Category: Problems

yes no puzzle
Humor & Fun StuffsLogicMathProblemsPuzzles

A Yes No Puzzle

This is not just math, but a very good test for linguistic reasoning. If you are serious about this test and think that you’ve a sharp [at least average] brain then read the statement (only) below –summarize it –find the conclusion and then answer that whether summary of the statement is Yes or No.
[And if you’re not serious about the test …then read the whole post to know what the stupid author was trying to tell you. 🙂 ]
STATEMENT: If the question you answered before you answered the question you answered after you answered the question you answered before you answered this one, was harder than the question you answered after you answered the question you answered before you answered this one, was the question you answered before you answered this one harder than this one? YES or NO?
 

Answer:

The answer is YES.

In other words, we could restate the statement as:

If the question you answered before this one was harder than THIS ONE, was the question you answered before this one harder than THIS ONE.

That makes the answer obvious. 

MathProblemsPuzzlesRecreations

Three Children, Two Friends and One Mathematical Puzzle

Two close friends, Robert and Thomas, met again after a gap of several years.
Robert Said: I am now married and have three children.$
Thomas Said: That’s great! How old they are?$
Robert: Thomas! Guess it yourself with some clues provided by me. The product of the ages of my children is 36.$
Thomas: Hmm… Not so helpful clue. Can you please give one more?$
Robert: Yeah! Can you see the number on the house across the street?$
Thomas: Yes! I can.$
Robert: The sum of their ages equal that number$ .
Thomas: Sorry! I still could not determine their ages.$
Robert: My oldest child has red hair.$
Thomas: OH.. Oldest one? Finally I got it. I know age of each of your children$ .

Question:

What were the ages of Robert’s children and how did Thomas know?

Discussion and probable answer

This is a very good logical problem. To do it, first write down all the real possibilities that the number on that building might have been. Assuming integer ages one get get the following which equal 36 when multiplied:

Read More

featured2
EducationMathNumber TheoryProblemsRecreations

Do you multiply this way!

Before my college days I used to multiply this way.

But as time passed, I learned new things. In a Hindi magazine named “Bhaskar Lakshya”, I read an article in which a columnist ( I can’t remember his name) suggested how to multiply in single line (row). That was a magic to me.  I found doing multiplications this way, very faster – easier and smarter. There may be many who already know this method, but many others will be seeing it for the first time.

The ‘only’ requirements for using this method is the quick summation. You should be good in your calculations. Smarter your calculations, faster you’re.
I’ll try to illustrate this method below. If you had any problems regarding language (it’s poor off-course) and understandings, please feel free to put that into comments.

Let we try to multiply 498 with 753.
$ 4 9 8 \ \times 7 5 3$

Step I

Multiply 8 and 3 and write the unit digit of result carrying other digits for next step. The same is to be done with each step.

Step 2

Step 3

Step 4

Step 5

The overall work looks like:

I don’t know if there is any algorithm behind it. The pattern of calculation is very simple, which is making crosses and adding numbers.

 

You can use this method, multiplying larger numbers too. Try this one at your own. Steps are marked for convenience. 🙂

Thanks for Reading!

MathNumber TheoryProblemsReal AnalysisRecreations

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.


Solution

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