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## Translating into algebra

Posted by tendstoinfinity on March 8, 2013

This is the worksheet I have made for the class VIII students for teaching algebraic equation

Translate in algebra Write an algebraic expression for each phrase.

the difference between ninety and a number ________________________________

the difference between twenty-four and a number ________________________________

fifty-two more than a number ________________________________

the difference between a number and sixty-eight ________________________________

the product of a number and ninety-six ________________________________

the sum of a number and ten ________________________________

a number increased by nineteen ________________________________

forty-three more than a number ________________________________

seven less than a number ________________________________

fourteen times a number ________________________________

a number increased by seven ________________________________

a number decreased by seventy-one ________________________________

the product of twenty-four and a number ________________________________

thirty-six times a number ________________________________

the quotient of ninety-five and a number _________________________________

Ram is 12 years older than his sister, his sister is a years old. Ram’s age _______________________

Posted in Algebra | Leave a Comment »

## Sach ya Juth

Posted by tendstoinfinity on June 13, 2012

So once again I am here after a very long break for almost one and half year. Actually I was busy and more lethargic at times. Almost everyday I think I should share some interesting facts(there are so many!!!).I do not know the trustfulness of the sentence in the bracket. There is a very interesting riddle about persons telling truth and false. Here it is……

Imagine that you are visiting an island on which there are only two kinds of people (other than yourself): *satyavadi*, who always tell the truth, and *jutha*, who always lie. There are two villages – one where all the *satyavadis* live, and another where all the juthas live. Although they live in separate villages, liars and truthers frequently roam about the island together and generally get along just fine. Talking to islanders is a bit difficult because they all observe the peculiar custom of not answering more than one question in a conversation and generally don’t elaborate on any statements they make. Another interesting feature of these islanders is that although outsiders can’t distinguish between *satyavadis * and juthas by how they look.

**1. Going to the Village**

You are on the island and see a village on the road ahead of you, and you are not sure whether it is the *satyavadi* village or the *jutha* village. An islander, who may be a *satyavadi or jutha*, is standing on the side of the road. What one question do you ask her to find out if the village is the *satyavadi* village or the *jutha* village

**2. Looking for the Ferry**

You’ve decided to leave the island and are trying to find the ferry that will take you back to the mainland. There is a fork in the road that splits off in two directions. Two islanders, Anand and Bhaskar, are standing at the fork. Anand and Bhaskar are from different villages; you don’t know who is from the truther village and who is from the liar village, and Anand and Bhaskar won’t answer questions about their villages. What question do you ask one of them to find out how to get to the ferry?

**3. On the Ferry**

It’s a slow day, and you are the only passenger on the Ferry: it is just you and the captain. As it pulls out into the harbour you realize that you might have boarded the wrong ferry – is this really the boat that is going to the mainland? You can ask the captain, an islander himself, one question to find out.

Posted in Reasoning, Recreational | 2 Comments »

## Congruence modulo

Posted by tendstoinfinity on March 15, 2011

Recall that (mod m) when a – b is divisible by m. Another way to understand this a and b leaves same remainder when divided by m. We have learnt that this relation is equivalent relation. So we can write

If (mod m) and (mod m) then (mod m). (transitivity). Actually congruence modulo m splits the set of natural number in m distinct classes. Those are classes of remainders 0,1,2,…n-1.

you can surely prove this. There are some more interesting relations and more interesting uses of congruence.

- (mod m) and (mod m) then (mod m).
- (mod m) and (mod m) then (mod m). we can prove this by noting ac – bd = ac – bc + bc – bd = c(a-b) + b(c -d). both the bracketed expressions are divisible by m.

one special case of this is

- (mod m) and n is any natural number then (mod m)

Some applications of the above results

**What will be the remainder if is divided by 7**

Since (mod 7) so (mod 7) that is (mod 7). therefore leaves remainder same as that of 1 when divided by 7. Which is clearly 1

**Prove that + is divisible by 31.**

Since (mod 31) so (mod 31) and (mod 31) so (mod 31)

therefore (mod31) . hence the result.(when anything leaves remainder 0 it is divisible by the given)

You can do a lot of interesting findings with the help of congruence modulo, divisibility tests are surely an important class.

Before discussion of divisibility tests we introduce a notation taken from the book *‘Elementary Mathematics*‘ by **G. Dorofeev, M. Potapov and N.Rozov** (originally published by MIR publishers, in India CBS publishes this book)

=

**Divisibility by 2, 5 and 10**

Any natural number is congruent to its last digit modulo 2(5 or 10) for

= + but 10 is divisible by 2(or 5 or 10) so

(mod 2). The number itself will have same remainder as that of last digit when divided by 2 , similar result for 5 or 10

**Divisibility by 4 and 8**

you can prove similarly (mod 4).

and (mod 8).

so any number leaves a remainder same as that of the last two digits when divided by 4, and of the last three digits when divided by 8.

I will share divisibility test by 3,7,9,11,13 in my later posts.

Posted in Number theory, Uncategorized | Leave a Comment »

## salesman problem

Posted by tendstoinfinity on February 10, 2011

pichak@scientist.com

Perhaps you have remembered Dr. A N Mishra, KV Jorhat one of the teachers blessed with splendid presence of mind. He has asked me a very beautiful question.

*One salesman used to carry 40 different weights 1kg,2kg,….40kg for weighing articles upto 40kgs, until somebody suggested why doesn’t he use less number of weights so that he can weigh all the kgs upto 40kg in one turn. Now can you help the salesman to choose minimum number of weights in the following cases*

* i. **weights can be placed on either of the pans of the balance.(this is Dr. Mishra’s question)*

* ii. **weights can be placed only on left pan and object on right pan.( this is my variation)*

before presenting the solutions directlyI want to see the Result: *All the naturals up to * *can be obtained by adding all or some of the element of the set {1=2 ^{0},2^{1},2^{2},2^{3},…2^{n}} only once. (Why?) *For example by adding some or all of the set only once you can get all naturals upto 7 using all or some {1,2,4}. You can see following table for illustaration

No of 4 | No of 2 | No of 1 | sum |

0 | 0 | 1 | 1 |

0 | 1 | 0 | 2 |

0 | 1 | 1 | 3 |

1 | 0 | 0 | 4 |

*1 | 0 | 1 | 5* |

#1 | 1 | 0 | 6 |

1 | 1 | 1 | 7 |

Are you able to see that in the above table (*no of 4,no of 2 and no of 1)* form the binary ( number system of base 2)representation of the respective sum. *Just as (101)_{2} =(5)_{10} . Now binary representation of 8 to 15 =2^{4}-1 contains 4digits. So the sum of any natural number from 1 to 15 can be represented as a sum of all or some of {2^{0}=1, 2, 4, 8=2^{3}} only once. I think my variation i.e (ii) can now be solved. Because you can only use one pan to place weights so addition only should be used so 6 weights of {1,2,4,8,16,32} will be necessary for 2^{5}-1<40<2^{6}-1. In fact the salesman can weigh any weights upto 61kg in one turn using these weights.

Another approach of thinking

There are two places for the weight units either in the balance pan or in bag of salesman. Let 0 represents the bag and 1 represents pan so weighing a weight of 6kg we have representation 110# means 4kg, 2kg on the pan and 1 kg in the bag.

Before we turn to Dr. Mishra’s problem we want to see the a wonderful property of ternary system(number system of base 3) : every natural can be represented as the sum or difference of two numbers whose ternary representation contain only 0’s and 1’s or otherwise every natural can be expressed as the sum or differences of numbers of the form 3^{n}(n is a natural no.). If the natural number is a power of 3(of the form 3^{n}) the prove is obvious. Now for the other case see the following example

(4)_{10}=(11)_{3}=(11)_{3}-0_{3} = 3^{1}+3^{0}

(5)_{10}= (12)_{3}

= 1.3^{1}+2.3^{0}

= 1.3+(3-1)3^{0}

= 1.3+3-1

=2.3-1

=(3-1)3-1

=3^{2}-3^{1}-3^{0}

= (100)_{3}-(10)_{3}-1_{0}

I hope you can generalize the facts.

So every natural number upto can be expressed by adding or subtracting all or some elements of the set {1=3^{0},3^{1},3^{2},…3^{n}}

So every number less than 40 can be expressed as sum or difference of all or some of {1,3,9,27}. These are weights required by the salesman.

For base 4, Similarly every natural number can be expressed by sum or difference of the powers of 4 taking any specific element not more than twice.

5 = 4^{1}+4^{0}

6 = 4+1+1 (1 =4^{0})

7 = 16 – 4 – 4 – 1

53 = 64 – 16 + 4 +1

Its proof is so simple. But giving a little time you can prove it.

Hopefully in some post I will discuss some more interesting fact of the bases of number system.

Posted in Recreational, Uncategorized | 3 Comments »

## Another Geometrical construction…

Posted by tendstoinfinity on February 5, 2011

pichak@scientist.com

I in previous post posed a geometric problem

*An angle and a point D lying inside it are given. Construct a line segment with end points on arms of the angle and bisected at D. *

let us consider a set of points on a line BA and all the segments having one end on these points and mid point at D. then the other end of each line segment will be just the reflection of the former points. hence the other ends will always lie on a line parallel to BA.

then the above construction can be easily done.

suppose the given angle is <APQ. Image of the vertex P of the angle with respect to D is drawn (draw a line through P and D. cut an arc of length equal to PD from centre D. where it cuts the line is image P’ of P with respect to D

draw a line parallel to PA through P’ . suppose this parallel cuts PQ at R then RS will be required line which has mid- point at D and ends on the arms of the angle

surely you can justify this. I will give a very good use of this construction

*An angle and a point D lying inside it are given. Draw a straight line through D which makes a triangle with the arms of the angle of minimum possible area.*

Then obviously the triangle RPS (with reference to the above figure) will have minimum area amongst all the triangles of this type !! got it.

Posted in Geometry | Leave a Comment »

## A Geometrical construction

Posted by tendstoinfinity on January 25, 2011

## An easy challenge ” Draw a circle of radius r centered at A and another point C is given. Draw a line through C so that chord cut off by circle along this line is of length d“

I request you to try on your own first then see my solution. you can have better solution than me. if you got another one pl. share.

First this simple geometrical construction ” A circle is given and a point B outside it. Draw a tangent to the circle through B.” You, I hope will be able to draw the required construction.

Draw the circle “c” with centre A and the point B. Join the points A and B. Bisect the line segment AB at C. draw a circle “d” with centre C and radius BC. let the circle “d” intersects circle “c” at D.

the line through B and D will be required tangent.

Now the original construction

Draw the given circle “c” with radius r centered at A

Draw another circle “d” concentric with the given circle with radius

draw the tangent through C to d. this line will cut a chord of length d of circle “c”

hope the construction is justified by you. Again a * challenge* to you which I will solve in another post.

## An angle and a point D lying inside it are given. Construct a line segment with end points on arms of the angle and bisected at D.

Posted in Geometry | Leave a Comment »

## How you can determine a value of pi,the mathematical wonder.

Posted by tendstoinfinity on January 23, 2011

“PI” is one of the beauty mathematics has given the world. can you find the value of this mathematics marvel.

first I am giving a simple sentence which will help you to memorize the value of to13 places of decimal. Simply you have to count the number of letters in the following sentence.

**Why I want 2 learn calculate pi, for I have a large spherical shape**

= 3.1415923141595

But have you ever thought how we can calculate the value?

We want to find the area of a circular region of radius 1cm. Obviously we can easily find it out by the formula

“Area of the circular region = “so the area of our region will be

Now can we use a little bit of “*probability theory*” and the above to determine the value .

Suppose we throw a tiny particle in square of side 2 units in which a circle of radius 1 unit is inscribed(see the fig. below) What is the probability that the particle will fall inside the circle, it is supposed that falling of the stone at every point of square is equally likely.

Required probability = area of circular region/area of square region

=

Suppose the stone falls 75 times in the circular space and 25 times in the empty space in hundred throws. So by definition of probability will be . These two values should be equal.

i.e =

We can calculate the value for from this. Surely the value will be more accurate if we throw the stone a great number of times. Obviously the experiment will take a lot of time. So we want to speed it up. we construct a mathematical model of the above experiment. considering a quadrant of the circle (you can use b’coz circle is symmetric) we will have the same effect. so we are taking only the 1st quadrant of . let the stone falls on an arbitrary point P(x,y) in the the part of the square in 1st quadrant,surely and . If then the point will be inside the quadrant of circle otherwise exterior to the circle.

this is the mathematical model of the experiment. Now instead of performing the actual experiment we can calculate the results of the experiments directly using pen and paper, all we need a set of random numbers between 0 and 1. but it is very tough to calculate all these manually. instead use a computer programme. you can use any computer programming language. I have used the C programming language. copy the following and compile and run in any C compiler.

**C programme for calculate the pi**

#### #include<stdio.h>

#### #include<conio.h>

#### #include<stdlib.h>

#### double pi()

#### {

#### long int z;

#### double x,y,P,N,n=0,b=0,a;

#### printf(“How many times you want to throw?\n”);

#### scanf(“%ld”,&z);

#### for(N=1;N<=z;N++)

#### {

#### x=rand();

#### x=x/32767.0;

#### y=rand();

#### y=y/32767.0;

#### a=(x*x)+(y*y);

#### if(a>1)

#### b=b+1;

#### else

#### n=n+1;

#### }

#### N=N-1;

#### P=(n/N)*4;

#### return P;

#### }

#### void main()

#### {

#### double p;

#### clrscr();

#### p=pi();

#### printf(“\n Value of pi is %lf”,p);

#### getch();

#### }

*you can throw as many times as you wish. hope this is working on your machine. Good day.*

Posted in Probability | Leave a Comment »

## Who has taken what?

Posted by tendstoinfinity on January 17, 2011

pichak@scientist.com

I recently attended in-service training programme of teachers. There Dr. A N mishra of kendriya vidyalaya, ONGC Jorhat demonstrated a very beautiful magic to begin the chapter *Permutation*. I am presenting the magic. Hope you’ll like it.

I have three friends say A, B, C whom I invited. They took my three important possession say my watch, ring and my pen one each when I was away from my room. When I returned to the room they told if I can correctly say “** who has taken what**?” then they would return otherwise I should forget those three things!! I had 24 toffees( you can take any other thing like ground nut etc.)in my pocket. I gave one toffee to A, 2 toffees to B and 3 toffees to C and place the rest on the table. I told them “I am going out of the room, the person who taken watch should take same number of toffee as he has, the person with ring will take double of what he has and he with the pen will take four times of his possession.” I went out of the room and then returned and gazed the remaining toffees on the table. Hurray! I found

**. Are you astonished? How did I find.**

*who has taken what*I simply wrote the total arrangement possibility and remember initially A has 1, B has 2 and C has 3 toffees.

A | B | C | Toffees taken | Rest toffees |

W(1+1) | R(2+4) | P(3+12) | 23 | 1 |

W(1+1) | P(2+8) | R(3+6) | 21 | 3 |

R(1+2) | W(2+2) | P(3+12) | 22 | 2 |

R(1+2) | P(2+8) | W(3+3) | 19 | 5 |

P(1+4) | W(2+2) | R(3+6) | 18 | 6 |

P(1+4) | R(2+4) | W(3+3) | 17 | 7 |

hope you understood how I guessed by seeing the rest number toffees.

Posted in combinatorics | 2 Comments »

## A SPECIAL YEAR 2011

Posted by tendstoinfinity on January 13, 2011

I recently came across Prof. Gary E. Davis‘s blog Republic of mathematics. He pointed out 2011 is itself a prime number and it can be represented as the sum of 11 consecutive prime numbers.

2011=157+163+167+173+179+181+191+193+197+199+211

or sum of 3 consecutiprime numbers 2011 = 661+673+677

The next year after 2011 that is a prime number and a sum of consecutive prime numbers is 2027 = 29+31+37+41+43+47+53+59+61+67+71+73+

79+83+89+97+101+103+107+109+113+127+131+137+139

but that year is not a sum of a prime number of consecutive prime numbers.

However, 2027 does have the curious and interesting property that it is prime and the sum of its digits 2+0+2+7=11 is also a prime number. It is the first year after 2003 that has this property.

The next year that is a prime number and a sum of a prime number of consecutive prime numbers is 2081 =401+409+419+421+431.

Posted in Uncategorized | Leave a Comment »