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Just another way to Multiply

Multiplication is probably the most important elementary operation in mathematics; even more important than usual addition. Every math-guy has its own style of multiplying numbers. But have you ever tried multiplicating by this way?
Exercise: $ 88 \times 45$ =?
Ans: as usual :- 3960 but I got this using a particular way:
88            45
176          22
352           11
704            5
1408          2
2816          1

Sum of left column=3960

Thus, $ 88 \times 45=3960$ (as usual).
You might be thinking that what did I do here. Okay, let we understand this method by illustrating another multiplication, of 48 with 35.

Step 1. Write the numbers in two separate columns.

$ 48 \\ 35$

Step 2. Now, double the number in left column and half the number in right column such that the number in right column reduces to 1. If the number [remaining] in right column is odd, then leave the fractional part and only write integer part.

$ 48 35 \\  96 17 \\192 8 \\ 384 4 \\  768 2 \\ 1536 1$

Step 3: Cancel out any number in the left column whose corresponding number in the right column is even.

48                       35
96                       17
192                      8
384                       4
768                       2
1536                      1

Step 4:Sum all the numbers in the left column which are not cancelled. This sum is the required product.

$ =1680$

I agree this method of multiplying numbers is not easy and you’re not going to use this in your every day math. It’s a bit boring and very long way of multiplication. But you can use this way to tease your friends, teach juniors and can write this into your own NOTEBOOK for future understandings. Remember, knowing more is getting more in mathematics. Have Fun.

How Genius You are- Splash
Humor & Fun StuffsLogicMathProblemsPuzzles

How Genius You Are?

Let have a Test:

You need to make a calculation. Please do neither use a calculator nor a paper. Calculate everything “in your brain”.

Take 1000

and add 40.

Now, add another 1000.

Now add 30.

Now, add 1000 again.

Add 20.

And add 1000 again.

And an additional 10.


So, You Got The RESULT! 

Quicker you see the answer, sharper you are!

MathProblemsPuzzlesStudy Notes

A Problem On Several Triangles

A triangle $ T $ is divided into smaller triangles such that any two of the smaller triangles either have no point in common, or have a vertex in common, or actually have an edge in common. Thus no two smaller triangles touch along part of an edge of them.
For an illustration let me denote the three vertices of T by 1, 2 and 3. Now number each of the vertices of the small triangles by 1, 2, 3. Do this in an arbitrary way, but such that vertices lying on an edge of T must not be numbered by the same number as the vertex of T opposite to that edge.

Show that among the small triangles there is always one whose vertices are numbered by 1, 2 and 3.


To show that among the small triangles there is always one whose vertices are numbered by 1, 2 and 3, we show that the number of small triangles whose vertices are labeled with $ 1,2,3$ is odd and thus actually $ >0$ !

We enumerate all small triangles in the picture as $ T_1$ , $ T_2, \ldots, T_n$ and denote by $ a_i$ the number of edges with endpoints $ 1$ and $ 2$ in each triangle $ T_i$ . Thus, if say the vertices of $ T_i$ are labeled by $ 1,1,2$ , then $ a_i=2$ , and so on …

Observe now that obviously we have

$ \displaystyle a_1+a_2+a_3+\cdots +a_n= A+2B, $

where $ A$ is the number of triangles whose vertices are labeled $ 1,2,3$ , while $ B$ is the number of those triangles labeled by $ 1,1,2$ or $ 1,2,2$ . (Actually it is easily seen that $ a_i=2$ for such triangles, while $ a_i=1$ if the vertices of $ T_i$ are $ 1,2,3$ and $ a_i=0$ otherwise.) All we have to show is that $ A$ is odd.

Let $ C$ denote the number of $ 12$ -edges lying inside the original triangle $ T$ and let $ D$ be the number of $ 12$ -edges lying on the boundary of $ T$ . Every interior $ 12$ -edge lies in two triangles $ T_i$ and thus it is counted twice in the sum $ a_1+a_2+a_3+\cdots +a_n$ , while every boundary $ 12$ -edge is counted only once. In conclusion we get

$ \displaystyle a_1+a_2+a_3+\cdots +a_n= 2C+D, $

which yields

$ \displaystyle A+2B=2C+D. $

Hence $ A$ is odd if and only if $ D$ is odd. It is therefore enough to show that $ D$ is odd.

According to the hypothesis of the problem, edges labeled $ 12$ or $ 21$ can occur only on the $ 12$ -edge of the large triangle $ T$ . We start walking along the edge $ 12$ of the triangle $ T$ starting at the vertex $ 1$ toward the vertex $ 2$ . Now, only when we first pass an edge labeled $ 12$ will we arrive at the first vertex labeled $ 2$ . A number of vertices labeled $ 2$ may now follow, and only after we have passed a segment $ 21$ do we reach a label $ 1$ , and so on. Thus after an odd number of segments $ 12$ or $ 21$ we arrive at vertices labeled $ 2$ , and after an even number of such segments we arrive at vertices labeled $ 1$ . Since the last vertex we will reach is the vertex $ 2$ of the big triangle $ T$ , it follows that the total number of segments $ 12$ or $ 21$ lying on the side $ 12$ of the big triangle $ T$ must be odd! The same reasoning applies for each of the other edges of the big triangle $ T$ , so we deduce that $ D$ , the total number of $ 12$ or $ 21$ -edges lying on the boundary of $ T$ , must be odd. Proved

Graphical Proof

It is obvious. As a result of this numbering we get following diagram:

Problem Image

Chess Problems

  1. In how many ways can two queens, two rooks, one white bishop, one black bishop, and a knight be placed on a standard $ 8 \times 8$ chessboard so that every position on the board is under attack by at least one piece?
    Note: The color of a bishop refers to the color of the square on which it sits, not to the color of the piece.
  2. Can you attack every position on the board with fewer than seven pieces?


  1. Two ways as follow:
  2. No. (…and I’m not pretty sure about this No)

Improved chessboard images via this website using Creative Commons.


How many apples did each automattician eat?

Česky: Jablka jsou všeamericky úspěšná potravi...

Image via Wikipedia

Four friends Matt, James, Ian and Barry, who all knew each other from being members of the Automattic, called Automatticians, sat around a table that had a dish with 11 apples in it. The chat was intense, and they ended up eating all the apples. Everybody had at least one apple, and everyone know that fact, and each automattician knew the number of apples he ate. They didn’t know how many apples each of the others ate, though. They agreed to ask only questions that they didn’t know the answers to:

Matt asked: Did you eat more apples than I did, James?

James: I don’t know. Did you, Ian, eat more apples than I did?

Ian: I don’t know.

Barry: Aha!! I figured out..

So, Barry figured out how many apples each person ate. Can you do the same?


Matt: 1 Apple

James: 2 Apples

Ian: 3 Apples

Barry: 5 Apples

The Logic

Matt could not have eaten 5 or more. James could not have eaten only one or he would have known that he hadn’t eaten more than Ian. Neither could he have eaten 5 or more. He could have eaten 2 or 3 or 4 apples. Ian figures this out, although he still doesn’t know if he ate more than James. This mean that Ian must have eaten 3 or 4 apples. Barry can only deduce the other amounts if he ate 5 apples. And the rest, in order to add up to 11 , must have eaten 1, 2 and 3.

Inspired from a childhood heard puzzle.

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Fox – Rabbit Chase Problems

Part I:

A fox chases a rabbit. Both run at the same speed $ v$ . At all times, the fox runs directly toward the instantaneous position of the rabbit , and the rabbit runs at an angle $ \alpha $ relative to the direction directly away from the fox. The initial separation between the fox and the rabbit is $ l$ .

When and where does the fox catch the rabbit (if it does)? If it never does, what is their eventual separation?

Part II:

Similarly think about the same situation, except now let the rabbit always move in the straight line of its initial direction in above part of the question.

When and where does the fox catch the rabbit (if it does)? If it never does, what is their eventual separation?


Part I

The relative speed of the fox and the rabbit, along the line connecting them, is always $ v_{\text{rel}}= v- v \cos \alpha$ . Therefore, the total time needed to decrease their separation from $ l$ to zero is $ T=\dfrac{l}{v-v \cos \alpha} =\dfrac{l}{v(1-\cos \alpha)} \ \ldots (1)$ which is valid unless $ \alpha=0$ , in which case the fox never catches the rabbit.
The location of their meeting is a little trickier to obtain. We have two methods to do :


Imagine that the rabbit chases another rabbit, which chases another rabbit, etc. Each animal runs at an angle $ \alpha$ relative to the direction directly away from the animal chasing it. The initial positions of all the animals lie on a circle, which is easily seen to have radius $ R=\dfrac{l/2}{\sin (\alpha/2)} \ \ldots (2)$ .
The center of the circle is the point, O, which is the vertex of the isosceles triangle with vertex angle $ \alpha$ , and with the initial fox and rabbit positions as the othe two vertices. By symmetry, the positions of the animals at all times must lie on a circle with center O. Therefore, O, is the desired point where they meet. The animals simply spiral into O.


An equivalent solution is the the following:

At all times, the rabbit’s velocity vector is obtained by rotating the fox’s velocity vector by angle $ \alpha$ . The meeting point O, is therefore the vertex of the above mentioned isosceles triangle,


The speed of the rabbit in the direction orthogonal to the line connecting the two animals in $ v \sin \alpha$ . Therefore, during a time $ dt$ , the direction of the fox’s motion changes by an angle $ d\theta =\dfrac {v \sin \alpha}{l_t} dt$ , where $ l_t$ is the separation at time $ t $ . Hence the change in the fox’s velocity has magnitude $ |d\overrightarrow{v}|=v d\theta =v (v \sin \alpha dt/l_t)$ . The vector $ d\overrightarrow{v}$ is orthogonal to $ \overrightarrow{v}$ , therefore, to get the $ x$ -component of $ d\overrightarrow{v}$ , we need to multiply $ |d\overrightarrow{v}|$ by $ v_y/v$ . Similar reasoning holds for $ y$ -component of $ d\overrightarrow{v}$ , so we arrive at the two equations $ \dot{v_x}= \frac{vv_y \sin \alpha}{l_t} \ \ldots (3)$ $ \dot{v_y}=- \frac{vv_x \sin \alpha}{l_t} \ \ldots (4)$
Now, we know that $ l_t =\{ l-v(1-\cos \alpha) t \}$ . Multiplying the above equations (3) and (4) by $ l_t$ , and integrating from the initial to final times, yields $ v_{x,0}l+v(1-\cos \alpha)X=v \sin \alpha \, Y \ \ldots (5)$ $ v_{y,0}l+v(1-\cos \alpha)Y=-v \sin \alpha \, X \ \ldots (6)$
where (X,Y) is the total displacement vector and $ (v_{x,0},v_{y,0})$ is the initial velocity vector. Putting all the X and Y terms on the right sides, and squaring and adding the equations, we get $ l^2v^2=(X^2+Y^2)(v^2 \sin^2 \alpha +v^2{(1-\cos \alpha)}^2). \ \ldots (7)$ Therefore , the net displacement is
$ R=\sqrt{X^2+Y^2}=\dfrac{l}{\sqrt{2(1-\cos \alpha)}}=\dfrac{l/2}{\sin (\alpha/2)} \ \ldots (8)$
To find the exact location, we can, without loss of generality, set $ v_{x,0} =0$ , in which case we find $ Y/X=(1-\cos \alpha)/\sin \alpha =\tan \alpha/2$ . This agrees with the result of the first solution. $ \Box$

Part II:


Let $ A(t)$ and $ B(t)$ be the positions of the fox and the rabbit respectively. Let $ C(t)$ be the foot of the perpendicular dropped from $ A$ to the line of the rabbit’s path. Let $ \alpha_t$ be the angle, dependent to the time, at which the rabbit moves relative to the direction directly away from the fox (so at $ t=0, \ \alpha_0=\alpha$ and at $ t=\infty , \ \alpha_{\infty}=0$ ).

The speed at which the distance AB decreases is equal to $ v-v \cos \alpha_t$ . Therefore, the sum of the distances AB and CB doesn’t change. Initially, the sum is $ l+l \cos \alpha$ and in the end , it is $ 2d$ where $ d$ is desired eventual separation. Therefore, the desired eventual separation
$ d=\dfrac{l(1+\cos \alpha)}{2} \ \ldots (9)$


Let $ \alpha_t$ be defined as in the first solution, and let $ l_t$ be the separation at time $ t$ . The speed of the rabbit in the direction orthogonal to the line connecting the two animals is $ v \sin \alpha_t$ . The separation is $ l_t$ , so the angle $ \alpha_t$ changes at a rate $ \dot{\alpha_{t}}= – \dfrac{v \sin \alpha_t}{l_t} \ \ldots (10)$ . And $ l_t$ changes at a rate $ \dot{l_t}=-v(1-\cos \alpha_t) \ \ldots (11)$ .
Taking the quotient of the above two equations, separating variables, gives a differential equation $ \dfrac{\dot{l_t}}{l_t} = \dfrac {\dot{\alpha_t}(1-\cos \alpha_t)}{\sin \alpha_t} \ \ldots (12)$ which on solving gives $ \ln (l_t) = -\ln {(1+\cos \alpha_t)} + \ln (k) \ \ldots (13)$ . Where $ k$ is the constant of integration. Which gives $ k=l_t (1+\cos \alpha_t) \ \ldots (14)$ . Applying initial conditions $ k_0 = l_0 (1+\cos \alpha_0)= l(1+\cos \alpha) \ \ldots (15)$ . Therefore from (14), we get $ l(1+\cos \alpha) = l_t(1+\cos \alpha_t)$ , or $ l_t= \dfrac{l(1+\cos \alpha)}{1+\cos \alpha_t}$ .
Setting $ t= \infty$ and using $ \alpha_{\infty} =0$ , gives the final result $ l_{\infty} = \dfrac{l(1+\cos \alpha)}{2}$ . $ \Box$ .


The solution of Part II is valid for all $ \alpha$ except $ \alpha= \pi$ . If $ \alpha = \pi$ , the rabbit run directly towards the fox and they will meet halfway in time $ l/2v$ .

CalculusMathProblemsReal Analysis

D’ ALEMBERT’s Test of Convergence of Series


A series $ \sum {u_n}$ of positive terms is convergent if from and after some fixed term $ \dfrac {u_{n+1}} {u_n} < r < {1} $ , where r is a fixed number. The series is divergent if $ \dfrac{u_{n+1}} {u_n} > 1$ from and after some fixed term.

D’ Alembert’s Test is also known as ratio test of convergence of a series.

Definitions for Generally Interested Readers

(Definition 1) An infinite series $ \sum {u_n}$ i.e. $ \mathbf {u_1+u_2+u_3+….+u_n}$ is said to be convergent if $ S_n$ , the sum of its first $ n$ terms, tends to a finite limit $ S$ as n tends to infinity.
We call $ S$ the sum of the series, and write $ S=\displaystyle {\lim_{n \to \infty} } S_n$ .
Thus an infinite series $ \sum {u_n}$ converges to a sum S, if for any given positive number $ \epsilon $ , however small, there exists a positive integer $ n_0$ such that
$ |S_n-S| < \epsilon$ for all $ n \ge n_0$ .
(Definition 2)
If $ S_n \to \pm \infty$ as $ n \to \infty$ , the series is said to be divergent.
Thus, $ \sum {u_n}$ is said to be divergent if for every given positive number $ \lambda$ , however large, there exists a positive integer $ n_0$ such that $ |S_n|>\lambda$ for all $ n \ge n_0$ .
(Definition 3)
If $ S_n$ does not tends to a finite limit, or to plus or minus infinity, the series is called Oscillatory


Let a series be $ \mathbf {u_1+u_2+u_3+…….}$ . We assume that the above inequalities are true.

  • From the first part of the statement:
    $ \dfrac {u_2}{u_1} < r$ , $ \dfrac {u_3}{u_2} < r $ ……… where r <1.
    Therefore $ \mathbf {{u_1+u_2+u_3+….}= u_1 {(1+\frac{u_2}{u_1}+\frac{u_3}{u_1}+….)}}$
    $ =\mathbf {u_1{(1+\frac{u_2}{u_1}+\frac{u_3}{u_2} \times \frac{u_2}{u_1}+….)}} $
    $ < \mathbf {u_1(1+r+r^2+…..)}$
    Therefore, $ \sum{u_n} < u_1 (1+r+r^2+…..)$
    or, $ \sum{u_n} < \displaystyle{\lim_{n \to \infty}} \dfrac {u_1 (1-r^n)} {1-r}$
    Since r<1, therefore as $ n \to \infty , \ r^n \to 0$
    therefore $ \sum{u_n} < \dfrac{u_1} {1-r}$ =k say, where k is a fixed number.
    Therefore $ \sum{u_n}$ is convergent.
  • Since, $ \dfrac{u_{n+1}}{u_n} > 1$ then, $ \dfrac{u_2}{u_1} > 1$ , $ \dfrac{u_3}{u_2} > 1$ …….
    Therefore $ u_2 > u_1, \ u_3 >u_2>u_1, \ u_4 >u_3 > u_2 >u_1$ and so on.
    Therefore $ \sum {u_n}=u_1+u_2+u_3+….+u_n$ > $ nu_1$ . By taking n sufficiently large, we see that $ nu_1$ can be made greater than any fixed quantity.
    Hence the series is divergent.


  • When $ \dfrac {u_{n+1}} {u_n}=1$ , the test fails.
  • Another form of the test–

    The series $ \sum {u_n}$ of positive terms is convergent if $ \displaystyle {\lim_{n \to \infty}} \dfrac {u_n}{u_{n+1}}$ >1 and divergent if $ \displaystyle{\lim_{n \to \infty}} \dfrac {u_n}{u_{n+1}}$ <1.
    One should use this form of the test in the practical applications.

A Problem:
Verify whether the infinite series $ \dfrac{x}{1.2} + \dfrac {x^2} {2.3} + \dfrac {x^3} {3.4} +….$ is convergent or divergent.


We have $ u_{n+1}= \dfrac {x^{n+1}}{(n+1)(n+2)}$ and $ u_n= \dfrac {x^n} {n(n+1)}$
Therefore $ \displaystyle {\lim_{n \to \infty}} \dfrac{u_n} {u_{n+1}} = \displaystyle{\lim_{n \to \infty}} (1+\frac{2}{n}) \frac{1}{x} = \frac{1}{x}$
Hence, when 1/x >1 , i.e., x <1, the series is convergent and when x >1 the series is divergent.
When x=1, $ u_n=\dfrac{1} {n(n+1)}=\dfrac {1}{n^2} {(1+1/n)}^{-1}$
or, $ u_n=\dfrac{1}{n^2}(1-\frac{1}{n}+ \frac {1}{n^2}-…..)$
Take $ \dfrac{1}{n^2}=v_n$ Now $ \displaystyle {\lim_{n \to \infty}} \dfrac {u_n}{v_n}=1$ , a non-zero finite quantity.
But $ \sum {v_n}=\sum {\frac{1}{n^2}}$ is convergent.
Hence, $ \sum {u_n}$ is also Convergent.

MathProblemsPuzzlesStudy Notes

Applications of Complex Number Analysis to Divisibility Problems

  • Prove that $ {(x+y)}^n-x^n-y^n$ is divisible by $ xy(x+y) \times (x^2+xy+y^2)$ if $ n$ is an odd number not divisible by $ 3$ .
  • Prove that $ {(x+y)}^n-x^n-y^n$ is divisible by $ xy(x+y) \times {(x^2+xy+y^2)}^2$ if $ n \equiv \pmod{6}1$


1.Considering the given expression as a polynomial in $ y$ , let us put $ y=0 $ . We see that at $ y=0 $ the polynomial vanishes (for any $ x$ ). Therefore our polynomial is divisible by $ y$ . Similarly, it is divisible by $ x$ as well. Thus the polynomial is divisible by $ xy$ .
To prove that it is divisible by $ x+y $ , put $ x+y=0 \ {or} \ y=-x $ . It is evident that for odd n we have : $ {(x+(-x)}^n-x^n-{(-x)}^n = 0 $ for $ y=-x $ .
Consequently, our polynomial is divisible by $ x+y $ . It only remains to prove the divisibility of the polynomial by $ x^2 +xy+y^2$ , which also be written as $ (y-x\epsilon)(y-x{\epsilon}^2 ) $ where $ \epsilon^2+\epsilon+1=0 $ .
For this purpose it only remains to replace $ y $ first by $ x \epsilon $ and then by $ x\epsilon^2 $ to make sure that with these substitutions the polynomial vanishes. Since, by hypothesis, $ n$ is not divisible by 3, it follows that $ n=3l+1 \ or \ 3l+2 $ , for every $ l \in \mathbb{Z} $ , in which $ 3l+1$ is not acceptable since $ n$ is odd from the problem. At $ y=x\epsilon $ the polynomial attains the following value
$ {(x+x\epsilon)}^n-x^n-{(x\epsilon)}^n=x^n [{(1+\epsilon)}^n-1-\epsilon^n] \\ =x^n {(-\epsilon^2)}^n -1 -\epsilon^n …. $ since ($ 1+\epsilon + \epsilon^2=0 $ ) substituting $ n=3l+2 $ we get
$ 1+\epsilon+\epsilon^2 =0 $
Likewise we prove that at $ y=x\epsilon^2$ the polynomial vanishes as well, and consequently, its by divisibility by $ xy(x+y) \times (x^2+xy+y^2) $ is proved.

2.To prove the second statement, let us proceed as follows. Let the quantities $ {-x, -y, \, and \, x+y} $ be the roots of a cubic equation $ X^3-rX^2-pX-q=0 $ . Then by virtue of the known relations between the roots of an equation and its coefficients we have $ r=-x-y-(x+y)=0 \\ -p=xy-x(x+y)-y(x+y)$ or $ p=x^2+xy+y^2$ and $ q=xy(x+y)$ .
Thus, $ -x, \, -y \, x+y$ are the roots of the equation $ X^3-pX-q=0 $ where $ p=x^2+xy+y^2$ and $ q=xy(x+y) $
Put $ {(-x)}^n-{(-y)}^n+{(x+y)}^n=S_n$ . Among successive values of $ S_n$ , there exist the relationship $ S_{n+3}=pS_{n+1}+qS_n$ ,: $ S_1$ being equal to zero.
Let us prove that $ S_n$ is divisible by $ p^2$ if $ n \equiv 1 \pmod{6}$ using the method of mathematical induction. Suppose $ S_n $ is divisible by $ p^2 $ and prove that then $ S_{n+6} $ is also divisible by $ p^2$ .
So, using this relation we get that
$ S_{n+6}=p(pS_{n+2} + qS_{n+1}) + q(pS_{n+1}+qS_n) \\ =p^2S_{n+2}+2pqS_{n+1}+q^2S_n$ .
Since, by supposition, $ S_n$ is divisible by $ p^2$ , it suffices to prove that $ S_{n+1}$ is divisible by $ p$ . Thus we only have to prove than $ S_n={(x+y)}^n+(-x)^n+(-y)^n$ is divisible by $ p=x^2+xy+y^2$ if $ n \equiv 2 \pmod{6}$ , we easily prove our assertion. And so, assuming that $ S_n$ is divisible by $ p^2$ , we have proved that (from induction) $ S_{n+6}$ is also divisible by $ p^2$ . Consequently $ S_n ={(x+y)}^n+(-x)^n+(-y)^n={(x+y)}^n-x^n-y^n$ for any $ n \equiv 1 \pmod{6}$ is divisible by $ p^2={(x^2+xy+y^2)}^2 $ .
Now it only remains to prove its divisibility by $ x+y $ and by $ xy$ , which is quite elementary.



Essential Steps of Problem Solving in Mathematical Sciences

Learning how to solve problems in mathematics is simply to know what to look for.


Mathematics problems often require established procedures. To become a problem solver, one must know What, When and How to apply them. To identify procedures, you have to be familiar with the different problem situations. You must also be good in gathering information, extracting strategies and use them. But exercise is must for problem solving. Problem solving needs practice!! The more you practice, the better you become. Sometimes it may happen that you knew the formula for a problem, but as you haven’t tried it earlier — you failed to solve it by a minor margin. On the same topic, George Polya published “How to Solve It” in 1957. Many of his ideas that worked then, do still continue to work for us. The central ideas of Polya’s books were to identify the clues and to learn when & how to use them into solution strategies.
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