PLAYING WITH NUMBERS

BASIC CONCEPTS

The concept of playing with numbers is one of the most fundamental concepts in mathematics.

NATURAL NUMBERS (N): The most basic number system (i.e., counting numbers) is the system of natural numbers. N= {1, 2, 3,…}. This system is denoted by the symbol N.

WHOLE NUMBERS (W): when the number 0 is included with the system of natural numbers, we obtain the system of whole numbers. This system is denoted by the symbol w.W= {0,1,2,3,....}0 is the smallest whole number and there is no greatest whole number, and consequently W is an infinite collection.

CONTINUE..

INTEGERS (Z OR I): When the negatives of all the natural numbers are included with the system of whole numbers, we obtain the system of integers. Thus, the whole numbers, taken together with the negatives of natural numbers form the system of integers. This system is denoted by Z or I.INTEGERS (Z OR I) = {....-5,-4,-3,-2,-1,0,1,2,3,4,5,...}

The positive integers as Z+ = {1,2,3,...} The negative integers as Z- = {-1,-2,-3,...}

There are infinitely many integers.

NUMBERS IN GENERALISED FORM.

A number is said to be generalized form if it is expressed as the sum of the products of its digits with their respective place values.

A number can be written in expanded form by using the place values of its digits and a natural number can be expressed in exponential form by using powers of 10 and the digits of the number.

EXAMPLE

Expanded form 57=10x5+7

Exponential-form57=101x5+100x7

Now consider a two digit number having a and b respectively as tens and units digits. Using the above definitions the number can be written as10xa+b or 101xa+100xb.

THREE-DIGIT NUMBERS

Consider a 3-digit number having a, b, c as its hundreds digit, tens digit and units digit respectively.

This number is (100a+10b+c),where a can be whole number from 1to 9. b can be any whole number from 0 to 9.

EXAMPLE

157=100x1+10x5+7

850=100x8+10x5+0.

TEST OF DIVISIBILITY OF NUMBERS.

DIVISIBILTY BY 2 : In general, a given number is divisible by 2 only when its units digit is 0,2,4,6 or 8.

EXAMPLE: Each of the numbers 60,72,84,96,108 is divisible by 2.

CONTINUE.. DIVISIBILITY BY 3 : A given number is

divisible by 3 only when the sum of its digits is divisible by 3.

EXAMPLE : The given number is 16785.Sum of its digits = (1+6+7+8+5) = 27, which is divisible by 3.∴ 16785 is divisible by 3.

CONTINUE.. DIVISIBILTY BY 5 : A given number is

divisible by 5 only when the units digit is 0 or 5.

EXAMPLE : Each of the numbers 67230 and 83715 is divisible by 5.And , none of the numbers 136,247,158,373,419,241,514,632 is divisible by 5.

CONTINUE.. DIVISIBITY BY 9 : A given number is

divisible by 9 only when the sum of its digits is divisible by 9.

EXAMPLE : The given number is 27891. Sum of its digits = (2+7+8+9+1) = 27 ,

which is divisible by 9. ∴ 2789 is divisible by 9.

CRYPTARITHMS

Cryptarithms are puzzles , on various operations on numbers , in which letters take the place of digits and one has to find out which letter represents which digit. While solving cryptarithms involving addition and multiplication , we assume that each letter in a puzzle stands for just one digit and each digit is represented by just one letter. We also assume that the first digit of a number can not be zero.

EXAMPLE

Solve the following crytarithms 31A + 1A3 501

Solution: We have

31A + 1A3 501 Here , we have to find the value of A which

can take values from 0 to 9.

SOLUTION CONTINUE…

As A takes values from 0 to 9. Therefore , A+3 can take values from 3 to 12.

Since digit at the units place of the sum of two digits A and 3 is 1. Therefore either A+3 is equal to 1 or A+3 is a number between 3 and 12 whose units digit is 1. Clearly such a number between 3 and 12 is 11.∴ A+3 = 11 A = 8This value of A satisfies the addition in tens and hundreds columnsHence , A = 8.

ACTIVITY

GAMES WITH NUMBERS

REVERSING THE DIGITS – TWO DIGIT NUMBER

Alisha asks Amar to think of 2-digit number, and then to do whatever she asks him to do, to that number. Their conversation is shown in next slide….

CONVERSATIONS BETWEEN ALISHA AND AMAR : FIRST ROUND

Choose 2Digit

number

Alright [Chooses

49]

CONTINUE CONVERSATION…

Reverse the digits to get

a new number

Ok, I will…(gets 94)

CONTINUE CONVERSATION…

Add this to the number you started

with

Um…, Ok. (gets 143)

CONTINUE CONVERSATION…

Now divide the answer

by 11.

Um…,Ok. (gets 13))

CONTINUE CONVERSATION…

There won’t be any

remainder!

Yes, you are right. But

how did you know?

CONTINUE CONVERSATION…

It so happens that Amar choose the number 49. So , he got the reversed number 94 ; then he added these two numbers and got 49 + 94 = 143.

Finally he divided this number by 11 and got 143 ÷11 = 13, with no remainder. This is just what Alisha had predicted.

AGAIN, THE GAME BETWEEN ALISHA AND AMAR CONTINUES!

Alisha: Think of another 2-digit number,but don’t tell me what it is.

Amar: Alright. Alisha: Now reverse the digits of the number ,

and subtract the smaller number from the larger one.

Amar: I have done the subtraction. what next? Alisha: Now divide your answer by 9. I claim that

there will be no remainder! Amar: Yes, you are right. There is needed no

remainder! But this time I think I know how you are so sure of this!

In fact, Amar had thought of 29. So his calculations were: first he got the number 92; then he got 92-29 = 23;and finally he did (63÷9) and got 7 as quotient, with no remainder.

ACTIVITY

PLAYING WITH NUMBERS

OBJECTIVE: TO verify by activity method that the sum of first n odd natural numbers is n2 i.e.1+3+5+7+….n times = n2

PRE-ACQUIRED KNOWLEDGE: Concept of odd natural numbers. Concept of square of a number.

CONTINUE…

MATERIALS REQUIRED: Cubes of equal size in five different colours. A chart paper. A pencil. A ruler. An eraser.

CONTINUE ..

CONCLUSION: This verifies that the sum of first n odd natural numbers is n2.

1 = 1×1 = 12

CONTINUE ..

1+3 = 2×2 = 22

RESULT: The sum of first n odd natural numbers is n2.