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 And the justification for the neglect was obviously no longer valid, if it had ever been.}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.

https://www.amazon.in/INSTRUMENTS-TI-Nspire-CX-II-CAS/dp/B07XCM6SZ3/ref=sr_1_1?crid=3RNR2QRV1PEPH&keywords=ti+nspire+cx+cas&qid=1585782633&s=electronics&sprefix=TI+n%2Caps%2C253&sr=8-1

https://www.amazon.in/INSTRUMENTS-TI-Nspire-CX-II-CAS/dp/B07XCM6SZ3/ref=sr_1_1?crid=3RNR2QRV1PEPH&keywords=ti+nspire+cx+cas&qid=1585782633&s=electronics&sprefix=TI+n%2Caps%2C253&sr=8-1

https://www.amazon.in/INSTRUMENTS-TI-Nspire-CX-II-CAS/dp/B07XCM6SZ3/ref=sr_1_1?crid=3RNR2QRV1PEPH&keywords=ti+nspire+cx+cas&qid=1585782633&s=electronics&sprefix=TI+n%2Caps%2C253&sr=8-1

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. 1 + 1 \times 2 + 3 \times 4 + 5 \times 6 + 7

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.

9 \times 1 - 2 = 7 \times 3 -4 = 17 \times 5 - 6 = 79 \times 7 - 8 = 545 \times 9 - 10 = 4895

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 a \leq b 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 🙂 🙂 🙂

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 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.

Answer 1:

Without loss of generality, we may take 1<a_{1}\leq a_{2}\leq \ldots \leq a_{10}<55…..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: a+b>c, a+c>b and b+c>a). We will consider triplets a_{i}, a_{i+1}, a_{i+2} and 1 \leq i \leq 8. 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, a_{i}+a_{i+1} \leq a_{i+2}. Hence, we get the following set of inequalities:

i=1 gives a_{1}+a_{2} \leq a_{3} giving 2 < a_{3}.

i=2 gives a_{2}+a_{3} \leq a_{4} giving 3 < a_{4}

i=3 gives a_{3}+a_{4} \leq a_{5} giving 5 < a_{5}

i=4 gives a_{4}+a_{5} \leq a_{6} giving 8 < a_{6}

i=5 gives a_{5}+a_{6} \leq a_{7} giving 13 < a_{7}

i=6 gives a_{6}+a_{7} \leq a_{8} giving 21 < a_{8}

i=7 gives a_{7}+a_{8} \leq a_{9} giving 34 < a_{9}

i=8 gives a_{8}+a_{9} \leq a_{10} giving 55<a_{10}

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 \{ C_{1}, C_{2}\} be a partition of C. Let e(C_{1}) denote the total number of enemy pairs in C_{1}. Let e(C_{2}) denote the total number of enemy pairs in C_{2}.

Let e(C_{1}, C_{2})= e(C_{1})+e(C_{2}) denote the total number of enemy pairs corresponding to the partition \{ C_{1}, C_{2}\} of C. Note e(C_{1}, C_{2}) is an integer greater than or equal to zero. Hence, by Well-Ordering Principle, there exists a partition having the least value of e(C_{1}, C_{2}).

Claim: This is “the” required partition.

Proof: If not, without loss of generality, suppose there is a person P in C_{1} having at least 2 enemies in C_{1}. Construct a new partition \{D_{1}, D_{2}\} of C as follows: D_{1}=C_{1}-\{ P \} and D_{2}=C_{2}- \{P\}. Now, e(D_{1}, D_{2})=e(D_{1})+e(D_{2}) \leq \{ e(C_{1})-2\} + \{ e(C_{2})+1\}=e(C_{1}, C_{2})-1. Hence, e(D_{1}, D_{2})<e(C_{1}, C_{2}) contradicting the minimality of e(C_{1}, C_{2}). 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 N=2n \times 2n \times 2n=8n^{3}. To find the total number of favourable outcomes we proceed as follows:

Let a be any odd integer such that 1 \leq a \leq 2n-1 and let us count the sequences having a as least element.

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

(ii) There are 2n-a sequences of the form (a,a,b) with a<b \leq 2n. For each such sequence there are three distinct permutations possible. Hence, there are in all 3(2n-a) sequences with a repeated twice.

iii) When n>1, for values of a satisfying 1 \leq a \leq (2n-3), sequences of the form (a,b,c,) with a<b<c \leq 2n are possible and the number of such sequences is r=1+2+3+\ldots+(2n+a-1)=\frac{1}{2}(2n-a)(2n-a-1). For each such sequence, there are six distinct permutations possible. Hence, there are 6r=3(2n-a)(2n-a-1) sequences in this case.

Hence, for odd values of a between 1 and 2n-1, the total counts of possibilities S_{1}, S_{2}, S_{3} in the above cases are respectively.

S_{1}=1+1+1+\ldots+1=n

S_{2}=3(1+3+5+\ldots+(2n-1))=3n^{2}

3(2 \times 3 + 4 \times 5 + \ldots+ (2n-2)(2n-1))=n(n-1)(4n+1).

Hence, the total number A of favourable outcomes is A=S_{1}+S_{2}+S_{3}=n+3n^{2}+n(n-1)(4n+1)=4n^{3}. Hence, the required probability is \frac{A}{N} = \frac{4n^{3}}{8n^{3}} = \frac{1}{2}. QED>

Cheers,

Nalin Pithwa