted several interesting, non-trivial, yet do-able-with-some-effort problems for preRMO and RMO.

A student of mine, RI has submitted the following beautiful solution to the chess problem. I am reproducing the question for convenience of the readers:

Question:

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 integers on the neighbouring squares. (Two squares are neighbours if they share a common edge on vertex. Thus, a square can have 8, 5 or 3 neighbours depending on its position.) Show that all the sixty four entries are in fact equal.

Answer (by RI):

Let us denote the set of all integers on the chess board by S (assume they are distinct). [Now, we can use the Well-ordering principle: every non-empty set of non-negative integers contains a least element. That is, every non-empty set S of non-negative integers contains an element a in S such that for all elements b of S}. So, also let “a” be the least element of set S here. As it is the average of the neighbouring elements, it can’t be less than each of them. But it can’t be greater than all of them also. So, all the elements of S are equal.

QED.

Three cheers for RI ðŸ™‚ ðŸ™‚ ðŸ™‚

Regards,

Nalin Pithwa

# Miscellaneous questions: part II: solutions to tutorial practice for preRMO and RMO

Refer the blog questions a few days before:

**Question 1:**

Let 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.

**Answer 1:**

Without loss of generality, we may take …..call this relation (i).

Let, if possible, no three of the given numbers be the lengths of the sides of a triangle. (That is, no three satisfy the triangle inequality. Note that when we say three numbers a, b and c satisfy the triangle inequality —- it means all the following three inequalities have to hold simultaneously: , and ). We will consider triplets and . As these numbers do not form the lengths of the sides of a triangle, the sum of the smallest two numbers should not exceed the largest number, that is, . Hence, we get the following set of inequalities:

gives giving .

gives giving

gives giving

gives giving

gives giving

gives giving

gives giving

gives giving

contradicting the basic hypothesis. Hence, there exists three numbers among the given numbers which form the lengths of the sides of a triangle.

**Question 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 the collection into two parts such that each person has at most 1 enemy in his sub-collection.

**Answer 2:**

Let C denote the collection of given 1234 persons. Let be a partition of C. Let denote the total number of enemy pairs in . Let denote the total number of enemy pairs in .

Let denote the total number of enemy pairs corresponding to the partition of C. Note is an integer greater than or equal to zero. Hence, by Well-Ordering Principle, there exists a partition having the least value of .

Claim: This is “the” required partition.

Proof: If not, without loss of generality, suppose there is a person P in having at least 2 enemies in . Construct a new partition of C as follows: and . Now, . Hence, contradicting the minimality of . QED.

**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?

**Answer 3:**

The total number of possible outcomes is . To find the total number of favourable outcomes we proceed as follows:

Let a be any odd integer such that and let us count the sequences having a as least element.

(i) There is only one sequence with a repeated thrice.

(ii) There are sequences of the form with . For each such sequence there are three distinct permutations possible. Hence, there are in all sequences with a repeated twice.

iii) When , for values of a satisfying , sequences of the form with are possible and the number of such sequences is . For each such sequence, there are six distinct permutations possible. Hence, there are sequences in this case.

Hence, for odd values of a between 1 and , the total counts of possibilities , , in the above cases are respectively.

.

Hence, the total number A of favourable outcomes is . Hence, the required probability is . QED>

Cheers,

Nalin Pithwa

# Miscellaneous questions: part I : solutions: tutorial practice preRMO and RMO

The following questions were presented in an earlier blog (the questions are reproduced here) along with solutions. Please compare your attempts/partial attempts too are to be compared…that is the way to learn:

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 integers in 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 the sixty four squares are in fact equal.

Solution 1:

Consider the smallest value among the 64 entries on the board. Since it is the average of the surrounding numbers, all those numbers must be equal to this number as it is the smallest. This gives some more squares with the smallest value. Continue in this way till all the squares are covered.

Problem 2:

Let T be the set of all triples of integers such that . For each triple in T, take the product abc. Add all these products corresponding to all triples in T. Prove that the sum is divisible by 7.

Solution 2:

For every triple in T, the triple is in T and these two are distinct as . Pairing off with for each , 7 divides .

Problem 3:

In a class of 25 students, there are 17 cyclists, 13 swimmers, and 8 weight lifters and no one is all the three. In a certain math 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.

Solution 3:

Let S denote the set of all 25 students in the class, X the set of swimmers in S, Y the set of weight lifters in S, and Z the set of all cyclists. Since students in all get grades B and C, and six students get grades D or E, the number of students in . Now assign one point to each of the 17 cyclists, 13 swimmers and 8 weight lifters. Thus, a total of 38 points would be assigned among the students in . Note that no student can have more than 2 points as no one is all three (swimmer, cyclist and weight lifter). Then, we should have as otherwise 38 points cannot be accounted for. (For example, if there were only 18 students in the maximum number of points that could be assigned to them would be 36.) Therefore, and each student in is in exactly 2 of the sets X, Y and Z. Hence, the number of students getting grade , that is, no student gets A grade. Since there are students who are not weight lifters all these 11 students must be both swimmers and cyclists. (Similarly, there are 2 who are both swimmers and weight lifters and 6 who are both cyclists and weight lifters.)

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.

Solution 4:

Suppose E is wearing a white cap. Then, D is lying and hence must be wearing a white cap. Since D and E both have white caps, A is lying and hence, he must be wearing white cap. If C is speaking truth, then C must be having a black cap and B must be wearing a black cap as observed by C. But then B must observe a cap on C. Hence, B must be lying. This implies that B is wearing a white cap which is a contradiction to C’s statement.

On the other hand, if C is lying, then C must be wearing a white cap. Thus, A, C, D and E are wearing white caps which makes B’s statement true. But, then B must be wearing a black cap and this makes C statement correct.

Thus, E must be wearing a black cap. This implies that B is lying and hence, must be having a white cap. But then D is lying and hence, must be having a white cap since B and D have white caps. A is not saying the truth. Hence, A must also be wearing a white cap. These together imply that C is truthful. Hence, C must be wearing a black cap. Thus, we have the following: A: white cap; B: white cap; C:black cap; D:white cap; E: black cap.

Problem 5:

Let f be a bijective function from the set to itself. Show that there is a positive integer such that for each . Here denotes the composite function repeated M times.

Solution 5:

Let us recall the following properties of a bijective function:

a) If is a bijective function, then there is a unique bijective function such that the identity function on A. The function g is called the inverse of f and is denoted by . Thus,

b)

c) If f and g are bijections from A to A, then so are and .

d) If f, g, h are bijective functions from A to A and , then .

Apply at left to both sides to obtain .

Coming to the problem at hand, since A has n elements, we see that the there are only finitely many (in fact, n!) bijective functions from A to A as each bijective function f gives a permutation of by taking . Since f is a bijective function from A to A, so is each of the function in the sequence:

All these cannot be distinct, since there are only finitely many bijective functions from A to A. Hence, for some two distinct positive integers m and n, , say, we must have

If , we take , to obtain the result. If , multiply both sides by to get . We take to get the relation with . Note that this means for all . QED.

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.

Solution 6:

Let ABCDEF be an equiangular hexagon with side lengths 1,2,3,4,5,6 in some order. We may assume without loss of generality that . Let .

Since the sum of all angles of a hexagon is equal to , it follows that each interior angle must be equal to . Let us take A as the origin, the positive x-axis along AB and the perpendicular at A to AB as the y-axis. We use the vector method: if the vector is denoted by we then have:

This is because these vectors are inclined to the positive x axis at angles 0, 60 degrees, 120 degrees, 180 degrees, 240 degrees, 300 degrees respectively.

Since the sum of all these six vectors is , it implies that

and

That is, ….call this I

and ….call this II

Since , in view of (II), we have

….(i)

…(ii)

…(iii)

The possibility that in (i), for instance, need not be considered separately, because we can reflect the figure about and interchange these two sets.

Case (i):

Here . Since , this is not possible.

Case (ii):

Here . This is satisfied by

Case (iii):

Here

Case (iv):

This is satisfied by .

Hence, we have (essentially) two different solutions: and . It may be verified that I and II are both satisfied by these sets of values.

Aliter: Embed the hexagon in an appropriate equilateral triangle, whose sides consist of some sides of the hexagon.

*Solutions to the remaining problems from that blog will have to be tried by the student.Â *

Cheers,

Nalin Pithwa.

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

Problem 1:

Let 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 …

Regards,

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 . 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 to itself. Show that there is a positive integer such that for each . Note that denotes the composite function 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 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 !!

Regards,

Nalin Pithwa