MY DIGITAL NOTEBOOK

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Category Archives: Problems

Do you multiply this way!

Wednesday, August 31st, 2011 00:01 / 5 Comments

Before my college days I used to multiply orthodoxly by this way.

But as time passed, I learned new things. I remember, In a Hindi magazine named “Bhaskar Lakshya”, I read an article in which a lecturer (apology, I don’t remember his name) had suggested how to multiply in single line (row). Today I thought that I should share this method on MY DIGITAL NOTEBOOK too.
I know there are many, who already know this method, but I think maximum people wouldn’t have any idea about this method. I found multiplicating this way, very faster – easier and smarter. The ‘only’ requirements for using this method is quick summation. You should be good in calculation and addition. Smarter your calculations, faster you’re.
I’ll try to illustrate this method below. If you had any problems regarding language (poor off-course) and understandings, please feel free to put that into comments.
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Just another way to Multiply

Thursday, August 25th, 2011 18:30 / 13 Comments

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. [LOL] I don’t know who, silly else me, made this quote. Have Fun.

Do you multiply this way! (wpgaurav.wordpress.com)

How Genius You Are?

Saturday, August 20th, 2011 19:10 / 13 Comments

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! Be Quick! And Click here to check your result.

Quicker you see your result, sharper you are!

Do you think the result is 5000?
Actually, it is not. The correct result is 4100.

A Problem On Several Triangles

Saturday, August 13th, 2011 14:51 / 2 Comments

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

Two Interesting Math Problems

Wednesday, August 10th, 2011 21:00 / 1 Comment

Problem1: Smallest Autobiographical Number:

A number with ten digits or less is called autobiographical if its first digit (from the left) indicates the number of zeros it contains,the second digit the number of ones, third digit number of twos and so on.

For example: 42101000 is autobiographical.

Find, with explanation, the smallest autobiographical number.

Solution of Problem 1

Problem 2: Fit Rectangle:

A rectangle has dimensions $39.375$ cm $\times 136.5$ cm.

Find the least number of squares that will fill the rectangle.
Find the least number of squares that will fill the rectangle, if every square must be the same size and Find the largest square that can be tiled to completely fill the rectangle.

Solution of Problem 2

Solutions of Problem 1:

The restrictions which define an autobiographical number make it straightforward to find the lowest one. It cannot be 0, since by
definition the first digit must indicate the number of zeros in the number. Presumably then, the smallest possible autobiographical number will contain only one 0.If this is the case, then the first digit must be 1. 10 is not a candidate because the second digit must indicate the number of 1s in the number–in this case, 1. So If the
number contains only one zero, it must contain more than one 1.
(If it contained one 1 and one 0, then the first two digits
would be 11, which would be contradictory since it actually contains two 1s).
Again, presumably the lowest possible such number will contain the lowest
possible number of 1s, so we try a number with one 0 and two 1s. It will be of the form: 12-0–..
Now, there is one 2 in this number, so the first three digits must be 121. To meet all the conditions discussed above, we can simply take a 0 onto the end of this to obtain 1210, which is
the smallest auto-biographical number.

Solution of Problem 2:

We solve the second and third parts of the question
first:

We convert each number to a fraction and get a common denominator, then find the gcd (greatest common divisor) of the numerators.

That is, with side lengths $39.375$ cm and $136.5$ cm , we convert those numbers to fractions (with a common
denominator):
$39.375 = \dfrac{315}{8}$ .

$136.5 = \dfrac{273}{2} = \dfrac{1092}{8}$ .

Now we need to find the largest common factor of 315 and 1092.
Which is 21. So $\dfrac{21}{8}=2.625$ is the largest number that divides evenly into the two numbers $39.375$ and $136.5$ .
There will be $\dfrac{1092}{21} \times \dfrac{315}{21} = 52 \times 15 = 780$ squares, each one a $2.625$ cm $\times 2.625$ cm square needed to fill the rectangle (52 in each row,with 15 rows).

Now we shall solve the first part.
Number of squares lengthwise is 52 and breadthwise is 15. Now we will combine these squares in order to find least number of squares to fill the rectangle. First three squares would be of
dimension 15 by 15. In this way length of 45 units is utilized. Now the rectangle which is left with us excluding three squares is 7 by 15. Again in the same way we can make two squares of dimension 7 by 7. In this way breadth of 14 units is utilized.
Now we are left with the rectangle of dimension 7 by 1.
These can further be subdivided into seven squares each of
dimension 1 by 1. In this way the least number of squares to fill the
rectangle is 3 + 2+ 7 = 12. The required answer is 12.
Note that the three numbers 3, 2, and 7 are involved in the Euclidean Algorithm for finding the g.c.d.!

Source: Internet

Chess Problems

Monday, August 1st, 2011 06:35 / 1 Comment

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.
Can you attack every position on the board with fewer than seven pieces?

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How many apples did each automattician eat?

Friday, July 22nd, 2011 10:36 / 6 Comments

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?

Answer:

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.

MY DIGITAL NOTEBOOK

Category Archives: Problems

Do you multiply this way!

Just another way to Multiply

How Genius You Are?

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.

A Problem On Several Triangles

Solution

Graphical Proof

Two Interesting Math Problems

Problem1: Smallest Autobiographical Number:

Problem 2: Fit Rectangle:

Solutions of Problem 1:

Solution of Problem 2:

Chess Problems

How many apples did each automattician eat?

Answer:

The Logic

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Category Archives: Problems

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Related articles

Better You Share!

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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.

Better You Share!

Like this:

Solution

Graphical Proof

Better You Share!

Like this:

Problem1: Smallest Autobiographical Number:

Problem 2: Fit Rectangle:

Solutions of Problem 1:

Solution of Problem 2:

Better You Share!

Like this:

Better You Share!

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Answer:

The Logic

Better You Share!

Like this:

Post navigation

Follow “MY DIGITAL NOTEBOOK”

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.