A Recent Development

Late last month, around 23 October 2024, many newspapers carried the story that the State Council of Educational Research and Training (SCERT) in Maharashtra has decided to reduce the pass mark for mathematics and science from 35 to 20. The caveat is that students who have scored at least 20 marks but less than 35 will be declared as having passed that subject. However, their mark sheet will indicate that they no longer qualify to take further studies in mathematics or sciences. Quite naturally, there were mixed responses to this move. In this post we will look at some of these responses to which I will offer my own critique. In the next post, which will be two weeks from today, since I will not post next week, I will suggest what I think is the purpose of primary and secondary education. In that context I will raise two problems we currently face with mathematics in primary and secondary school. So let us look at the views expressed by some people about the lowering of the pass mark.

Introducing Common Pitfalls

According to Rahul Rekhawar, director of SCERT, “Failing in mathematics or science, and in effect in SSC, often leaves students with no opportunities to continue their education, even if their strength lies elsewhere. This change is designed to ensure that students are not unfairly pushed out of education system and can pursue their academic and career aspirations.” An educator called Heramb Kulkarni is reported to have said, “These subjects serve a purpose beyond mere academic scoring. Lowering standards might adversely affect the overall quality of education in the state.” However, another educator called Vasant Kalpande argued, “There are students who cannot understand mathematics. Many great writers like Munshi Premchand and Hari Narayan Apte dropped out of their respective courses because of mathematics. Hence, if a student wants to pursue arts or humanities, why force them to take up science and maths which they have no aptitude for?”

Pitfall 1: Reliance on Ad Hoc Reasoning

Each of these responses reflect a certain understanding of the purpose of secondary education and of the role of learning mathematics for a student. Rekhawar’s response has a prima facie logic to it until you realize that it is the determination of the exam board to indicate a pass or fail for a subject. This move seems to be helping students who are not doing too well in mathematics and sciences.

But what about students who do not do well in, say, history? If student A scored 34 in history, he would be declared as having failed the exams. Another student B, who scored 20 in mathematics, would be declared as having passed the exams. How would the first student get admission to a mathematics course, when in fact he does not need history beyond Grade 10? You see, if the goal is to ensure students are not unfairly pushed out of the education system, then such a determination must be made after a student decides what he/she wishes to study after Grade 10. So student A could decide, after the exams, that he is not suited to study history and should obtain a pass certification for that subject with an indicator that he should not be permitted to pursue history in the future.

We can see that this is simply an ad hoc approach. The SCERT is not doing anything to favor the students. Rather, in my view, it is simply attempting to jack up the pass percentage so that it appears to be doing a better job.

It is more likely that this move has been made because more students are finding it difficult to understand mathematics in the primary and secondary schools. This means that, either the curriculum is not suited for the purpose of primary and secondary education or the teachers are not teaching in a way that enables student learning. We will return to these twin ideas in the next post.

Pitfall 2: Teaching for Pass Statistics

Kulkarni’s claim that this will lower the quality of education in the state is appalling and my heart goes out to his students. In my view, it does not matter if the pass mark is 35 or 20 or 60. The goal for me when I teach is not to just push a student over some arbitrary line that some bureaucrats have drawn. No! My goal is to ensure that each student learns as much as he/she can from me. My goal is to ensure that each student’s understanding of mathematics is better today than it was yesterday. Teachers who teach based on an arbitrary pass mark are teachers I would not trust because their goal is not the student’s learning but the student’s passing. Lowering the passing mark from 35 to 20 may make it easier for a teacher to push a student over the line. However, if that determines the quality of teaching that a teacher delivers, then we are not doing right by our students.

It is likely that Kulkarni is simply telling us what would happen in some classrooms rather than what would happen in his classroom. In other words, may be he is simply warning us about a possibility he fears might become a reality in the classrooms of less dedicated teachers. But those teachers are anyway only teaching to the current pass mark of 35. Do we really think that their students are actually learning anything? Do we think that a student who is scoring 90 marks in a class taught by a teacher who teaches based on a pass mark is actually learning anything? Is it not more likely that the 90 marks are artificially inflated to prop up the teacher’s performance? And if the 90 marks are genuine, is it not more likely that this reflects the student’s innate mathematical competence?

In other words, if we are afraid that lowering the pass mark will lower the quality of teaching, then we must admit that we have a much bigger problem of accepting teachers who do not care about their student’s learning but more about their own pass statistics.

Pitfall 3: Appealing to the Absurd

Kalpande’s argument is completely an appeal to the absurd. Pointing out someone for whom the system did not work does not mean that the system should be jettisoned. While I do think the system needs a massive overhaul, mentioning great writers like Munshi Premchand and Hari Narayan Apte for whom the system did not work does nothing.

No system will work for everyone. Anyone who thinks he/she can make a system that will work for everyone is smoking something I would not recommend to my students! Every system, precisely because it forms a ‘box’, will result in being unfit for those who do not fit into that ‘box’. The solution is not to arbitrarily change the dimensions of the ‘box’ but to find another ‘box’ in which these people can fit. The very fact that Premchand and Apte succeeded as writers means that they found another ‘box’ that could accommodate their skills and talents.

Moreover, Premchand and Apte are part of a very tiny and exclusive minority. Not every who aspires to become a successful author succeeds in achieving that goal. Perhaps Premchand and Apte were forced to find their true calling precisely because they were pushed out of the ‘box’. In that case, one could possibly argue that, if the pass mark had been 20, neither Premchand nor Apte would have become authors because they would have passed mathematics and would have found some other run of the mill profession within the ‘box’.

Looking Ahead

We have taken a look at some of the common responses in addressing the issue of determining a pass mark for mathematics. It is quite likely that the newspapers, not known for their research rigor, have only interviewed people with loose lips who were willing to shoot from the hip. In other words, it is quite likely that the newspapers have not managed to interview any educator who is willing to think deeply about the issue.

Nevertheless, I am certain that the three responses above are not quite aberrant. However, all the responses reflect a failure to address three crucial issues. First, none of them address the issue of how the pass mark for a subject relates to the purpose of primary and secondary education. After all, if this pass mark is to be reflective of the culmination of the student’s learning over ten years of schooling, in what way does it provide us a measure of that learning? Second, assuming we understand what the purpose of primary and secondary education is, how does the mathematics curriculum reflect or fail to reflect this purpose? Third, in what way are teachers being equipped to teach the subject so that student learning is prioritized? We will look at these three questions in the next post, which, as I announced earlier, will be two weeks from today.

The Presenting Problem

There is one big problem with being a mathematics tutor. You aren’t the students’ main teacher and hence you get to hear what the students have actually learned from their main teacher. Why is this a problem? Well, it exposes me to the ways in which we mathematics teachers regularly let our students down. It is an indictment of the self and no one likes that.

Today, I wish to highlight one area in which we mathematics teachers often fall short – the area of terminology. I am a stickler for terminology. Technical terms in a field are essential for succinct and precise communication. Without these technical terms, we would have to use many more words. Not just that, when the number of words needed to refer to something increases, the possibility of paraphrasing also increases, thereby making the meaning dependent on the vicissitudes of the language in use. Hence, the communication becomes verbose and imprecise, leading to inefficiencies and possible lack of understanding. The use of correct terminology enables precise and succinct communication and facilitates understanding.

I am fed up, for example, of hearing students ask, “Do I minus the two?” Hello! Hello? What’s that? The word ‘minus’ is not a part of English grammar but a way of vocalizing the mathematical symbol that indicates subtraction. Similarly, hearing a student say that 1/x is the ‘inverse’ of x, has as bad an effect as it would have had if the student dragged his nails across the wall because there is a difference between an ‘inverse’ and a ‘reciprocal’. And finally, hearing a student tell me she wants to ‘bracket the terms’ when she means ‘factorize’ is pure agony.

Introduction to Nomenclature

I had a good early mathematics education. I don’t mean that all my mathematics teachers were capable teachers. I don’t remember most of them. And even the ones I remember weren’t all capable. But I remember being taught terminology from a very early age. While I do not remember ever using ‘addend’, ‘minuend‘, or ‘subtrahend‘, I clearly remember using ‘multiplicand‘, ‘dividend‘, ‘divisor‘, ‘quotient‘, and ‘remainder‘. The former set was probably never learned because addition and subtraction are covered very early. By the time multiplication and division were learned I guess I was old enough to learn bigger words.

But today I find that many students cannot recognize even the terms in the latter set. So instead of saying, “When you divide the dividend by the divisor, the result is the quotient with a remainder” we would have to say something like, “When you divide one number by another, the result is the largest number by which the second number should be multiplied without exceeding the first with the difference if any between the first number and the largest multiple of the second.” So instead of 16 words, in which most of the non-technical words are simply connectors of sorts, we would have to use 39 words, where many of the additional words describe the operation of division and the role of the different. And since others may phrase it differently, we would never be sure that we are communicating with accuracy.

Case in Point

One of the terms that causes me most pain is the ‘discriminant’. For those of you who do not know or do not remember or have conveniently suppressed the memory, when you attempt to solve a quadratic equation of the form ax² + bx + c = 0, the solutions are given by

Here the quantity under the radical sign (i.e. ‘√‘) is called the ‘discriminant’. Of late, many students come to me, having learned quadratic equations in grade 9 or 10, but having never heard the term. Why do I say this term is important? Because its name, like most mathematical terms, tells you what it does. The ‘discriminant’, you see, allows us to ‘discriminate’ between the kind of solutions a given quadratic equation has before we attempt to solve the equation. It can tell you if the equation has two different real solutions, only one real solution, or no real solutions. Give that this expression is so critical and that it is named to reflect what it does, a failure to teach students the term can only be viewed, at least according to me, as indicative of a failure of mathematics teachers to pass the baton of knowledge to the next generation. I even have textbooks written for high school students that do not have the term.

I can understand the reluctance that some teachers in the elementary school have concerning teaching some of the technical terms. Mathematical jargon can get to be quite heady. However, the failure to teach students these terms results in a lax approach to conversing about mathematics. However, even if we grant that many of these terms are too daunting to be taught in the elementary school, it is irresponsible to avoid teaching them at least in middle school.

Contextual Nomenclature

After all, soon after they enter high school, they will be introduced to contextual terminology. What do I mean? Consider the expression

Most people who have done some middle school mathematics would be able to say that, in the expression above, x is the base. But what do we call n? When this notation is first introduced, it is called a ‘power’. Later, the students are told that it is an ‘index’. Still later, they are told to call it an ‘exponent’. And after they are introduced to functions and their inverses, they are told that n can also be called a ‘logarithm’. Four different terms for the same quantity! But mathematics is an austere field. It does not do ‘synonyms’! So why do we have so many ways of naming the same thing? It has to do with the context within which the quantity is being referred to. A student who understands the contextual naming of the quantity will also understand the reason behind it. In other words, by introducing the student to contextual terminology, we facilitate his/her fluency in mathematical discourse and depth of mathematical understanding.

Giving No Quarter

But someone may wonder why I am so adamant that teachers should consistently use and teach their students to consistently use correct mathematical terminology. After all, would not a rose by any other name smell just as sweet? Indeed it would! But no one goes around describing a rose. We choose a name by convention and use it so that everyone understands what we mean. Hence, since the mathematical community has landed on a set of naming conventions, it behooves teachers to teach their students that convention of nomenclature.

Looking Back, Looking Ahead

The response to last week’s post on geometry was heartening. So I thought I would give the readers another small dose of geometry. Some of the ideas that arise from geometry are simple, yet so profound that it is hardly surprising that geometry was considered the pinnacle of human wisdom by many early mathematicians and philosophers. In the previous post I had expressed by consternation at the fact that, in some countries, mathematics is split into independent siloes, thereby rendering it difficult, if not impossible, for students to fully appreciate the subject. Here I wish to consider two geometry problems that demonstrate how connected the different branches of mathematics actually are, focusing today only geometry and algebra.

Areas in a Quadrilateral

Let ABCD be the convex quadrilateral, as shown, and let O be the point of intersection of its two diagonals. Suppose the area of △ABD is 1, the area of △BCA is 2 and the area of △DAC is 3. Find the areas of △CDB and △ABO.

As usual, I suggest the reader pause here and try to solve this question before proceeding.

We can start by assuming the area of △AOB is x. This would mean that the area of △AOD is 1 – x and that of △BOC is 2 – x. This would mean that the area of △COD is 2 + x. This is shown below.

We can quite easily see that the area of △CDB is the sum of the areas of △BOC and △COD. Hence, the area of △CDB is 2 – x + 2 + x = 4.

Now △AOB and △AOD have the same height. Their bases are OB and OD respectively. Similarly, △BOC and △COD have the same height and their bases are OB and OD respectively. Hence, the ratio of the areas of △AOB to △AOD is equal to the ratio of the areas of △BOC and △COD. This gives

Cross multiplying, expanding, and rearranging we get

Hence, the area of △AOB is 2/5.

Another way of solving the problem is to recognize that △ABC and △ADC have the same base AC. Hence, their areas (2 and 3 respectively) will be in the ratio of the lengths of BO and DO. Hence,

We can see that both solutions are relatively straightforward. However, it is clear that we have used algebra to solve the geometry question. Hence, if the mathematics curriculum is divided into separate study of algebra and geometry it will adversely affect the student’s ability to make both branches ‘speak’ to one another.

Stacked Trapeziums

The parallel sides of a trapezium have lengths 1 and 7, and the area of the trapezium is divided into two equal parts by a line segment parallel to those two sides. Find the length of that line segment.

I urge the reader to pause here and attempt the question before proceeding.

We proceed by extending the oblique sides until they meet at P. In this way they form triangles as shown below. We have also dropped a perpendicular from P intersecting AB, EF, and CD at points X, Y, and Z respectively.

Now it is clear that △PAB, △PEF and △PCD are similar. If we set EF = t, PX = x, PY = y, and PZ = z, we can obtain

Now the heights of the trapeziums ABFE and EFDC are XY = y – x and YZ = z – y respectively. We are given that the areas of the two trapeziums are equal. Hence, we can conclude

Using the earlier expressions for x, y, and z we get

Hidden Gems

In both problems we see that a little algebra gets us a long way. While it is certainly possible to solve both questions without algebra, given the ‘nice’ numbers involved, the use of algebra simplifies things quite a bit. More to the point, though I did not mention it, the two solutions to the first problem actually rely on the commutative property of multiplication. That is, when we say that the area of a triangle is half the product of the base and the height, it does not matter which side we take as the base and which measure as the height (as long as it is truly a height corresponding to the chosen base). Specifically, whether AC is the base and BO and DO proportional to the height or BD the base and AO and CO proportional to the height makes no difference.

In the case of the second problem, the completion of the triangle allowed the comparison of similar triangles. While this is a stock move in geometry, the result that involves the difference of squares is something that might not have been easy to predict. Despite this, the significance of the difference of squares, where the sums represent the lengths of the parallel sides and the differences represent the heights of the corresponding trapeziums is something that only the interplay between the geometry and the algebra could have revealed. If, however, the student is allowed to play with ideas from both branches, it is highly likely that she will make the links herself and understand the meaning of what she is doing.

These insights are examples of hidden gems of mathematical knowledge that will remain hidden if we do not allow the different branches to ‘speak’ to each other. If we treat the different branches as independent fields of inquiry, however, we will give students the impression that it is possible – or, God forbid, preferable! – to study the branches in isolation from each other and still be able to obtain deep insights about the subject and the ideas it contains.

The Presenting Problem

In response to the previous post, in which I had categorically declared that the use of calculators in mathematics education is a hindrance to students, a former student asked me, “Do you think it would be feasible or even beneficial to move mathematical assessments from the current manipulation & computation to something that focuses more on formulation and derivation from first principles from a given context (similar to IB DP AA HL paper 3)?”

To decipher the last few characters of the student’s question, ‘IB‘ stands for International Baccalaureate, ‘DP‘ for Diploma Programme, ‘AA’ for Analysis and Approaches, and HL for Higher Level.

In what follows, I will be addressing specifically what the IB does. However, this should not be taken as a critique of the IB alone, but of all the major exam boards around the world, like CAIE, Edexcel, AP, CBSE, and CISCE, just to name a few international and Indian boards. If you are reading in a country other than India, the UK, or the USA, please think of the exam boards that exist in your country.

What is Paper 3 Like?

Anyway, returning to my student, he was asking if I thought that the kind of questions that often appear in paper 3 for that syllabus would be something I considered ‘feasible’ and ‘beneficial’. Of course, many of you may be in the dark about what kind of questions might appear in this paper. If so, please click here to be taken to one example of paper 3.

If you read the paper, you will discover the following:

1. The question paper consists of 2 questions for a total of 55 marks and a duration of 1 hour.
2. The first question is worth 27 marks and combines geometry, sequences and series, and mathematical induction.
3. The second question is worth 28 marks and combines polynomials, complex numbers, coordinate geometry, and calculus.

As someone who loves mathematics, I must concede that I do find the Mathematics AA HL Paper 3 to be quite an interesting ‘animal’. As just mentioned, it gives the student problems in which different areas of mathematics are made to relate to each other. While the situations are contrived, the student is exposed to the possibilities that a little imagination can introduce to us.

Zoning Problems

Now, the IB has different papers for different time zones. In time zone 2 the students would have received this paper. If you read the time zone 2 paper you will discover the following:

1. The question paper consists of 2 questions for a total of 55 marks and a duration of 1 hour.
2. The first question is worth 28 marks and combines coordinate geometry, functions, and calculus.
3. The second question is worth 27 marks and combines polynomials, theory of equations, and complex numbers.

You may be wondering, “So what?” Of course the papers have to be different. Otherwise, the IB would not be able to administer different papers in different time zones, thereby risking the security of the question papers. And I fully agree.

But note that the time zone 1 paper has a significant number of marks devoted to mathematical induction, while the other one has marks devoted to the theory of equations. Mathematical Induction is a stand-alone topic that is somewhat weird to boot. So many students often decide to skip it and hedge their bets on it not being tested for too many marks. Theory of equations on the other hand is an integral part of mathematics and links with many other topics.

So consider four students, P, Q, R, and S. P and Q write exams in time zone 1, while R and S write in time zone 2. P and R regularly get a grade of 7 (the highest possible), while Q and S are borderline between 5 and 6 and have both chosen to ignore mathematical induction. Q, having ignored mathematical induction, will have a low score in paper 3 because he is writing a paper that tests that topic, while S, who also ignored the same topic, will not be affected since she is writing a paper that does not test that topic. P and R are not affected negatively in terms of their raw score. However, there will be little to distinguish between R and S since both are writing tests that do not include topics that S chose to ignore. Since many students writing exams in time zone 2 might have ignored mathematical induction without being negatively affected, the grade boundary for a 7 in time zone 2 will rise, thereby disadvantaging student R, who, for no fault of his own, is clumped with student S, who had hedged her bets. So what we have is some students, like S, in time zone 2 being unfairly favored, while others, like R, being unfairly disfavored.

The Problem of Grade Descriptors

Of course, the IB, like many exam boards around the world, claim to assess students on the basis of grade descriptors. But this is a hoax. There can be no grade descriptors when we are assigning final grades based on some numerical grade boundaries. If it were actually possible to relate grade descriptors to grade boundaries, we would actually not need both because everyone would automatically know how to translate from one to the other.

You see, I have experience marking IB mathematics exams and internal assessments. I know the difference between the two. The internal assessments are indeed marked according to a set of grade descriptors for each assessment criterion. However, the exam papers are not. In fact, if we are being honest, it is impossible to have a rigid mark scheme and also grade according to student achievement based on grade descriptors. You can do one or the other. You cannot do both. And the hoax that exam board the world over attempt to make us believe is that it is possible to do not just both, but also to be fair to all the students.

You see, the very idea of numerical grade boundaries that are determined after the exams are written and graded is that there is a post hoc determination of what separates the 6 from the 7. Why is this a fallacious approach?

The Problem of Grade Boundaries

It is impossible for every question to assess the student on every criterion. In other words, question 1 may assess a student on criteria A, B, and C, while question 2 may assess a student on criteria B, D, and E. However, when we say there is a grade boundary, we are saying that only the final mark matters and not where the student earned the mark. Hence, if criterion C is weighted more heavily than criterion D and E, then the student who gets a total of 40 marks with 25 marks question 1 and 15 in question 2 should receive a higher grade than a student who scores 15 in question 1 and 25 in question 2. All this is ignored when we assign grade boundaries based solely on the raw score.

What I am saying is that, for all their claims to assign grades based on some assessment criteria, the major exam boards are doing nothing of the sort. They have some highfalutin jargon that confuses people and dupes them into thinking the boards are actually assessing the students on criteria rather than relative to others. You see, what we all want is to know what a student has achieved because that tells us how much the student has learned. We do not want to know how she did relative to others because that tells us very little about her own learning. After all, if she is in the 95^th percentile when the average was 50 with a standard deviation of 10, then her raw score would be 66, meaning that she has ‘mastered’ about two-thirds of the syllabus. However, if she is in the 90^th percentile when the average was 70 with a standard deviation of 8, then her raw score would be 80, meaning that she had ‘mastered’ more than four-fifths of the syllabus. However, since the grade boundaries are based on the performance of the cohort, the board will determine that only a small fraction, say 8%, of the students should get the highest grade. Hence, the student who had only ‘mastered’ two-thirds of the syllabus would get a grade of 7 while the student who had ‘mastered’ four-fifths (i.e. 20% more) would get a grade of 6. You can check my working using the online inverse normal calculator here.

In other words, assigning grades based on numerical grade boundaries does precisely what the major international exam boards tell us they are not doing, that is assigning grades in a relative manner. If the grades are indeed criteria based, how is it that the criteria map so perfectly onto the raw marks that students achieve? Further, the practice of having different papers for different zones may mitigate against candidate malpractice for sure. But it seems it opens the door to ‘exam board malpractice’, something no one wishes to talk about! When I say ‘exam board malpractice’, I mean the practice of equating two things that appear equivalent only in the eyes of the exam board, but not to the students. If the exam boards are serious about grading the students on the basis of assessment criteria, then it is imperative that they move away from exam focused assessments.

The Proposed Revolution

Mind you, this is not the case of sour grapes. I am an exceptionally good test taker. And I have done well as a teacher to prepare my students not just with the knowledge to succeed at exams but also with the mental resilience that exams need.

However, for more than two decades now I have found that exams are dehumanizing. Yes, I have used a strong word. But I do this because no one ever finds themselves in ‘exam conditions’ outside school. The exam rooms are made pristine and silent, thereby placing at a disadvantage students who thrive in situations of controlled chaos and those who like to listen to music as they learn. Movement is prohibited, thereby disenfranchising those students who like to think as they walk or dance. Food is forbidden even though some students like a regular calorie burst to stimulate their minds. Everything that makes us human – food and movement and music – are forbidden from the exam room. Rather, the only acceptable involvement of the body is to hold a pen and move it along a piece of paper, something that humans have been doing only for the last 1% of their existence on this planet. The exam room forces humans to become just a bodiless head moving a torsoless hand. Exam rooms suit people like me – those who are able to concentrate in quiet environments and who are able to be seated for long stretches of time and who can go without food for ages.

Teaching and learning is a local affair. Even in this age of online classes, there is a ‘proximity’ of the student and the teacher. Exams strip the student of this support structure even though, in every job I can think of, one is free to consult with a more knowledgeable colleague or a mentor. Further, except in the most time sensitive jobs, one always has the freedom to ask for an extension, something that the very nature of an exam precludes.

But if teaching and learning are local affairs why should the assessment of learning not also be predominantly a local affair between a teacher and a student? That was how it was before the Industrial Revolution made us accept the production of graduates on an assembly line. That was how is was before we decided that large exam boards were better at gauging a student’s learning than the teachers who had taught the student.

I think it is time for a revolution in education. It is time we reaffirmed the agency and responsibility of the teacher not to teach to some predetermined syllabus but to teach what is relevant for the student’s plans and hopes for his/her future in accordance with the unique skills and talents and proclivities of each student in her/his care.

The Siege of Yodfat

In AD 67, in the middle of the first Jewish-Roman War, a rag tag band of forty Jewish revolutionaries managed to get themselves besieged in Yodfat. Realizing they were going to be captured and perhaps tortured, the group decided that they would commit mass suicide. However, given the Jewish aversion to committing suicide, none of them wanted to kill themselves. Also, some of them argued, if each person was given the responsibility of killing himself, there could be some squeamish rebels who would simply not carry through with the decision, leading to some of them losing their lives while others saved their own lives.

In order to avoid such rejection of their collective decision to kill themselves rather than being captured, they devised a plan where they cast lots and the lots told them how they were to arrange themselves. They would then arrange themselves in a circle in order of the lots. Then the first person would drive his spear through the second, killing him. Then the third person would drive his spear through the fourth, killing him. This would proceed until only one person was left and he was expected to kill himself. Hence, only the last person was expected to kill himself and end the mass suicide.

As it so happened, when thirty-eight of them had been killed, the last two decided that they would stop the killing and surrender to the Romans! One of the survivors was the Jewish historian Josephus, who narrates this story in The Wars of the Jews (Book 3, chapter 8, paragraph 7).

The Josephus Problem

This problem has come to be known as the ‘Josephus Problem’ because Josephus himself reports that he did not want to die but actually wanted to surrender to the Romans.

Hence, the problem can be framed as follows: Given forty people arranged in a circle, with each alternate person being killed by the person before him, what is the position you should be in to survive? We can generalize this as follows: Given n people arranged in a circle, with each alternate person being killed by the person before him, what is the position you should be in to survive? This involves skipping 1 person at each stage. However, the problem can be further generalized as follows: Given n people arranged in a circle, with k people being skipped and the person in position k+1 being killed by the person before him, what is the position you should be in to survive? Is there a solution to this general problem and, if so, how would we find it?

The Survivor Position

We start small! Suppose we have 3 people: A, B, and C. A kills B and C kills A. Hence, the survivor is in position 3. Suppose we have 4 people: A, B, C, and D. A kills B, C kills D, and A kills C. Hence, the survivor is in position 1. If we continue in this way we will obtain the following table

We should be able to observe some patterns here. No one in an even position is ever the survivor. This is because, when we skip just one person each time, all the even positions are killed in the first round itself. Next, whenever the number of people equals a power of 2 (i.e. 2, 4, 8, 16, etc.), the survivor is in position 1. Between two consecutive powers of 2 (i.e. 2 and 4, or 4 and 8 or 8 and 16, etc.), the survivor position proceeds according to successive odd numbers.

Hence, we can obtain the following algorithm:

1. Given, the number of people (n), determine the largest power (say m) of 2 that is less than or equal to n.
2. Calculate p = n – 2^m + 1.
3. The survivor will be in position s = 2p – 1.

Let’s test this algorithm. Say n = 13. Now 2³ = 8 < 13 < 16 = 2⁴. Hence, m = 3. This gives us p = 13 – 2³ + 1 = 6. This gives, s = 2(6) – 1 = 11, which is the survivor position according to the table.

Say n = 16. Now 2⁴ = 16. Hence, m = 4. This gives us p = 16 – 2⁴ + 1 = 1. This gives, s = 2(1) – 1 = 1, which is the survivor position according to the table.

For the situation in which Josephus found himself n = 40. Now 2⁵ = 32 < 40 < 64 = 2⁶. Hence, m = 5. This gives us p = 40 – 2⁵ + 1 = 9. This gives, s = 2(9) – 1 = 17. Hence, Josephus would have been safe if he had drawn the lot for the 17^th position.

The Second Last Survivor Position

However, remember that Josephus was one of two who remained and who decided to renege on their commitment to the fallen comrades. So he need not have been the last man standing. He could have been the second last man, the one whose fate is would have been to be killed by the man in the 17^th position. What position would this second last man have held? We can generate a similar table as before to obtain

The first line is in red because of the strange case in which a person occupying an even numbered position survives. However, this is only because we are starting with 2 people. If we observe the third column, we observe once again that, apart from the anomalous ‘2’, there are only odd numbers. Further, we can see that the position resets to 1 whenever we reach 6, 12, 24, etc. This is a geometric sequence and the general term can be express as

Between successive terms of this series, the third column merely goes through all the odd numbers. So we can obtain the following algorithm:

1. Given, the number of people (n), determine the largest term the series u_k that is less than or equal to n.
2. Calculate q = n – u_k + 1.
3. The second last survivor will be in position t = 2q – 1.

Let’s test this algorithm. Say n = 13. Now 6×2^2-1 = 12 < 13 < 24 = 6×2^3-1. Hence, k = 2. This gives us q = 13 – 6×2^2-1 + 1 = 2. This gives, t = 2(2) – 1 = 3, which is the second last survivor position according to the table.

Say n = 24. Now 6×2^3-1 = 24. Hence, k = 3. This gives us q = 24 – 6×2^3-1 + 1 = 1. This gives, t = 2(1) – 1 = 1, which is the second last survivor position according to the table.

For the situation in which Josephus found himself n = 40. Now 6×2^3-1 = 24 < 13 < 48 = 6×2^4-1. Hence, k = 3. This gives us q = 40 – 6×2^3-1 + 1 = 17. This gives, t = 2(17) – 1 = 33. Hence, Josephus would have been the second last survivor if he had drawn the lot for the 33^rd position.

Throwing Down the Gauntlet

So we have now obtained positions for the last and second last survivors. However, what we have done is look at the tables and use them to obtain the ‘algorithms’ for the positions. We have not used any rigorous mathematics. What we have done is play around with the situations and keep an eye out for patterns. While there are certainly ways of deriving these ‘algorithms’, we can see that we do not need any rigorous mathematics to pull off something quite remarkable. For example, suppose we have n = 1000. Imagine, a circle of one thousand people. What would we obtain?

Using the first algorithm we would get: 2⁹ = 512 < 1000 < 1024 = 2¹⁰. Hence, m = 9. This gives us p = 1000 – 2⁹ + 1 = 489. This gives, s = 2(489) – 1 = 977. Using the second algorithm we would get: 6×2^8-1 = 768 < 1000 < 1536 = 6×2^9-1. Hence, k = 8. This gives us q = 1000 – 6×2^8-1 + 1 = 233. This gives, t = 2(233) – 1 = 465. Hence, the last survivor is the person in position 977 and the second last survivor the one in position 465.

Now you know how to set up the problem and draw up the tables, can you obtain the ‘algorithms’ is we skip not one but two people each time? What kind of patterns would we see?

Confession

Playing with numbers is one of my favorite ways to pass time. Of course, we can get to numbers per se in a variety of ways. One can be from simply doodling on a page. I remember drawing grids of dots on a page and playing all kinds of games with my classmates when I was in school. I’d like to say that this was only during breaks or free periods, but I know I won’t be able to pull the wool over your eyes, at least in this regard. Those of you who know me personally also know the glint that appears in my eyes when I am up to no good. Oh well!

As the well known saying goes, “An idle mind is the devil’s workshop.” And I often found myself bored stiff in many of my classes. In some of them I did take a perverse pleasure in disrupting the class. This was in classes where I had some kind of antagonism toward the teacher. But in other classes, where I found the teacher likable, I would engage in my random play, one kind of which was drawing grids and either contriving new games to play on the grid or discovering intriguing patterns.

Anyway, during my dot grid days I would attempt to come up with different types of grids. And of course, as mentioned earlier, I had to count the number of dots and recognize patterns. It was only much later that I realized that my (successful) efforts at distracting myself in some classes were actually leading to a common recreational mathematics idea – the centered square numbers. Let me explain.

Introduction to Centered Square Numbers

The centered square starts with a single dot. Each successive centered square is obtained then by surrounding the existing centered square with another layer of dots to complete a new square. The first four members of this sequence are shown in the figure below.

Allow me to explain the notation. In the term C_4,3, the ‘4’ refers to the number of sides in the regular polygons we are dealing with. In this case, since we are considering centered squares, the first number is a ‘4’, indicating a quadrilateral. The ‘3’ refers to the third member in the sequence. The C indicates the count of the number of dots.

In the figure, apart from the first member, each figure has both red dots and grey dots. If you focus on either the red or gray dots, you will notice that the set of dots of that color form a square. So in the third member, the red dots form a square of side 2 dots while the gray dots form a square of side 3 dots. Similarly, in the fourth member, the gray dots form a square of side 3 dots and the red dots form a square of side 4 dots. This is what is indicated on the right side of each equation. For the third member, 4 is the number of dots in a square of side 2 while 9 is the number of dots in a square of side 3. Similarly, for the fourth member, the square of side 3 has 9 dots while the square of side 4 has 16 dots.

We can now add the numbers on the right side of the equations to get

From this it is clear that the number of dots in the n^th member is equal to the sum of the squares of n and n – 1. Hence, we can write

Fruits of Algebraic Massaging

With a little bit of algebraic ‘massaging’ we can show that

What this tells us is that the n^th centered square number is equal to half the sum of the square of the n^th odd number and 1. This can be illustrated as in the diagrams below:

Here the dots in gray indicate the count for the centered square while the total number of dots is the square of the n^th odd number.

We can also ‘massage’ the expression for C_4,n as follows:

Now, if we consider the sum of the first n – 1 natural numbers we can see that

This means that the n^th centered square number is one more than four times the sum of the first n – 1 natural numbers. We can depict this as follows:

Here, the central black dot represents the 1 that is added at the end. The remaining dots are divided into 4 triangles, 2 in gray and 2 in red. Each of these triangles represents the sum of the first n – 1 natural numbers.

Further, since

we can conclude that each centered square number is necessarily odd. We could obtain this intuitively from the fact that the centered square number is the sum of consecutive square numbers. Since one number must be odd and the other even, this must mean that we are adding an odd number, since the square of an odd number is odd, and an even number, since the square of an even number is even.

Of course, given the m^th odd number, we can obtain

This means that the square of an odd number is always 1 more than a multiple of 4. This leads easily to the conclusion that all centered square numbers are 1 more than a multiple of 4. This was depicted in the previous diagram since there are 4 identical triangles and the single black dot.

Absolution

As I played around with the grid of dots in my early days, I did recognize some of these patterns. Not all, of course. But it was quite intriguing that the simple endeavor of making a grid of dots could yield so many interesting properties. This began out of boredom in the classroom. I am confident that countless many of the discoveries and inventions we now know about and enjoy had their beginnings in the boredom of a child in a class he/she was forced to attend. If this is the case, I think I can be absolved for my boredom fueled explorations!

Inhale

Ok. It’s time for a little breather. On 17 March 2024 we began a series of posts on e. Then we moved to another series on principles of counting. Then last week we completed a series on π. All these were quite heavy and I’m sure some of you are still reeling from the shock caused by the posts on π. So for a few weeks I think we should lighten up a bit. I will dedicate the next few posts to recreational mathematics.

From as long as I can remember, I have enjoyed playing with numbers. I enjoyed performing all sorts of operations on them in the attempt to recognize patterns. At an early age, and I believe independent of any external influence, I noticed that when you multiply single digit numbers by nine, the digit in the units place keeps reducing by 1 while the digit in the tens place increases by 1.

Playing with numbers is exhilarating. The joy that one can get from a few simple operations cannot be expressed. In this post, I wish to give you one way I have occupied myself for endless hours and one easy but mathematically based trick. The challenge is to use four copies of the same number (from 1 to 9) and any of the mathematical operations (+, −, ×, ÷, a^b, and √a), with as many sets of parentheses to make as many numbers as possible.

Massaging the Numbers

For example, we can choose to use four ‘2’s. In that case, we can have:

(2 ÷ 2) ÷ (2 ÷ 2) = 1

(2 ÷ 2) + (2 ÷ 2) = 2

(2 + 2 + 2) ÷ 2 = 3

(2 + 2) ÷ 2 × 2 = 4

2 × 2 + 2 ÷ 2 = 5

2 × 2 × 2 – 2 = 6

2^{(2 + 2 ÷ 2)} = 8

(2 + 2 ÷ 2)² = 9

2 × 2 × 2 + 2 = 10

You get the idea. Is there a way of getting 7 as the answer? What about 11? What’s the largest number you can get using four ‘2’s?

We could repeat the same with another number. Try it on your own. Believe me, it keeps your mind engaged and, when you make a breakthrough, there is a serious dopamine rush.

A Constant Result

Now concerning the trick I mentioned earlier. Ask someone to choose a three digit number where the hundreds digit and the units digit as not the same. Now ask them to reverse the order of the digits. So if they choose 356, after reversing, they get 653.

Now ask them to subtract the small from the larger. With the above numbers 653 – 356 = 297. (If they get a two digit number ask them to place a 0 in the hundreds place. So if they had chosen 433, they would have 433 – 344 = 099.) Ask them to reverse the digits again and add the two numbers. So with the original number we chose we would get 297 + 792 = 1089. (If they had 099, they would now get 099 + 990 = 1089.) But tell them not to give you any of these results.

Now, the answer is always 1089. So you could keep a book with around 200 pages and memorize the 9^th line on page 108. So when they have finished adding, tell them to turn to the page indicated by the first three digits (i.e. 108) and check the line indicated by the units digit (i.e. 9). Now you read out the line you have memorized and bamboozle everyone. Of course, since the answer is always 1089, you cannot do this trick with the same person multiple times.

But why does it work? Say we have a in the hundreds place, b in the tens place, and c in the units place. For now let’s assume that a > c. Then the value of the original number is 100a + 10b + c

When we reverse the order, we get a number whose value is 100c + 10b + a. When you subtract this from the original number you will get 100(a – c) + (c – a).

Now since a > c, it followed that c – a is negative. Hence, when we subtracted, we would have had to borrow. However, from the form 100(a – c) + (c – a), there is nothing that represents the tens place. So we will have to borrow 1 from the hundreds place, and then a 1 from the tens place, giving

100(a – c) + (c – a) = 100(a – c – 1) + 10 × 9 + (10 + c – a)

When we reverse this number we get 100(10 + c – a) + 10 × 9 + (a – c – 1)

Now when we add the last two numbers we get

100(a – c – 1 + 10 + c – a) + 10 × 9 + 10 × 9 + (10 + c – a + a – c – 1)

which on simplification gives 900 + 180 + 9 = 1089

Exhale

As we have seen, numbers provide us with a lot of entertainment. We will continue to look at this lighter side of mathematics in the few posts that follow. Don’t forget to try your hand with four ‘3’s or four ‘4’s. And astound someone who does not read this blog with the trick. And tell them to come over here and read the other posts. I’ll see you next week.

Recapitulation

We are in the middle of a series of posts on π. Our journey began with A Piece of the Pi which was followed by Off On a Tangent. Last week, I posted The Rewards of Repetition in which I promised that we would break down the Gauss-Legendre algorithm into the distinct parts. Of course, as mentioned in the previous post, the algorithm uses mathematics that is way above the level of the blog. I will not be explaining any of that. I will only explain those parts that are at the level of this blog. So let’s proceed.

The Gauss-Legendre Iteration

First, let us recall the algorithm. The Gauss-Legendre algorithm begins with the initial values:

The iterative steps are defined as follows:

Then the approximation for π is defined as

The question, of course, is why this process even works. Surely this can’t be some random method that Gauss and Legendre independently stumbled upon! So what is the rationale behind this strange algorithm? After all, if we are honest, there is nothing in the steps that lead self-evidently to a conclusion that the iterations would converge to π. So let us consider the steps involved and why these initial values and iterative method leads to the approximation of π.

Comparison of Means

The method begins with the realization that, given to distinct positive numbers, their arithmetic mean (AM) is always greater than their geometric mean. Of course, the AM and GM are defined as

So we can proceed as follows since a and b are distinct

Convergence of a_n and b_n

Now the iterative formulas

Assure us that

and

This means that the sequence b_n is steadily (monotonically) increasing while the sequence a_n is steadily (monotonically) decreasing. Since we start with the fixed values of a₀ and b₀, this means that the two sequences will necessarily converge.

Now if we define

we can obtain

which, further yields

This means that the sequence c_n also converges. And since c_n cannot be negative, it must converge to 0. This means that a_n and b_n converge to the same value. Since this common value is obtained by converging series of arithmetic and geometric means, it is called the arithmetic-geometric mean (AGM). Let us designate it with m.

The Magic of Calculus

The next stage involves some calculus. Since I haven’t introduced calculus in the blog yet, let me just present the results. Those who wish to read in detail can look at the proofs here. Anyway, the method begins by defining

Then with some nifty substitutions and change of limits, we are able to prove

The observant reader will now realize why we had done all the stuff related to AM and GM and the convergence of the same. Now it is a trivial step to conclude that

From this we can conclude that

where, m, as defined earlier, is the AGM of a₀ and b₀. What this means is that

and

Now using the Gamma and Beta functions it is possible to prove

What this means is that if we start with the initial values as stated earlier, each iteration of

gets us closer to the value of π.

The iterative process of the Gauss-Legendre method converges so quickly that, after only 25 iterations, the algorithm generates 45 million digits of π.

Conclusion

I apologize to the reader for not including all the calculus behind the algorithm. I have done that in the interests of not making this post too heavy. I know that it is still quite heavy. But that is the nature of the beast! There are no rapidly converging methods for approximating the value of π that do not involve a lot of calculus. Indeed, some algorithms derived using calculus are really quite strange – so strange that even the Gauss-Legendre algorithm would seem quite sensible! We will tackle some of them in the next post, after which I will lay this π to rest.