Number Theory: Historical Introduction: Tom Apostol

Reference: Introduction to Analytic Number Theory by Tom Apostol.

The theory of numbers is that branch of mathematics which deals with properties of the whole numbers: 1, 2, 3, 4, 5, …

also called the counting numbers, or positive integers.

The positive integers are undoubtedly man’s first mathematical creation. It is hardly possible to imagine human beings without the ability to count, at least within a limited range. Historical record shows that as early as 5700 BC the ancient Sumerians kept a calendar, so they must have developed some form of arithmetic.

By 2500 BC, the Sumerians had developed a number system using 60 as a base. This was passed on to the Babylonians, who became highly skilled calculators. Babylonian clay tablets containing elaborate mathematical tables have been found, dating back to 2000BC.

When ancient civilizations reached a level which provided some leisure time to ponder about things, some people began to speculate about the nature and properties of numbers. This curiosity developed into a sort of number-mysticism or numerology, and even today numbers such as 3,7, 11 and 13 are considered omens of good or bad luck.

Numbers were used for keeping records and for commercial transactions for over 5000 years before any one thought of studying numbers themselves in a systematic way. The first scientific approach to the study of integers, that is, the true origin of the theory of numbers, is generally attributed to the Greeks. Around 600 BC, Pythagoras and his disciples made rather thorough studies of the integers. They were the first to classify integers in various ways: Even numbers, odd numbers, prime numbers, composite numbers.

A prime number is a number greater than 1 whose only divisors are 1 and the number itself. Numbers that are not prime are called composite numbers, except that the number 1 is neither prime nor composite.

The Pythagoreans also linked numbers with geometry. They introduced the idea of polygonal numbers: triangular numbers, square numbers, pentagonal numbers, etc. The reason for this geometrical nomenclature is clear: when the numbers are represented by dots arranged in the form of triangles, squares, pentagons, etc. …E.g.: Triangular numbers: 1, 3, 6, 10, 15, 21, 28, ….; square numbers: 1, 4, 9, 16, 25, 36, ….; pentagonal numbers: 1, 5, 12, 22, 35, 51, 70, ….

Another link with geometry came from the famous Theorem of Pythagoras which states that in any right triangle the square of the length of the hypotenuse is the sum of the squares of the lengths of the two legs. The Pythagoreans were interested in right triangles whose sides are integers. Such triangles are called Pythagorean triangles. The corresponding triple of numbers (x,y,z) representing the lengths of the sides is called a Pythagorean triple. Example (3,4,5)

A Babylonian tablet had been found dating from about 1700 BC, which contains an extensive list of Pythagorean triples, some of the numbers being quite large. The Pythagoreans were the first to give a method for determining infinitely many triples. In modern notation, it can be described as follows:

x=n, y=\frac{1}{2}(n^{2}-1), \frac{1}{2}(n^{2}+1);

the resulting triple (x,y,z) will always be a Pythagorean triple with z=y+1.

Here are some examples:

(3,4,5), (5,12,13), (7,24,25), (9,40,41), (11,60,61), (13,84,85), (15,112, 113), (17,144, 145), (19,180, 181)

There are other Pythagorean triples besides these, for example:

(8,15,17), (12,35,37), (16,63, 65), (20,99,101)

In these examples, we have z=y+2. Plato had found a method for determining all these triples : in modern notation, they are given by the formulas:

x=4n, y=4n^{2}-1, z=4n^{2}+1

Around 300 BC, an important event occurred in the history of mathematics. The appearance of Euclid’s Elements,a collection of 13 books, transformed mathematics from numerology into a deductive science. Euclid was the first person to present mathematical facts along with rigorous proofs of these facts. Three of the thirteen books were devoted to theory of numbers (books VII, IX and X). In book IX Euclid proved that there are infinitely many primes. His proof is still taught in the classroom today. In Book X, he gave a method for obtaining all Pythagorean triples although he gave no proof that his method did, indeed, give them all. The method can be summarized by the formulas:

x=t(a^{2}-b^{2}), y=2tab, z=t(a^{2}+b^{2}),

where t, a, b are arbitrary positive integers such that a>b, a and b have no prime factors in common, and one of a or b is odd, the other even.

Euclid also made an important contribution to another problem posed by the Pythagoreans —- that of finding all perfect numbers. The number 6 was called a perfect number because 6=1+2+3, the sum of all its proper divisors (that is, the sum of divisors less than 6). Another example of a perfect number is 28 because 28=1+2+4+7+14, and 1, 2, 4, 7 and 14 are the divisors of 28 less than 28. The Greeks referred to the proper divisors of a number as its “parts.” They called 6 and 28 perfect numbers because in each case the number is equal to to the sum of all its parts.

In Book IX, Euclid found all even perfect numbers. He proved that an even number is perfect if it has the form:

2^{p-1}(2^{p}-1)

where both p and 2^{p}-1 are primes.

Two thousand years later, Euler proved the converse of Euclid’s theorem. That is, every even perfect number must be of Euclid’s type. For example, for 6 and 28, we have:

6=2^{2-1}(2^{2}-1) = 2 \times 3 and 28=2^{3-1}(2^{3}-1)= 4 \times 7/

The first five even perfect numbers are

6,28,496,8128 and 33,550,336.

Perfect numbers are very rare indeed. At the present time (1975) only 24 perfect numbers are known. They correspond to the following values of p in Euclid’s formula:

2,3,5,7,13,17,19,31,61, 89,107, 127, 521, 607, 1279, 2203, 2281, 3217, 4253, 4423, 9689, 9941, 11213, 19937.

Numbers of the form 2^{p}-1, where p is a prime, are now called Mersenne numbers and are denoted by M_{p} in honour of Mersenne, who studied them in 1644. It is known that M_{p} is prime for the 24 primes listed above and composite for all other values of p \leq 257, except for possibly

p=157,167, 193, 199, 227, 229

for these it it not yet (1975) known whether M_{p} (PS: you can google and find out the status now) is prime or composite.

No odd perfect numbers are known, it is not even known if any exist. But, if any do exist they must be very large; in fact, greater than 10^{50}.

We turn now to a brief description of the history of theory of numbers since Euclid’s time.

After Euclid in 300 BC, no significant advances were made in number theory until about 250 AD when another Greek mathematician, Diophantus of Alexandria, published 13 books, six of which have been preserved. This was the first Greek work to make systematic use of algebraic symbols. Although his algebraic notation seems awkward by present-day standards, Diophantus was able to solve certain algebraic equations involving two or three unknowns. Many of his problems originated from number theory and it was natural for him to seek integer solutions of equations. Equations to be solved with integer values are now called Diophantine equations, and the study of such equations is known as Diophantine analysis. The equation x^{2}+y^{2}=z^{2} for Pythagorean triples is an example of a Diophantine equation.

After Diophantus, not much progress was made in the theory of numbers until the seventeenth century, although there is some evidence that the subject began to flourish in the Far East — especially in India — in the period between AD 500 and AD 1200.

In the seventeenth century the subject was revived in Western Europe, largely through the efforts of a remarkable French mathematician, Pierre de Fermat (1601-1665), who is generally acknowledged to be the father of modern number theory. Fermat derived much of his inspiration from the works of Diophantus. He was the first to discover really deep properties of the integers. For example, Fermat proved the following surprising theorems:

Every integer is either a triangular number or a sum of 2 or 3 triangular numbers; every integer is either a square or sum of 2, 3, or 4 squares; every integer is either a pentagonal number or the sum of 2, 3, 4 or 5 pentagonal numbers, and so on.

Fermat also discovered that every prime number of the form 4n+1 such as 5, 13, 17, 29, 37, 41, etc. is a sum of two squares. For example,

5=1^{2}+2^{2}; 13=2^{2}+3^{2}; 17=1^{2}+4^{2}; 29=2^{2}+5^{2}; 37=1^{2}+6^{2}; 41=4^{2}+5^{2}

Shortly, after Fermat’s time, the names of Euler (1707-1783), Lagrange (1736-1813), Legendre (1752-1833), Gauss (1777-1855), and Dirichlet (1805-1859) became prominent in the further development of the subject. The first textbook in number theory was published by Legendre in 1798. Three years later Gauss published Disquisitiones Arithmeticae, a book which transformed the subject into a systematic and beautiful science. Although he made a wealth of contributions to other branches of mathematics, as well as to other sciences, Gauss himself considered his book on number theory to be his greatest work.

In the last hundred years or so since Gauss’s time there has been an intensive development of the subject in many different directions. It would be impossible to give in a few pages a fair cross-section of the types of problems that are studied in the theory of numbers. The field is vast and some parts require a profound knowledge of higher mathematics. Nevertheless, there are many problems in number theory which are very easy to state. Some of these deal with prime numbers, and we denote the rest of this introduction to such problems.

The primes less than 100 have been listed above. A table listing all primes less than 10 million was published in 1914 by an American mathematician, D. N. Lehmer. There are exactly 664579 primes less than 10 million, or about 6.5 %. Later, D. H. Lehmer (the son of D. N. Lehmer) calculated the total number of primes less than 10 billion; there are exactly 455052512 such primes, or about 4.5% although all these primes are not known individually.

A close examination of a table of primes reveals that they are distributed in a very irregular fashion. The tables show long gaps between primes. For example, the prime 370261 is followed by 111 composite numbers. There are no primes between 20831323 and 20831533. It is easy to prove that arbitrarily large gaps between prime numbers must eventually occur.

On the other hand, the tables indicate that consecutive primes, such as 3 and 5, or 101 and 103, keep recurring. Such pairs of primes, which differ only by 2 are known as twin primes. There are over 1000 such pairs below 100000 and over 8000 below 1000000. The largest pair known to date (around 1975 AD) is 76.3^{139}-1 and 76.3^{139}+1. Many mathematicians think there are infintely many such pairs, but no one has been able to prove (I think the advanced reader can refer to Google to see the progress in work on twin primes: the work of Dr. Yitang Zhang; I think Dr. Terence Tao also did some further work based on it. Refer to terrytao.wordpress.com)

One of the reasons for this irregularity in distribution of primes is that no simple formula exists for producing all the primes. Some formulas do yield many primes. For example, the expression:

x^{2}-x+41

gives a prime for x=0, 1, 2, \ldots, 40, whereas

x^{2}-79x+1601

gives a prime for x=0,1,2, \ldots, 79. However, no such simple formula can give a prime for all x, even if cubes and higher powers are used. In fact, in 1752 Goldbach proved that no polynomial in x with integer coefficient can be prime for all x, or even for all sufficiently large x.

Some polynomials represent infinitely many primes. For example, as x runs through the integers 0, 1, 2, 3, ….the linear polynomial

2x+1

gives all the odd numbers, hence, infinitely many primes. Also, each of the polynomials

4x+1 and 4x+3

represent infinitely many primes. In a famous memoir published in 1837, Dirichlet proved that, if a and b are positive integers with no prime factors in common, the polynomial

ax+b

gives infinitely many primes as x runs through all the positive integers. The result is now known as Dirichlet’s theorem on the existence of primes in a given arithmetical progression.

To prove the theorem, Dirichlet went beyond the realm of integers and introduced tools of analysis such as limits and continuity. By so doing, he laid the foundations for a new branch of mathematics called analytic number theory, in which ideas and methods of real and complex analysis are brought to bear on problems about integers.

It is not known if there is any quadratic polynomial ax^{2}+bx+c with a \neq 0 which represents infinitely many primes. However, Dirichlet had used his powerful analytic methods to prove that, if a, 2b, and c have no prime factor in common, the quadratic polynomial in two variables

ax^{2}+2bxy+cy^{2}

represents infinitely many primes as x and y run through positive integers.

Fermat thought that the formula 2^{2^{n}}+1 would always give a prime for n=0,1,2,.... These numbers are called Fermat numbers and are denoted by F_{n}. The first five are

F_{0}=1, F_{1}=5, F_{2}=17, F_{3}=257, F_{4}=65537

and they are all primes. However, in 1732 A.D., Euler found that F_{5} is composite. In fact,

F_{5} = 2^{32}+1= (641)(6700417)

These numbers are also of interest in plane geometry. Gauss had proved that if F_{n} is prime, say F_{n}=p, then a regular polygon of p sides can be constructed with straight edge and compass.

Beyond F_{5}, no futher Fermat primes have been found. (Please check this in Google. This is the author’s claim as of 1975 A.D.) In fact, for 5 \leq n \leq 16 each Fermat number F_{n} is composite. Also, F_{n} is known to be composite for the following further isolated values of n:

n=18,19,21, 23,25,26,27,30, 32, 36, 38, 39, 42, 52, 55, 58, 63,73,77,81,117, 125, 144, 150, 207, 226, 228, 260, 267, 268, 284, 316, 452, 1945.

The greatest known Fermat composite F_{1945} (till 1975 AD) has more than 10^{582} digits, a number larger than the number of letters in the Los Angeles and New York telephone directories combined (then in 1975 AD).

It was mentioned earlier that there is no simple formula that gives all the primes. In this connection, we should mention a result discovered in 1947 A.D. by an American mathematician, W. H. Mills. He proved that there is some number A, greater than 1 but not an integer, such that

[A^{3^{x}}] is prime for all x = 1, 2, 3…

Here, [A^{3^{x}}] means the greatest integer \leq A^{3^{x}}. Unfortunately no one knows what A is. (Again the reader can check the latest about this in Google).

The foregoing results illustrate the irregularity of the distribution of the prime numbers. However, by examining large blocks of primes one finds that their average distribution seems to be quite regular. Although there is no end to the primes, they become more widely spread, on the average, as we go further and further in the table. The question of the diminishing frequency of primes was the subject of much speculation in the early nineteenth century. To study this distribution, we consider a function, denoted by \pi{(x)}, which counts the number of primes less than or equal to x. Thus,

\pi{(x)} = the number of primes p satisfying 2 \leq p \leq x.

Here is a brief table of this function and its comparison with \frac{x}{\ln{x}}.

\begin{array}{cccc} x & \pi{(x)} & \frac{x}{\ln{(x)}} & \frac{\pi{(x)}}{\frac{x}{\ln{(x)}}} \\ 10 & 4 & 4.3 & 0.93 \\ 100 & 25 & 21.7 & 1.15 \\ 1000 & 168 & 144.9 & 1.16 \\ 10000 & 1229 & 1086 & 1.11 \\ 100000 & 9592 & 8686 & 1.10 \\ 1000000 & 78498 & 72464 & 1.08 \\ 10000000 & 664579 & 621118 & 1.07 \\ 100000000 & 5761455 & 5434780 & 1.06 \\ 1000000000 & 50847534 & 48309180 & 1.05 \\ 10000000000 & 455052512 & 434294482 & 1.048 \end{array}

By examining a table like this for x \leq 10^{6}, Gauss and Legendre proposed independently that for large x, the ratio

\frac{\pi{(x)}}{\frac{x}{\ln{(x)}}}

was nearly 1 and they conjectured that this ratio would approach 1 as x approaches \infty. Both Gauss and Legendre attempted to prove this statement but did not succeed. The problem of deciding the truth or falsehood of this conjecture attracted the attention of eminent mathematicians for nearly 100 years.

In 1851 A.D. the Russian mathematician Chebyshev made an important step forward by proving that if the ratio did tend to a limit, then this limit must be 1. However, he was unable to prove that the ratio does tend to a limit.

In 1859, Riemann attacked the problem with analytic methods, using a formula discovered by Euler in 1737 which relates the prime numbers to the function

\zeta{(s)} = \Sigma_{n=1}^{\infty}\frac{1}{n^{s}} for real s greater than 1.

Riemann considered complex values of s and outlined an ingenious method for connecting the distribution of primes to properties of the function \zeta{(s)}. The mathematics needed to justify all the details of his method had not yet been fully developed and Riemann was unable to completely settle the problem before his death in 1866.

Thirty years later the necessary analytic tools were at hand and in 1896 J. Hadamard and and C. J. de la Vallee Poussin independently and almost simultaneously succeeded in proving that

\lim_{n \rightarrow \infty} \frac{\pi{(x)}\ln{(x)}}{x}=1.

This remarkable result is called the prime number theorem, and its proof was one of the crowning achievements of analytic number theory.

In1949, two contemporary mathematicians, Atle Selberg and Paul Erdos caused a sensation in the mathematical world when they discovered an elementary proof of the prime number theorem. Their proof, though very intricate, makes no use of \zeta{(s)} nor of complex function theory and in principle is accessible to anyone familiar with elementary calculus.

One of the most famous problems concerning prime numbers is the so-called Goldbach conjecture (again please refer to Google for the latest reseach results on this). In 1742 A.D., Goldbach wrote to Euler suggesting that every even number greater than or equal to 4 is a sum of two primes. For example,

4=2+2; 6=3+3; 8=3+5; 10=3+7=5+5; 12=5+7

This conjecture is undecided to this day (again, please google for twin prime conjecture, Yitang Zhang and Terence Tao etc.), although it might be true as per the progress suggested. Now, why do mathematicians think it is probably true if they haven’t been able to prove it? First of all, the conjecture has been verified by actual computation for all even numbers less than 33 \times 10^{6}. It has been found that every even number greater than 6 and less than 33 \times 10^{6} is, in fact, not only the sum of two odd primes but the sum of two distinct odd primes. But, in number theory verification of a few thousand cases is not enough evidence to convince mathematicians that something is probably true. For example, all the odd primes fall into two categories, those of the form 4n+1 and those of the form 4n+3. Let \pi_{1}{(x)} denote all the primes less than or equal to x that are of the form 4n+1 and let \pi_{3}{(x)} denote the number that are of the form 4n+3. It is known that there are infinitely many primes of both types. By computation, it was found that \pi_{(x)} \leq \pi_{3}{(x)} for all x \leq 26861. But, in 1957, J. Leech found that for x=26861, we have \pi_{1}{(x)}=1473 and \pi_{3}{(x)}=1472, so the inequality was reversed. In 1914, Littlewood had proved that this inequality reverses back and forth infinitely often. That is, there are infinitely many x for which \pi_{1}{(x)} < \pi_{3}{(x)} and infinitely many x for which \pi_{3}{(x)} < \pi_{1}{(x)}. Conjectures about prime numbers can be erroneous even if they are verified by computations in thousands of cases.

Therefore, the fact that Goldbach’s conjecture has been verified for all even numbers less than 33 \times 10^{6} is only a tiny bit of evidence in its favour.

Another way that mathematicians collect evidence about the truth of a particular conjecture is by proving other theorems which are somewhat similar to the conjecture. For example, in 1930 the Russian mathematician Schnilrelmann proved that there is a number M such that every number n from some point on is a sum of M or lower primes:

n=p_{1}+p_{2}+p_{3}+\ldots+p_{n} for sufficiently large n

If we know that M were equal to 2 for all even n, this would prove Goldbach’s conjecture for all sufficiently large n. In 1956 A.D., the Chinese mathematician Yin Wen-Lin proved that M \leq 18. That is, every number n from some point on is a sum of 18 or fewer primes. Schnilrelmann’s result is considered a giant step toward a proof of Goldbach’s conjecture. It was the first real progress made on this problem in nearly 200 years.

A much closer approach to a solution of Goldbach’s problem was made in 1937 by another Russian mathematician, I. M. Vinogradov, who proved that from some point on every odd number is the sum of three primes:

n=p_{1}+p_{2}+p_{3} when n is odd, n is sufficiently large

In fact, this is true for all odd n greater than 3^{3^{15}}. To date (1975 A.D.) this is the strongest piece of evidence in favour of Goldbach’s conjecture. For one thing, it is easy to prove that Vinogradov’s theorem is a consequence of Goldbach’s statement. That is, if Goldbach’s conjecture is true, then it is easy to deduce Vinogradov’s statement. The big achievement of Vinogradov was that he was able to prove his result without using Goldbach’s statement. Unfortunately, no one has been able to work it the other way around and prove Goldbach’s statement from Vinogradov’s.

Another piece of evidence in favour of Goldbach’s conjecture was found in 1948 by the Hungarian mathematician Renyi who proved that there is a number M such that every sufficiently large even number n can be written as a prime plus another number which has no more than M prime factors:

n=p+A

where A has no more than M prime factors (n even, n sufficiently large). If we know that M=1, then Goldbach’s conjecture would be true for all sufficiently large n. In 1965 A.D., A. A. Buhstab and A. I. Vinogradov proved that M \leq 3, and in 1966 Chen Jing-run proved that M \leq 2.

We conclude this introduction with a brief mention of some outstanding unsolved problems (till date 1975 A.D.) concerning prime numbers:

  1. Goldbach’s problem: Is there an even number greater than 2 which is not the sum of two primes?
  2. Is there an even number greater than 2 which is not the difference of two primes?
  3. Are there infinitely many twin primes?
  4. Are there infinitely many Mersenne primes, that is, primes of the form 2^{p}-1 where p is prime?
  5. Are there infinitely many composite Mersenne numbers?
  6. Are there infintely many Fermat primes, that is, primes of the form 2^{2^{p}}+1?
  7. Are there infinitely many composite Fermat numbers?
  8. Are there infinitely many primes of the form x^{2}+1, where x is an integer? (It is known that there are infinitely many of the form x^{2}+y^{2}, and of the form x^{2}+y^{2}+1, and of the form x^{2}+y^{2}+z^{2}+1.)
  9. Are there infinitely many primes of the form x^{2}+k, where k is given?
  10. Does there always exist at least one prime between n^{2} and (n+1)^{2} for every integer n \geq 1?
  11. Does there always exist at least one prime between n^{2} and n^{2}+n for every integer n>1?
  12. Are there infinitely many primes whose digits (in base 10) are all ones? (Here are two examples: 11 and 11,111,111,111,111,111,111,111)

The professional mathematician is attracted to number theory because of the way all the weapons of modern mathematics can be brought to bear on its problems. As a matter of fact, many important branches of mathematics had their origin in number theory. For example, the early attempts to prove the prime number theorem stimulated the development of the theory of functions of a complex variable especially the theory of entire functions. Attempts to prove that the Diophantine equation x^{n}+y^{n}=z^{n} has no non-trivial solution if n greater than or equal to three (Fermat’s conjecture or Fermat’ s Last Theorem (this has been proved by Sir Andrew Wiles in 1995, Princeton University) led to the development of algebraic number theory, one of the most active areas of modern mathematical research. (1975 A.D. comment by Prof Tom Apostol: Even though Fermat’s conjecture is still undecided, this seems unimportant by comparison to the vast amounts of valuable mathematics that has been created as a result of work on this conjecture.) Another examples has been the theory of partitions which has been an important factor in the development of combinatorial analysis and in the study of modular functions.

There are hundreds of unsolved problems in number theory. New problems arise more rapidly than the old ones are solved, and many of the old ones have remained unsolved for centuries. As the mathematician Sierpinski once said, “…the progress of our knowledge of numbers is advanced not only by what we already know about them, but also by realizing what we yet do not know about them.”

Note: More References:

  1. History of the Theory of Numbers Volumes I, II, III: Dickson
  2. Reviews in Number Theory by LeVeQue, Volumes I, II, III, IV, V, VI.

Happy reading. I hope it arouses tremendous passion/curiosity among some readers to consider number theory as a possible professional option.

Regards,

Nalin Pithwa

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

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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

 

 

 

 

 

 

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 (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 T. Prove that the sum is divisible by 7.

Solution 2:

For every triple (a,b,c) in T, the triple (7-c,7-b,7-a) is in T and these two are distinct as 7 \neq 2b. Pairing off (a,b,c) with (7-c,7-b,7-a) for each (a,b,c) \in T, 7 divides abc-(7-a)(7-b)(7-c).

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 X\bigcup Y \bigcup Z all get grades B and C, and six students get grades D or E, the number of students in X\bigcup Y \bigcup Z \leq 25-6=19. 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 X \bigcup Y \bigcup Z. 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 X \bigcup Y \bigcup Z \geq 19 as otherwise 38 points cannot be accounted for. (For example, if there were only 18 students in X \bigcup Y \bigcup Z the maximum number of points that could be assigned to them would be 36.) Therefore, X \bigcup Y \bigcup Z=19 and each student in X \bigcup Y \bigcup Z is in exactly 2 of the sets X, Y and Z. Hence, the number of students getting grade A=25-19-6=0, that is, no student gets A grade. Since there are 19-8=11 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 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. Here f^{M} denotes the composite function f \circ f \circ \ldots \circ f repeated M times.

Solution 5:

Let us recall the following properties of a bijective function:

a) If f:A \rightarrow A is a bijective function, then there is a unique bijective function g: A \rightarrow A such that f \circ g = g \circ f=I_{A} the identity function on A. The function g is called the inverse of f and is denoted by f^{-1}. Thus, f \circ f^{-1}=I_{A}=f^{-1}\circ f

b) f \circ I_{A} = f = I_{A} \circ f

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

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

Apply f^{-1} at left to both sides to obtain g=h.

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 \{ 1,2,3,\ldots, n\} by taking \{ f(1),f(2), \ldots, f(n)\}. Since f is a bijective function from A to A, so is each of the function in the sequence:

f^{2}, f^{3}, \ldots, f^{n}, \ldots

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, m > n, say, we must have f^{m}=f^{n}

If n=1, we take M=m, to obtain the result. If n>1, multiply both sides by (f^{-1})^{n-1} to get f^{m-n+1}=f. We take M=m-n+1 to get the relation f^{M}=f with M>1. Note that this means f^{M}(i)=f(i) for all i \in A. 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 AB=1. Let BC=a, CD=b, DE=c, EF=d, FA=e.

Since the sum of all angles of a hexagon is equal to (6-2) \times 180=720 \deg, it follows that each interior angle must be equal to 720/6=120\deg. 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 (x,y) we then have:

\overline{AB}=(1,0)

\overline{BC}=(a\cos 60\deg, a\sin 60\deg)

\overline{CD}=(b\cos{120\deg},b\sin{120\deg})

\overline{DE}=(c\cos{180\deg},c\sin{180\deg})=(-c,0)

\overline{EF}=(d\cos{240\deg},d\sin{240\deg})

\overline{FA}=(e\cos{300\deg},e\sin{300\deg})

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 \overline{0}, it implies that 1+\frac{a}{2}-\frac{b}{2}-c-\frac{d}{2}+\frac{e}{2}=0

and (a+b-d-e)\frac{\sqrt{3}}{2}=0

That is, a-b-2c-d+e+2=0….call this I

and a+b-d-e=0….call this II

Since (a,b,c,d,e)=(2,3,4,5,6), in view of (II), we have

(a,b)=(2,5), (a,e)=(3,4), c=6….(i)

(a,b)=(5,6), (a,e)=(4,5), c=2…(ii)

(a,b)=(2,6), (a,e)=(5,5), c=4…(iii)

The possibility that (a,b)=(3,4), (a,e)=(2,5) in (i), for instance, need not be considered separately, because we can reflect the figure about x=\frac{1}{2} and interchange these two sets.

Case (i):

Here (a-b)-(d-e)=2c-2=10. Since a-b=\pm 3, d-e=\pm 1, this is not possible.

Case (ii):

Here (a-b)-(d-e)=2c-2=2. This is satisfied by (a,b,d,e)=(6,3,5,4)

Case (iii):

Here (a-b)-(d-e)=2c-2=6

Case (iv):

This is satisfied by (a,b,d,e)=(6,2,3,5).

Hence, we have (essentially) two different solutions: (1,6,3,2,5,4) and (1,6,2,4,3,5). 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. 

Cheers,

Nalin Pithwa.

 

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 …

Regards,
Nalin Pithwa