Deepak's Classes

Real Numbers

Chapter 01/ Day 01

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𝐂𝐡𝐞𝐜𝐤 𝐰𝐡𝐞𝐭𝐡𝐞𝐫 𝟔^𝐧 𝐜𝐚𝐧 𝐞𝐧𝐝 𝐰𝐢𝐭𝐡 𝐭𝐡𝐞 𝐝𝐢𝐠𝐢𝐭 𝟎 𝐟𝐨𝐫 𝐚𝐧𝐲 𝐧𝐚𝐭𝐮𝐫𝐚𝐥 𝐧𝐮𝐦𝐛𝐞𝐫 𝐧.

End with 0 means number must be divisible by 2 & 5
Why?
Since 10 = 2 X 5
10 is divisible by 2 & 5 so any number that is divisible by 10 or say that any number ends with number 0 must be divisible by both 2 & 5
Multiples of 6 always end with 6 mo matter how many times we multiply 6 with itself. Also
6 is not divisible by 5
So it is not possible to get any number out of 6^n that ends with 0.

𝐄𝐱𝐩𝐥𝐚𝐢𝐧𝐰𝐡𝐲𝟕×𝟏𝟏×𝟏𝟑+𝟏𝟑𝐚𝐧𝐝𝟕×𝟔×𝟓×𝟒×𝟑×𝟐×𝟏+𝟓𝐚𝐫𝐞𝐜𝐨𝐦𝐩𝐨𝐬𝐢𝐭𝐞𝐧𝐮𝐦𝐛𝐞𝐫𝐬.

Composite Numbers are all numbers divisible by more than 2 numbers (In simple words divisible by at least 3 numbers out of that two numbers are 1 and the number itself.).

In the mentioned cases outcome is definitely divisible by more than 3 numbers
7X11X13 + 13
13 ( 7X11 + 1)
So the outcome is definitely divisible by 13 apart from 1 and number self.

Similarly
7 X 6 X 5 X 4 X 3 X 2 X 1 + 5
5 ( 7 X 6 X4 X 3 X 2X 1 + 1)
Result is definitely divisible by 5

LCM and HCF by prime factors

𝑺𝒕𝒆𝒑 01 - Find prime factors of given numbers like for 35 and 75
35 = 5 X 7
75 = 5 X 5 X 3

𝑺𝒕𝒆𝒑 02 - For LCM pen down all multiples along with the common multiple with the highest power. like in the above case
5^2 X 7 X 3
So LCM is 25X7X3 = 525

𝑺𝒕𝒆𝒑 03 - For HCF Pen down lowest power of common prime multiple i.e.
5 in the above example

𝐑𝐞𝐦𝐞𝐦𝐛𝐞𝐫 𝐚 𝐅𝐚𝐜𝐭

For two integers A&B

(LCM of A&B) X (HCF of A&B) = AXB

In above case

35 X 75 = 525 X 5 = 2625

NCERT
Class Xth
CBSE Board
Chapter 01 - PART 05 - LCM
Exercise 1.1

LCM - Lowest Common Multiple
Let’s take an example of two numbers
18 & 27
18= 2X3X3 = 2X3^2
27=3X3X3 = 3^3
The common multiple is 3 and common power is 2
The Highest Common Factor is 9
LCM = Product of Given Numbers
_______________________
HCF
LCM = 18X27 / 9 = 54

Theorem 1.2 (Fundamental Theorem of Arithmetic): Every composite number can be expressed (factorized) as a product of primes, and this factorization is unique, apart from the order in which the prime factors occur.

Composite Number:- A composite number is a positive integer that can be formed by multiplying two smaller positive integers.
or
A number that is not a prime number.
Like 4,8,9,10,12,14,15,16,18,20,22,24,25,26,27………..
and every number can be expressed as multiple prime numbers like
4 = 2X2
8= 2X2X2
9=3X3
10 = 2X5
14=2X7
15-3X5
26=2X13
27=3X3X3

NCERT
Class Xth
CBSE Board
Chapter 01 - PART 04

Exercise 1.1

1. Use Euclid’s division algorithm to find the HCF of :
(i) 135 and 225 (ii) 196 and 38220 (iii) 867 and 255
Solution:
225 = 135 X 1 + 90
135 = 90 X 1 + 45
90 = 45 X 2 + 0 HCF is 45

(ii) 38220 = 196 X 195 + 0 HCF is 195
(iii) 867 = 255 X 3 + 102
255 = 102 X 2 + 51
102 = 51 X 2 + 0 HCF is 51

2. IF q is an integer. Show that any positive odd integer is of the form 6q+1or, 6q+3 or 6q+5

Solution:

Let’s check what we get if we multiply any integer (q) with 6
6X1 = 6, 6X2=12, 6X3=18, 6X4=24, 6X5=30, 6X999 = 5994 So it is seen that no matter what q’s value is, even/odd, if we multiply it with 6 we will get an even number.

Always remember

Any integer X Any even integer = An even Integer
Odd Integer X Odd Integer = Odd Integer
Even Integer X Even Integer = Even Integer

Even Integer + Odd Integer = Even Integer
Odd Integer + Odd Integer = Even Integer
Even Integer + Even Integer = Even Integer

Every number we get is an even number ( Divisible by 2 completely)
Let’s check what we get when we add any odd number in an even number

2 + 1 = 3
2 + 5 = 7
2 + 3 = 5

Hence it is seen that any positive odd integer is of the form 6q+1or, 6q+3 or 6q+5.
3. An army contingent of 616 members is to march behind an army band of 32 members in a parade. The two groups are to march in the same number of columns. What is the maximum number of columns in which they can march? 

Solution:
616 = 32 X 19 + 8
32 = 8 X 4 + 0
8 Columns

4. Use Euclid’s division lemma to show that the square of any positive integer is either of the form 3m or 3m + 1 for some integer m. 

Solution:
Let “x” be an integer 3q, 3q+1, 3q+2
for q a positive integer
x = 3X1=3
x = 3X1+1 = 4
x = 3X1+2 = 5
Square of x = 9,16,25
Square of x can be written as = (3X3), (3X5+1), (3X8+1)
We can conclude that square of any positive integer can be written as 3m or 3m+1.

5. Use Euclid’s division lemma to show that the cube of any positive integer is of the form 9m,9m+1or9m+8.

Solution:

If “n” is a positive integer it’s cube will be n^3
Euclid’s division lemma says
n = qb+r
for r = 0 & b = 9
n^3 = (q + 0)^3
= 729q^3
= 9 (81q^3)
Here m = 81q^3
for r = 1 & b = 9
n^3 = (9q + 1)^3
= 729q^3 + 1 + 27q(9q+1)
= 729q^3 + 1 + 729 q^2 + 27 q
= (729q^3 + 729 q^2 + 27 q ) + 1
= 9(81q^3+81q^2+3q) + 1
Here m = 81q^3+81q^2+3q
for r = 2 & b = 9
n^3 = ( 9q+2)^3
= (9q)^3+ 8+ 54q(9q+2)
=(9q)^3+8+486q^2+108q
=9(81q^3+54q^2+12q)+8
Here m = (81q^3+54q^2+12q)

NCERT Book
CBSE Board
Class Xth
Chapter 01- Part 03

Show that any positive odd integer is of the form 4q + 1 or 4q + 3, where q is some integer.

Positive Odd integer?
an integer value can be +ve or -ve that is why it is important to mention Positive odd /even integer

So let's prove it using Euclid’s Division Lemma

a = bq+r
if a is an odd number it can be
1,3,5,7,9,11,13,17,99,199,201 any number that is not divisible by 2

if b = 4 as said in the conditions
possible values of r can be 1 or 3

why not 5 or 7 or 9?
since these numbers can also be represented as 4q+r.
see how?
5=4X1+1
7=4X1+3
9=4X2+1
________________________________________________________

Practical application of HCF

A sweet-seller has 420 Chocolate donuts and 130 cream donuts. She wants to stack them in such a way that each stack has the same number, and they take up the least area of the tray. What is the number that can be placed in each stack for this purpose?

To know this we need to get HCF

We will use Euclid’s Division Lemma way
420 = 130 X 3 + 30
130 = 40 X 3 + 10
40 = 10 X 4 + 0

So HCF is 10
Seller can make sacks of 10 donuts.

NCERT Book
CBSE Board
Class Xth
Chapter 01- Part 02

Show that every positive even integer is of the form 2q, and that every positive odd integer is of the form 2q + 1, where q is some integer.

Method 01

q can be any positive integer we will check the given conditions for a sequence of possible q values
suppose q = 1 or 2 or 3 or 4 or 5 ………..
As said
For even numbers

2 = 2 X 1 + 0
4 = 2 X 2 + 0
6 = 2 X 3 + 0



998 = 2 X 499 + 0
It is clearly visible that every even number can be represented as multiple of 2 only i.e.2q

For odd numbers

1 = 2 X 0 + 1
3 = 2 X 1 + 1
5 = 2 X 2 + 1



11111 = 2 X 5555 + 1
So we can also say that every odd number can be presented in 2q + 1 form

Method 02

Euclid’s Division Lemma

a = bq+r

it is given that value of b is 2
r must be greater or equal to 0 and
smaller than b so
r is either 1 or 0

so for any even number/integer a must be 2q since even number is completely divisible by 2 like 28 = 2 X 14, 8 = 2X 4 so on
and for any odd integer r = 1
since a can not be completely divisible by 2
like 3 = 2X1+1
5 = 2X2+1
So every odd number can be represented as 2q+1
and every even number can represented as 2q

NCERT Book
CBSE Board
Class Xth
Chapter 01

Real Numbers

What are Real numbers?
All rational and irrational numbers.

What are rational numbers?

Those numbers we can present in p/q form and where Denominator (q) can divide the numerator ( p ) completely example

3/2 = 1.5
7/5 = 1.4

What are irrational numbers?

Those numbers we can not define in p/q form. But in general, any number can be defined in p/q form.
There are 2 main conditions

Denominator must not be 0 or 1
Denominator must not be able to divide numerator completely

Like value of pi = 3.14159 26535 89793 23846 264…
we represent pi as 22/7 also but that docent makes it a rational number since 7 can not divide 22 completely.

Calculating Highest Common Factor

Euclid’s Division Lemma

Given positive integers a and b, there exist unique integers q and r satisfying a = bq + r, 0 ≤ r < b.

means?

let's take an example

we have two integers
8 and 5

Consider a = 8 and b = 5
can we represent them in
a = bq + r
can we say that

8= 5x1+3 here q =1 and r =3
This also satisfy the condition
0 ≤ r < b.
since r, i.e. 3 is greater than 0 and smaller than b i.e. 5

Now represent 1 & 3

3 = 1x3+0
We need to continue with the process until we get r =0
When we get an r-value of 0 We will find the highest common factor of the two digits
i.e. value of “b” in the last equation like here in the above example b=1

Let's take another example
79 & 55
79 = 55 X 1 + 24
55 = 24 X 2 + 7
24 = 7 X 3 +3
7 = 3 X 2 +1
3 = 1 X 2 +1
1 = 1 X 1 + 0
Final Value of b is 1, So HCF of 55 and 79 is 1

45 & 600
600 = 45 X 13 + 15
15 = 13 X 1 + 2
13 = 2 X 6 +1
2 = 1 X 2 + 0
HCF is 2

STATISTICS - 06

Madhu purchased a Gold coin in 2017 for 44000, in year 2018 its value fall by 10% but in 2019 it increased by 20%.
in 2020 it again increases by 15%
But in 2021 it falls by 17%
What is the average growth rate on Gold?

When numbers have an exponential effect we need a more accurate way to find the mean or average. That way is called
"Geometric Mean"

We will discuss it but first, understand what this Geometric Sequence mean
To understand the power of Geometric sequence or exponentially find answers to minister Birbal question

Once King Akbar in India got impressed with his chief minister Birbal. He asked him to ask for anything he wants.

Birbal wanted to show Akbar that he (Akbar) is not the almighty God who can give anything to anybody. So he asked him

My King, please bring chess. Put a sack of wheat in the first box and 2 in the second 4 in 3rd, 8 in 4th, and 16 in 5th keep multiplying the sacks of the preceding box with 2 to find the following box numbers and then

Ask your treasury to give me the number of wheat sacks he will be able to put in the 64th box of chess.

1,2,4,8,16,32,64,128,256,512,1024,2048,4096,8912,17824,35648,71296,142592,285184,570368,1140736,2281472,4562944,9125888,18251776,36503552,73007104,146014208,292028416,584056832,....

30th Box only and all ministers exhausted. Forget about the 64th box King does not have half of the wheat that Birbal asked as a reward.

This is the power of Geometric Progression.
Compounding interest is another and very common application of geometric progression.

We will discuss the Geometric mean soon.