Tests of divisibility - 2

I asked last time, "What is the significance of $2$ as a multiplier in the test for divisibility by $7$?'' I also asked, ``Are there tests similar to this for testing divisibility by other divisors - like $9$, $11$, $13$, $17$, $19$, ..."


The answer to the latter question is a loud "Yes!". Indeed, one can find such a test for any divisor which has no factors in common with $10$; that is, any divisor whose units digit is $1$, $3$, $7$ or $9$.


The logic behind this gets revealed when we understand the role played by the multiplier $2$ in the test for divisibility by $7$.


One way to understand the role played by $2$ is to note that $7$ has $21$ as a multiple (note that $21$ has $2$ as its tens digit), and if we apply the transformation $10a + b \mapsto a - 2b$ to $21$, we get $0$ right away. We also get $0$ if we apply it to multiples of $21$ like $42$, $63$, $84$, $105$, $126$, ... Try it out and you'll see this for yourself.


Noting this, we are led to invent a new test of divisibility. Let $k$ be any positive integer. Consider the following function $f_k$ which acts on the positive integers $x$ thus: If $x = 10a + b$, where $0 \le b \le 9$, then $f_k (x) = a - k b$. Then:

• $x$ is divisible by $10k + 1$ if and only if $f_k (x)$ is divisible by $10k + 1$.

To see that this claim is valid, observe that
\[kx + f_k (x) = k(10a + b) + (a - kb) = a(10k + 1).\]
Now:

• If $x$ is divisible by $10k + 1$, then so is $kx$, and so also is $f_k (x)$ from the above relation.
• If $f_k (x)$ is divisible by $10k + 1$, then so also is $kx$, from the above relation. And since $k$ and $10k + 1$ are coprime (they must be, because $10k + 1$ leaves remainder $1$ when divided by $k$), this means that $x$ itself is divisible by $10k + 1$.

It follows that $x$ is divisible by $10k + 1$ if and only if $f_k (x)$ is divisible by $10k + 1$.


From this observation we quickly get many different tests of divisibility, all at the same time (in fact, infinitely many of them):

• Putting $k = 3$, we get a test of divisibility by $31$: The integer $10a + b$ is divisible by $31$ if and only if $a - 3b$ is divisible by $31$. So the test is: Subtract three times the units digit from the rest of the number, and proceed as earlier.
• Putting $k = 4$, we get a test for divisibility by $41$: The integer $10a + b$ is divisible by $41$ if and only if $a - 4b$ is divisible by $41$. So the test is: Subtract four times the units digit from the rest of the number, and proceed as earlier.
• Similarly, $k = 6$ gives a test for divisibility by $61$, $k = 7$ gives a test of divisibility by $71$, and so on.

What about $k = 2$? This gives us a test for divisibility by $21$. But $21 = 3 \times 7$ is a composite number. So the same test serves as a test for divisibility by both $3$ and $7$. Hence:

• The number $10a + b$ is divisible by $3$ if and only if $a - 2b$ is divisible by $3$.
• The number $10a + b$ is divisible by $7$ if and only if $a - 2b$ is divisible by $7$.

Similarly, if $k = 5$ we get another composite number, $51 = 3 \times 17$. Hence:

• The number $10a + b$ is divisible by $17$ if and only if $a - 5b$ is divisible by $17$.

(Note that from $k = 5$ we get another test for divisibility by $3$; however a bit of thought will show that it is equivalent to the test obtained by taking $k = 2$. So we need not list it here.)


If $k = 9$ we get yet another composite number, $91 = 7 \times 13$. So from this we get a test for divisibility by $13$:

• The number $10a + b$ is divisible by $13$ if and only if $a - 9b$ is divisible by $13$.

Two values of $k$ of particular interest are $k = 1$, and $k = 8$. Let us see why.


If $k = 1$ then $10k + 1 = 11$. The test now is:

• The number $10a + b$ is divisible by $11$ if and only if $a - b$ is divisible by $11$.

Now the mapping $10a + b \mapsto a - b$ amounts to this: Subtract the units digit from the rest of the number.


I wonder if you see that this test is just a disguised form of the usual test for divisibility by $11$.


If $k = 8$ then $10k + 1 = 81 = 9 \times 9$; so this should give us a test for divisibility by $9$. The test is:

• The number $10a + b$ is divisible by $9$ if and only if $a - 8b$ is divisible by $9$.

Now, it is obvious that $a - 8b$ is divisible by $9$ if and only if $a + b$ is divisible by $9$ (because $(a + b) - (a - 8b) = 9b$, which is a mutiple of $9$). Hence:

• The number $10a + b$ is divisible by $9$ if and only if $a + b$ is divisible by $9$.

Now the mapping $10a + b \mapsto a + b$ amounts to this: Add the units digit to the rest of the number.


I wonder if you see that this test too is just a disguised form of the usual test for divisibility by $9$.


In this manner we can get a test for divisibility by any number with units digit $1$, and by any number which divides a number with units digit $1$.


A very similar line of reasoning gives tests for divisibility by numbers with units digit $9$; i.e., divisors like $19$, $29$, .... But we'll leave this to the next post.

Tests of divisibility - 1

Everyone knows the tests for divisibility by $2$ and $5$; they are obvious. The tests for divisibility by $4$, $8$, $3$, $6$, $9$ and $11$ are less obvious but still well known; each one is based in some way on the digits of the given number. (The logic behind these tests may be less clear - particularly for divisibility by $11$. But we expect that most students have at least an inkling of the underlying logic. I hope so anyway!) 

What about divisibility by $7$? Some of you may know of this test: 

Given an integer $N$, let $b$ be its units digit, and let $a$ be the rest of the number, obtained by deleting the units digit; this means that $N = 10a + b$. Now replace $N$ by $a - 2b$. In other word, subtract twice the units digit from the rest of the number. Now repeat the same step for the new number, and keep doing this till you get a small number $M$ for which you can check divisibility by $7$ mentally. If this check works out (i.e., $M$ is divisible by $7$), then the original number $N$ is divisible by $7$; else it is not. 

Thus this is an ``if and only if'' test. Here are two examples:

• If $N = 123456$, then we do $12345 - 12 = 12333$. Next we do: $1233 - 6 = 1227$. Next: $122 - 14 = 108$, and $10-16 = -6$. The last number ($-6$) is clearly not a multiple of $7$; hence, neither is the original number, $123456$.
• Or try $N = 1504055$. We get: $150405 - 10 = 150395$. Next: $15039 - 10 = 15029$. Next: $1502 - 18 = 1484$. Next: $148 - 8 = 140$. We can stop now, since $140$ is quite visibly a multiple of $7$. We conclude that $1504055$ too is a multiple of $7$.

Why does this work? Here is a proof; it uses two simple facts: 

• The sum and difference of two multiples of $7$ are also multiples of $7$.
• If $y$ is an integer such that $3y$ is a multiple of $7$, then $y$ itself is a multiple of $7$. (Do you see why? The point is that $3$ and $7$ are relatively prime to each other.)

Let us now show that for any two integers $a$ and $b$, the quantities $x = 10a+b$ and $y = a-2b$ are either both divisible by $7$ or both indivisible by $7$. For we have:

\[x - 3y = (10a + b) - 3(a - 2b) = 7a + 7b = 7(a + b),\]

which tells us that $x-3y$ is a multiple of $7$. Now:

• If $x$ is a multiple of $7$, then $3y$ is a multiple of $7$, and hence so is $y$.
• And if $y$ is a multiple of $7$, then so is $3y$, and hence so is $x$.

This reasoning carries right through the entire working of the algorithm. 


Conclusion. The initial number is a multiple of $7$ if and only if the final number is a multiple of $7$.

So this is why the test works, but the proof raises several more questions:

1. What is the significance of $2$ as a multiplier in the above test (i.e., "twice the units digit")? How could we have found this multiplier?
2. Are there tests similar to this for testing divisibility by $9$? $11$? $13$? $17$? $19$?

We'll try to answer these in the next entry! Cheers till then.

The problem of the four 2's

I am basing today's entry on an entry I saw in another blog: Kolipakkam's Is there a PAM Dirac amongst us?  This in turn was brought to my attention by my colleague and good friend B Sankararaman of The Valley School.


A type of puzzle frequently seen is:

• Using exactly four $2$'s (no less and no more), make the number $5$ (or $6$ or $10$ or $50$ any other given number).

Or:

• Using exactly four $4$'s (no less and no more), make the number $5$ (or $6$ or $10$ or $50$ any other given number).

Or:

• Using the digits $1$, $9$, $4$, $7$ exactly once each (no less and no more), make the number $5$ (or $6$ or $10$ or $50$ any other given number).

Obviously one can make an unlimited number of such puzzles. A popular pastime is to do this while on the road, using the license plate number of the car or bus just ahead of you. Or one can use it to set an exercise for one's students, using the digits of a particular year. This can be a lot of fun at, say, the 7th or 8th standard level.


Here I will make some remarks on the first problem (where we try to make the successive positive integers using just four $2$'s). A lot of fun can be had generating the answers:

• $$1 = \frac{2}{2} \times \frac{2}{2}$$
• $$2 = \frac{2}{2} + \frac{2}{2}$$
• $$3 = 2^2 - \frac{2}{2}$$
• $$4 = 2^2 \times \frac{2}{2}$$
• $$5 = 2^2 + \frac{2}{2}$$
• $$6 = 2 + 2 + \sqrt{2 \times 2}$$
• $$8 = 2^2 + 2^2$$
• $$9 = \left(2 + \frac{2}{2}\right)^2$$

One can go on like this, generating the positive integers one after another (but note that we skipped the entry for $7$). At some point one is bound to ask, 

• Can this go on for ever? Can we express every integer using precisely four $2$'s? Or are there some integers for which the representation cannot be found?

Unfortunately this problem has been solved once for all! - there is a simple algorithm which yields the answer for every positive positive integer $n$. Here is how it works. Let

\[x = 2^{1/2^n}.\]

Note that $x$ depends on $n$, and can be written using just one $2$ and $n$ square root signs:

\[x = \sqrt{\sqrt{\sqrt{ \cdots \sqrt{2}}}},\]

with $n$ square root signs nested inside each other. By taking logarithms to base $2$ we get:

\[\log_2 x = \frac{1}{2^n}.\]

Inverting this relation we get $2^n = 1/\log_2 x$, and hence:

\[2^n = \log_{x} 2.\]

Now take logarithms to base $2$ yet again; we get:

\[n = \log_2 \left( \log_{x} 2 \right).\]

Now putting in the expression we had for $x$ (recall that it used just one $2$ and $n$ square root signs), we get an expression for $n$ using just three $2$'s!

For example, for $n = 5$ we have:

\[5 = \log_2 \left( \log_{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{2}}}}}} 2 \right).\]
And for $n = 7$:

\[7 = \log_2 \left( \log_{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{2}}}}}}}} 2 \right).\] 
Wow!

Note that this solution uses just three $2$'s. But we can very easily convert this to a solution using four $2$'s by (wastefully) writing one of the $2$'s using two $2$'s, say as $\sqrt{2 \times 2}$.


I am told that this solution was given by the great physicist Paul Dirac (who in spirit was a mathematician rather than a physicist; or maybe I should say that he liked to think of the world in purely mathematical terms rather than in physical terms). The story is told more fully in the blog I referred to above (Kolipakkam's Is there a PAM Dirac amongst us?). Do have a look at it.


Why do I call this "unfortunate"? Well, you don't have to take that seriously! On the one hand, Dirac's solution is neat and beautiful. At the same time it is always a matter of regret when a problem gets solved so completely that there is nothing for anyone to add any more ...! Would you agree?

A pattern of signs

Take any four consecutive integers $a, b, c, d$. Then it is obvious that $a+d = b+c$. Let us write this relation as 
\[+a-b-c+d = 0.\]
Note the sequence of signs on the left side: $+,-,-,+$. We shall write the above statement in the following form: 

The sign sequence $+--+$ if applied to any $4$ consecutive integers yields a sum of $0$.

For example, if the 4 consecutive integers are $12$, $13$, $14$, $15$ we have: 
\[+ 12 - 13 - 14 + 15 = 0.\]
Now reverse all the signs in this string; we get the string $-++-$. If we concatenate the two strings together we get the string $+--+-++-$ (which is now of length $8$). Now here is a surprising fact:

The sign sequence $+--+-++-$ if applied any $8$ consecutive squares yields a sum of $0$.

For example, take the $8$ consecutive squares $9$, $16$, $25$, $36$, $49$, $64$, $81$, $100$. We have now:
\[+ 9 - 16 - 25 + 36 - 49 + 64 + 81 - 100 = 0.\]
Please verify this statement. Then try it out on other collections of $8$ consecutive squares.


Now take the string $+--+-++-$ and reverse all its signs; we get the string $-++-+--+$. If we concatenate these two strings together, we get the string $+--+-++--++-+--+$ (which is now of length $16$). Here is the next surprising fact:

The sign sequence $+--+-++--++-+--+$ if applied any $16$ consecutive cubes yields a sum of $0$.

Here is an example: take the $16$ consecutive cubes $27$, $64$, $125$, $216$, $343$, $512$, $729$, $1000$, $1331$, $1728$, $2197$, $2744$, $3375$, $4096$, $4913$, $5832$. Applying the sign sequence just given to these numbers, you will find you get a sum of $0$. I have not written the full "sum" here as it does not fit into this space properly: $+ 27 - 64 - 125 + 216 - 343 + 512 + 729 - 1000 - 1331 + 1728 + 2197 -2744 + 3375 - 4096 - 4913 + 5832 = 0$. But please check it out.

Then try this out for other collections of $16$ consecutive cubes.


I suppose you will now be able to guess what comes next in this sequence of statements .... 


But why do we have such a pattern?

Update on the 120 degree triangle

In the piece that I uploaded to my website (mathcelebration) a few days back, about a property that triangles with a $120^{\circ}$ angle possess, I made the following remark: 


"The pure geometry proof given above is very finely dependent on the hypotheses, and is difficult to generalize (this is a common feature of many such geometric proofs)."


I have just found that this somewhat off the cuff remark was not justified; in fact I have found a pure geometry proof of the converse proposition of that article.


That is, I have found a "pure geometry" way of showing this proposition: if $\angle A$ is not equal to $120^{\circ}$ then $\angle QPR$ is not a right angle. And to my surprise it is a very easy proof!


Here are the links:

• Triangle120degree2Screen
  
• Triangle120degree2PDF

Do have a look at them!

Catalog of MAA Publications 2010 Annual

Catalog of MAA Publications 2010 Annual

Trial math entries

After a bit of web search I found a blog which had something to say precisely about writing math text in blogspot using LaTeX. 


I have immediately tried to implement what I saw, and the results may be seen in the entry just below this post, on "Non-elementary integrals". It looks good on my screen (but I use the latest version of Firefox) but I do not know how it will look on other browsers, or on those without math fonts installed. Feedback wanted from readers!

Non-elementary integrals

It is a strange fact that many natural looking integrals turn out to defy solution. Students often wonder how they can be done, and are baffled by them. Unlike the case with differentiation, we seem to run into a stone wall very easily while doing integration.


There is a good reason for this bafflment: some of these integrals can be shown to be "non-elementary"! This means they cannot be expressed in terms of the usual functions we know (polynomials, rational functions, trigonometric, logarithmic and exponential functions, and all possible combinations of these). And these claims can actually be proved! Consequently, try as we might, we will not be able to do these integrals in the way we have gotten used to. The only way is to introduce entirely new functions.


Here are some integrals that fit this description: 

• $$ \int e^{x^2} \, dx $$
• $$ \int \frac{\sin x}{x} \, dx $$
• $$ \int \sqrt{\sin x} \, dx $$

On the other hand, there are some definite integrals involving these very same functions which can be done in an elementary way. For example, we have the following very famous identities involving definite integrals: 

• $$ \int_{-\infty}^{\infty} e^{-x^2} \, dx = \sqrt{\pi} $$

(this integral arises in the study of the normal distribution in probability theory) and 

• $$ \int_{-\infty}^{\infty} \frac{\sin x}{x} \, dx = \pi. $$

But the anti-derivatives of both $e^{-x^2}$ and $\sin x/x$ are non-elementary! So the proofs of these two identities (and others like them) involve completely different ideas, not involving anti-differentiation.


Here is another such example - a rather spectacular identity (first discovered by one of the Bernoullis, I think): 

• $$ \int_0^1 \frac{1}{x^x} \, dx = \sum_{n=1}^{\infty} \frac{1}{n^n}. $$

The last one may be done using integration by parts, after first writing $x^x$ as $$e^{x \ln x}$$.


Here are some links for those wishing to read more on this topic:

• Non-elementary integral (Wolfram)
• Non-elementary integral (Wikipedia)
• Non-elementary integrals

Note on notation

As I observed once earlier, it is tricky to write mathematical text in a blog; there is no good way to write mathematical symbols (I do not think the blog software allows it; if any reader knows a way out of this, please let me know!). So I have to innovate. 

Here are the symbols I will use in this and future entries. (Some of you may recognize that I am merely following the input protocol of a computer algebra software which I use a great deal: Derive. I find it a very simple and logical notation.)

1. To denote the derivative of a function f(x) with respect to x, I write dif(f(x), x). For example, dif(x^2, x) = 2x.
2. The symbol dif(f(x), x, 2) denotes the second derivative of f(x) with respect to x; dif(f(x), x, 3) denotes the third derivative, and so on. For example, dif(x^3, x, 2) = 6x.
3. To denote the indefinite integral of f(x) with respect to x, I write int(f(x), x). For example, int(x^2, x) = x^3/3.
4. To denote the definite integral of f(x) with respect to x, evaluated between two given limits a and b, I write int(f(x), x, a, b). For example, int(x^2, x, 0, 1) = 1/3.
5. To denote the limit of f(x) as x tends to a, I write lim(f(x), x, a). For example, lim((x^2-a^2)/(x-a), x, a) = 2a.
6. If I want to specify the direction of approach, I use an additional symbol: lim(f(x), x, a, -1) indicates that x approaches a from the left, and lim(f(x), x, a, 1) indicates that x approaches a from the right.

More such notation will follow in due course. I think the systematic use of such notation will allow us to communicate more easily with each other.

Two proofs uploaded to my website

I have just uploaded the proofs to the statements made in two earlier posts:

• Curious occurrence of the powers of 4
• Property of a 120 degree triangle. 

Here are the links:

• Powers of 4 (screen)
• Powers of 4 (pdf)
• 120 degree triangle (screen)
• 120 degree triangle (pdf)

Comments are welcome! I am particularly interested in generalizations or variations.

Book lists

I often get asked about lists of books for studying mathematics at various levels. I have compiled such a list, and display it below. I will keep adding to it, as and when I think of suitable titles. 


Readers are very welcome to suggest their favorite titles - I will add these too to the list. 


Remark. 
It is quite difficult to keep such a list to a modest, manageable size - there are so many books worth looking at, and worth studying. One has to decide for oneself how much one wants to spend on books. 


Some of these books may well be available on the Web, free, as pdf or djvu files. But for serious study, it is difficult to study off a computer screen, at least for me :-). 


So either one must print the file in its entirety, or take the trouble to purchase a regular copy.


Here is the list, subdivided into various categories. Some are given in the form of web links.



I. Books by Dan Pedoe

1. Geometry: A Comprehensive Course (Dover)
2. Circles: A Mathematical View (Cambridge)
3. Geometry and the Visual Arts (Dover)
4. Geometry and the Liberal Arts
5. The Gentle Art of Mathematics (Pelican)

See this web page too:

1. [Dan Pedoe]

II. Books by W W Sawyer

1. Mathematicians Delight (Penguin)
2. Prelude to Mathematics
3. Introducing Mathematics: Vision in Elementary Mathematics (Penguin)
4. Introducing Mathematics: A Path to Modern Mathematics (Penguin)
5. Introducing Mathematics: The Search for Pattern (Penguin)
6. A Concrete Approach to Abstract Algebra
7. What is Calculus About? (MAA)

III. Books by Ross Honsberger
See this web page:

1. [Ross Honsberger] 

IV. Books by Martin Gardner

1. Mathematical Puzzles and Diversions
2. Mathematics Magic and Mystery

See these web pages too:

1. [Martin Gardner1]
2. [Martin Gardner2]
3. [Martin Gardner3]

V. Books by Shailesh Shirali

1. A Primer on Logarithms (Univ Press)
2. A Primer on Number Sequences (Univ Press)
3. First Steps in Number Theory: A Primer on Divisibility (Univ Press)
4. Adventures in Problem Solving (Univ Press)
5. Adventures in Iteration, Volumes I and II (Univ Press)

VI. Books for “Plus Two” and IIT-JEE Mathematics

1. Higher Algebra, Hall and Knight
2. Higher Algebra, Barnard and Child
3. Elements of Coordinate Geometry, S L Loney
4. Plane Trigonometry (two volumes), S L Loney
5. Challenge and Thrill of Pre-College Mathematics, Pranesachar, Krishnamurthy et al
6. Calculus (two volumes), Tom Apostol
7. An Elementary Course Of Infinitesimal Calculus, Horace Lamb
8. Combinatorics: Including Concepts Of Graph Theory (Schaum), V Balakrishnan
9. Introduction to Enumerative Combinatorics, Miklos Bona
10. Outline Of Vector Analysis (Schaum), Spiegel, Lipschutz, Spellman 

VI. Other Books

1. One Two Three Infinity, George Gamow
2. Geometry Revisited, HSM Coxeter and SL Greitzer (MAA)
3. Mathematical Circles, Dmitri Fomin & Sergey Genkin & Ilia Itenberg (Univ Press)
4. The Enjoyment of Mathematics, Hans Rademacher & Otto Toeplitz (Dover)
5. One Hundred Problems in Elementary Mathematics, Hugo Steinhaus (Dover)
6. From Zero To Infinity, Constance Reid

VII. Books on History and Culture of Mathematics

1. Mathematics in Western Culture, Morris Kline
2. Great Moments in Mathematics, Volumes I and II, Howard Eves (MAA)
3. Journey Into Genius, William Dunham (MAA)
4. A Mathematician's Apology, G H Hardy
5. The Mathematical Experience, Philip Davis and Reuben Hersh

VIII. Problem Books and Puzzle Books

1. Fun With Mathematics, Ya Perelman
2. The Moscow Puzzles
3. Mathematical Circles, Dmitri Fomin + Sergey Genkin + Ilia Itenberg; Universities Press
4. Problem-Solving Strategies, Arthur Engel, Springer Verlag 
5. The Enjoyment of Mathematics, Hans Rademacher & Otto Toeplitz, Dover
6. One Hundred Problems in Elementary Mathematics, Hugo Steinhaus, Dover
7. Challenging Mathematical Problems with Elementary Solutions, A M Yaglom & I M Yaglom; Dover (in two volumes)
8. USA Mathematical Olympiads 1972-1986, M S Klamkin, MAA

IX. Biographies

1. The Man Who Knew Infinity, Robert Kanigel

X. Journals

1. Crux with Mayhem (Canadian Mathematical Society, CMS)
2. Mathematics Magazine (Mathematical Association of America, MAA)
3. College Mathematics Journal (Mathematical Association of America, MAA)
4. Mathematical Gazette (Mathematical Association, UK)
5. Mathematics Teacher (National Council for Teachers of Mathematics, USA)
6. Arithmetic Teacher (National Council for Teachers of Mathematics, USA)
7. Resonance (Indian Academy of Sciences)

XI. Websites

1. [Mathworld - Wolfram]
2. [Art of Problem Solving]
3. [Mathematical Excalibur]
4. [Math Forum]
5. [Math Pro Press]
6. [Plus]
7. [Cut the Knot!]
8. [Euclid's Elements]
9. [Math History 1]
10. [Math History 2]
11. [Math Celebration]

A curious occurrence of the powers of 4

Here is a curious way in which the powers of 4 turn up in a recursively defined sequence. 


Let the sequence of positive integers a(1), a(2), a(3), ... be defined using the following recursive formula:


a(1) = 1,
a(n) = a(n-1) + Floor(Sqrt(a(n-1)), for n > 1.


The "Floor" function is defined as follows: If x is any real number, then Floor(x) = the largest integer that does not exceed x. For example, Floor(3.1) = 3, Floor (10.7) = 10, Floor(-1.8) = -2, and so on. And, of course, Sqrt simply means "square root". 


I have used these notations because normal mathematical notation does not display properly in a blog.


Using the definition  one can compute the entire sequence recursively. Thus: 


a(2) = 1 + Floor(Sqrt(1)) = 2,
a(3) = 2 + Floor(Sqrt(2)) = 3,
a(4) = 3 + Floor(Sqrt(3)) = 4,
a(5) = 4 + Floor(Sqrt(4)) = 6,


and so on. Continuing this way, we can enumerate the entire sequence (it is particularly easy to do it using a computer). Here are the first 50 terms:


1, 2, 3, 4, 6, 8, 10, 13, 16, 20, 24, 28, 33, 38, 44, 50, 57, 64, 72, 80, 88, 97, 106, 116, 126, 137, 148, 160, 172, 185, 198, 212, 226, 241, 256, 272, 288, 304, 321, 338, 356, 374, 393, 412, 432, 452, 473, 494, 516, 538, ...


From this list, let us sift out just those numbers which are perfect squares; we get the following:


1, 4, 16, 64, 256, ...


Why, these are just the powers of 4. How very curious!


How is one to explain this?


Can we modify the defining rule so that we get some other power sequence (or some other sequence of interest)? Yes we can!  - if we replace Sqrt by CubeRoot we get something equally striking. 


So let us define the sequence of positive integers b(1), b(2), b(3), ...  using the following formula:


b(1) = 1,
b(n) = b(n-1) + Floor(CubeRoot(b(n-1)), for n > 1.


As earlier, we can enumerate the entire sequence, recursively. Here are the first 50 terms:


1, 2, 3, 4, 5, 6, 7, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 31, 34, 37, 40, 43, 46, 49, 52, 55, 58, 61, 64, 68, 72, 76, 80, 84, 88, 92, 96, 100, 104, 108, 112, 116, 120, 124, 128, 133, 138, 143, 148, ...


The perfect cubes in this sequence turn out to be these numbers:


1, 8, 64, 512, ...


and these are all the powers of 8.


Many variations are possible, within the same theme. 


Proofs, anyone? Here is a link to one possible line of analysis, written by my friend Ramana: http://ramana.posterous.com/an-intriguing-sequence-contain. Do have a look at it.


I'll write the proofs and post them to my website, later this week.

Triangle with a 120 degree angle

Here is a beautiful result concerning triangles in which one angle measures 120 degree.


Let ABC be a triangle in which angle A measures 120 degree, and let the internal bisectors of all three angles be drawn; let them meet the opposite sides at P, Q, R (so that AP is the bisector of angle A, BQ is the bisector of angle B, and so on).

Now a beautiful fact emerges: angle QPR is a right angle! 


There are many nice proofs of this fact. Here is a figure showing the result.

Just as beautiful is the fact that the statement has a converse: 


If points P, Q, R are constructed as described above, starting with an arbitrary triangle ABC, and angle QPR is a right angle, then angle A measures 120 degree. 


But this is less easy to prove.


I invite the reader to find both the proofs!

Powers of 2 and 3

Exploring power sequences can be a lot of fun! Many famous problems have arisen from playing with such sequences; for example, Fermat's Last Theorem (proved three and a half centuries after it was first stated by Fermat), Catalan's problem ("The only two perfect powers that differ by 1 are 8 and 9"; see http://mathworld.wolfram.com/CatalansDiophantineProblem.html; the theorem was proved very recently, in 2002, by the Swiss mathematician Preda Mihăilescu), and Beal's problem (which is still open).

On my website www.mathcelebration.com, which I launched more than a year back, and which I use to upload articles I have written for teachers and students, I have an article all about the powers of 2 and 3. Here are two links to it:

• http://www.mathcelebration.com/PDF/PowerTwoScreen.pdf
• http://www.mathcelebration.com/PDF/PowerTwoPDF.pdf.

Do have a look at it!