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Numbers. The basic pillar of Mathematics, the language of the universe. The thing which we learned in our first mathematics class. The thing we deal with, in our day-to-day life. The thing which is almost everywhere.

So, just be with me in the entire session and see how these numbers are going to amaze you every time. So, be ready to enjoy as well as learn.

CLASSIFICATION:

Here is the chart which classifies numbers.

REAL NUMBERS: These are the numbers whose squares are positive numbers.

IMAGINARY NUMBERS: These are the numbers whose squares are negative numbers. They are represented by iota ‘i’.

e.g. If the number 6i is given, it means it’s value is 6 × √(-1) = 6√(-1) = √(-36)

learning by analogy. We’ll approach imaginary numbers by observing its ancestor, the negatives. Here’s your guidebook:

RATIONAL NUMBERS: The numbers which can be written in the form p/q where p and q are co-prime integers and q ≠ 0. i.e. In decimal form, they are either terminating or recurring numbers.

Let’s find out how we can convert a rational number from decimal form to fraction form.

e.g. 1. Express 0.82828282…. in the form of a fraction.

Ans- Let x = 0.82828282…            _____(1)

As the period containing 2 digits, we multiply by 102 = 100,

∴ 100x = 82.82828282….             ______(2)

Now, (2) – (1), we get,

⇒ 99x = 82

∴ x = 82/99.

e.g. 2. Express 0.024024024024024024….. in the form of a fraction.

Ans- Let x = 0.024024024024….      ____(1)

As the period is containing 3 digits, we multiply with 103 = 1000

∴ 1000x = 24.02404024….      ____(2)

Now, (2) – (1), we get,

⇒ 999x = 24

∴ x = 24/999 = 8/333.

IRRATIONAL NUMBERS: The numbers which cannot be written as p/q. i.e. In decimal form, these numbers are non-terminating and non-recurring numbers. e.g. √2

INTEGERS: The numbers whose value after the decimal is 0.

DECIMALS: The numbers whose value after the decimal is non-zero. e.g. 3.456

NEGATIVES: The numbers whose value is less than 0.

POSITIVES: The numbers whose value is more than 0.

NATURAL NUMBERS: The integers more than 0.

WHOLE NUMBERS: Whole numbers are basically natural numbers, including 0. These are also called non-negative integers.

PRIME NUMBER: The number which has exactly two factors (1 and the number itself).

COMPOSITE NUMBER: The number which has more than two factors.

[Note: 1 is neither Prime nor Composite number]

PRIME NUMBERS:

As we have discussed earlier, these are the numbers that have exactly two factors. And these two factors are 1 and the number itself.

From 1 to 50, we have 15 prime factors.

From 1 to 100, we have 25 prime factors. And these are,

2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89 and 97.

Prime numbers have some properties like-

i) No integer can divide them completely other than 1 and the number itself (that’s why they have only two factors).

ii) Every prime number should be in the form of [4n + 1 or 4n – 1] for less than 5 and [6n + 1 or 6n – 1] for more than or equal to 5. But the reverse of this statement is not true. e.g. 25 can be written as 6(4) + 1 but it is not a prime number.

iii) Every integer, greater than 1, can be written as the product of prime numbers.

Since there is no formula to find out whether a number is prime or not. But for small numbers, we can use this trick,

Take the square root of the number (approximately) and check its divisibility by the prime numbers smaller than its square root.

e.g. 2. Check whether 97 is a prime number or not.

Ans- Square root of 97 will be between 9 and 10. So the prime numbers smaller than 9 are 2, 3, 5 and 7. Since 97 is not divisible by any of these, hence it is a prime number.

e.g. 3. Check whether 91 is a prime number or not.

Ans- Square root of 91 will be between 9 or 10. So the prime numbers smaller than 9 are 2, 3, 5 and 7. Since 91 is divided by 7 (13 × 7 = 91). 91 is not a prime number.

DIVISIBILITY RULE:

DIVISIBILITY OF 2: A number is always divisible by 2 if it is an even number i.e. its last digit is 0, 2, 4, 6 or 8.

DIVISIBILITY OF 3: A number is divisible by 3 if the sum of its digits is divisible by 3.

DIVISIBILITY OF 4: A number is divisible by 4 if the last two digits of the number are divisible by 4.

DIVISIBILITY OF 5: A number is divisible by 5 if its last digit is 0 or 5.

DIVISIBILITY OF 6: A number is divisible by 6 if it is even and the sum of its digits is divisible by 3.

DIVISIBILITY OF 8: A number is divisible by 8 if its last 3 digits are divisible by 8.

DIVISIBILITY OF 10: A number is divisible by 10 if its last digit is 0.

DIVISIBILITY OF 12: A number is divisible by 12 if it is divisible by both 3 and 4.

DIVISIBILITY RULE OF 9 AND 11:

DIVISIBILITY RULE OF 9: If sum of the digits of the number is 9 at the end (or a multiple of 9), hence it is divisible by 9.

 e.g. 356211 has sum of the digits, 3 + 5 + 6 + 2 + 1 + 1 = 18 and 18 had sum of the digits, 8 + 1 = 9, so 356211 is divisible by 9.

DIVISIBILITY RULE OF 11: If the difference between the sum of the digits at odd places (from right) and the sum of the digits at even place (from right) is 0 or a multiple of 11, hence it is divisible by 11.

e.g. 356213 has sum of the digits at odd places 3 + 2 + 5 = 10 and the sum of the digits at even places, 1 + 6 + 3 = 10. So 10 – 10 = 0. Hence it is divisible by 11.

The most interesting fact is, the divisibility rule of 9 and 11 is also be applicable for 99, 999, 101, 1001, etc. but in a slightly other way.

Let’s understand this thing by taking an example of a number ‘abcdefg’

  • This number is divisible by 9 if: (a + b + c + d + e + f + g) is 9 at the end (or a multiple of 9)
  • This number is divisible by 99 if: (a + bc + de + fg) is divisible by 99. [Pairing from right side]
  • This number is divisible by 999 if: (a + bcd + efg) is divisible by 999. [Tripling from right side]

And so on…

Also,

  • This number is divisible by 11 if: (g + e + c + a) – (d + f + b) is 0 or 11
  • This number is divisible by 101 if: (fg + bc) – (de + a) is 0 or a multiple of 101 [Pairing from right side]
  • This number is divisible by 1001 if: (efg + a) – (bcd) is 0 or a multiple of 1001 [Tripling from right side]

And so on…

Let us take some examples to understand it clearly.

e.g. 4. Check whether 4636665 is divisible by 9, 99 and 999 or not.

Ans- Divisibility by 9: 4 + 6 + 3 + 6 + 6 + 6 + 5 = 36 and 3 + 6 = 9

So, yes it is divisible by 9

Divisibility by 99: (Pairing from right side) 4 + 63 + 66 + 65 = 198

So, yes it is divisible by 99

[Note: Since the number is divisible by 99. Hence, it is also divisible by all the factors of 99. And so, this number is divisible by 9]

Divisibility by 999: (Tripling from right side) 4 + 636 + 665 = 1305

So, no it is not divisible by 999.

e.g. 5. Check whether 8064143 is divisible by 11, 101, 1001.

Ans- Divisibility by 11: (3 + 1 + 6 + 8) – (4 + 4 + 0) = 18 – 8 = 10

So, it is not divisible by 11

Divisibility by 101: (Pairing from right side) (43 + 06) – (41 + 8) = 49 – 49 = 0

So, it is divisible by 101

Divisibility by 1001: (Tripling from right side) (143 + 8) – (064) = 151 – 64 = 87

So, it is not divisible by 1001.

DIVISIBILITY BY 7 AND 13:

DIVISIBILITY BY 7: We have a rule for 7 that if the difference between the number of tens in the number and twice the unit digit is divisible by 7.

But it has some limitations. It is good for some small numbers like 651, 994, 1078 etc. but it is a very time-consuming process for numbers like 456745.

So, we can do some additional stuff here.

We know how to find whether a number is divisible by 1001 or not. And,

1001 = 7 × 11 × 13

So, if a number is divisible by 1001 it will also be divisible by 7 and 13.

If there is a number, which is not divisible by 1001 but divisible by 7 then how we can check it.

We will discuss it too in the coming topics. (For now, just remember that 1001 is the product of three prime numbers 7, 11 and 13)

FACTORS AND MULTIPLES:

Firstly, let’s just differentiate between the meaning of factors and multiples.

FACTORS: Factors of a number are the numbers by which when the number is divide, the remainder will be 0.

MULTIPLES: Multiples of a number are the numbers which are if divided by the number, the remainder will be 0.

e.g. 8 has factors 1, 2, 4 and 8 and has the multiples 8, 16, 24, etc.

Now, let’s talk about factors.

Any number greater than 1 can be written as the product of prime factors. Let’s generalize this term then we will explore this topic.

Let N is a composite number. So, by the above statement, N can be written as,

N = ap × bq × cr × …

(Where a, b, c are distinct prime factors of N and p, q, r is integers)

So, the general formula to calculate the number of factors of N is,

  • Total number of factors of N = (p + 1) (q + 1) (r + 1) …

e.g. 6. Find the total number of factors of 756.

Ans- 756 can be written as, 756 = 22 × 33 × 7

So, total number of factors of 756 = (2 + 1) (3 + 1) (1 + 1) = 3 × 4 × 2 = 24.

e.g. 7. How many factors of 756 are multiple of 12?

Ans- 756 = 22 × 33 × 7 and 12 = 22 × 3

⇒ 756 = 22 × 31 [32 × 71]

Since we have to find out the multiples of 12 in it. So, this [32 × 71] is important for us.

No. of factors = (2 + 1) (1 + 1) = 3 + 2 = 6.

  • Sum of all factors of N =   https://hranker.com/admin/assets/uploads/ckeditor/Screenshot%202020-02-06%20at%201.02.54%20AM.png

Or we can simply write it as,

Sum of all factors of N = (a0 + a1 + … + ap) (b0 + b1 + … + bq) (c0 + c1 + … + cr) …

e.g. 8. Find the sum of all factors of 48.

Ans- We can write it as, 48 = 24 × 3

So, Sum of all factors of 48 = (20 + 21 + 22 + 23 + 24) (30 + 31) = (1 + 2 + 4 + 8 + 16) (1 + 3) = 31 × 4 = 124

Or, [(24+1 – 1)/(2 – 1)] [(31+1 – 1)/(3 – 1)] = 31 × 8/2 = 124.

  • Product of all factors of N = (N)number of factors/2

e.g. 9. Find the product of all factors of 756.

Ans- 756 = 22 × 33 × 7.

So, number of factors = 3 × 4 × 2 = 24

Product of all factors of 756 = (756)24/2 = (756)12.

[Note: The number of factors of a perfect square is always odd because if we write the factors of any number in ascending order then the product of corresponding terms will give the number itself.

e.g. Factors of 100 = 1, 2, 4, 5, 10, 20, 25, 50, 100.

Here, 1 × 100 = 100, 2 × 50 = 100, 4 × 25 = 100, 5 × 20 = 100. But since only 10 is left so it should be multiply by itself (that’s why 100 is perfect square). So that’s why, the number of factors of a perfect square is always odd. And since it is perfect square, in the product of all factors’ formula, 1/2 will be cancelled by square.]

  • Number of ways of expressing N as a product of two factors = ½(p + 1) (q + 1) (r + 1) …

e.g. 10. In how many ways can 3420 be written as a product of two factors?

Ans- 3420 = 22 × 32 × 5 × 19,

So, the number of factors of 3420 = (2 + 1) (2 + 1) (1 + 1) (1 + 1) = 3 × 3 × 2 × 2 = 36

So, the required answer = 36/ = 18 ways.

CO-PRIME NUMBERS: Co-Prime numbers are the two numbers that have no common factor other than 1. e.g. 24 and 25 are co-prime numbers because 24 has factors 1, 2, 3, 4, 6, 8, 12, 24 and 25 has factors 1, 5, 25. No factor is common other than 1.

  • Number of ways of writing N as a product of two co-prime numbers = 2n-1

(Where n is the number of distinct prime factors)

e.g. Let us take an example of 48. 48 = 24 × 3. The value of ‘n’ is 2 because only two distinct prime factors are involved (i.e. 2 and 3)

Hence, the number of ways = 22-1 = 2. i.e. 48 can be written as a product of two co-primes, in two different ways. Precisely, they are (1, 48) and (3, 16).

  • Number of co-primes to N, that are less than N =   https://hranker.com/admin/assets/uploads/ckeditor/Screenshot%202020-02-06%20at%201.03.10%20AM.png

This is denoted by φ(N).

e.g. 11. Find the number of co-primes to 48, which are less than 48.

Ans- 48 = 24 × 3

So, φ(48) = 48 (1 – ½) (1 – 1/3) = 48 × ½ × 2/3 = 16

Here, 16 means, if the numbers less than 48 are listed and co-primes to 48 are spotted, the count of co-primes will be 16.

  • Sum of co-primes to N that are less than N = N/2 × φ(N)

e.g. For 48, the sum of co-primes, which are less than 48, will be 48/2 × φ(48) = 24 × 16 = 384.

 Now let us take a look at some interesting points.

1) The product of two consecutive integers is always even. Because one of them is even and one of them is odd, so the product will always be even.

2) The product of three consecutive integers is always divisible by 6. Because one of them is always even and one of them is always divisible by 3.

3) Like that, the product of four consecutive integers is always divisible by 24.

4) The product of n consecutive integers will always be divisible by n!

5) 1 digit numbers are 9.

    2 digit numbers are 90 (from 10 to 99). So, they have 90 × 2 = 180 digits.

   3 digit numbers are 900 (from 100 to 999). So, they have 900 × 3 = 2700 digits.

   4 digit numbers are 9000 (from 1000 to 9999). So, they have 9000 × 4 = 36000 digits. And so on.

e.g. 12. Divya typed the first n natural numbers on a keyboard without any spaces. If she had to press keys 1692 times, find n.

Ans- 1 digit numbers have 9 digits. 2 digit numbers are 90 and have 180 digits. So, 1692 – (180 + 9) = 1503.

Since 3 digit numbers are 900 have 2700 digits but we left only 1503 digits, it means the number is 3 digit number. Now, 1503/3 = 501, which means it is the 501st 3 digit number.

So, the number is, 501 + 90 + 9 = 600.

Before we bind up this section, let us look at two examples and if you are now confident about this section, just try these questions by yourself before looking at my solution. It would be fun.

e.g. 13. How many natural numbers divide 35999 but do not divide 35998?

Ans- Just read question 2 or 3 times and find what this question wants us to calculate. It wants us to find the factors of 35999 which are not the factors of 35998. In simple words,

(No. of factors of 35999) – (No. of factors of 35998),

Because these are the only factors which are greater than 35998 and, so that, they do not divide 35998.

Now,

⇒ 35999 = (5 × 7)999 = 5999 × 7999

⇒ No. of factors = (999 + 1) (999 + 1) = 1000 × 1000 = 10002

Also,

⇒ 35998 = (5 × 7)998 = 5998 × 7998

⇒ No. of factors = (998 + 1) (998 + 1) = 999 × 999 = 9992

∴ required answer = 10002 – 9992 = (1000 + 999) (1000 – 999) = 1999 × 1 = 1999.

e.g. 14. In an NGO, there are 70 workers. All the workers visited an old age home having 70 senior citizens. The first worker donated Rs. 1000 to each senior citizen. The second worker donated Rs. 1000 to every second senior citizen starting from the second senior citizen. The third worker donated Rs. 1000 to every third senior citizen starting from the third senior citizen and so on. How many senior citizens received donations from an odd number of workers?

Ans- Read the question twice and thrice and try to understand what this question wants to say. The first worker gives rupees to all 70 citizens. But 2nd worker gives only 2nd, 4th, 6th, 8th, etc. citizens. The same thing is done by 3rd, he gives rupees to only 3rd, 6th, 9th, 12th, etc. citizens. Same for 4th, 5th, 6th worker till 70th.

It means a worker whose number is x is giving money to all the citizens whose number is a multiple of x.

And hence, the citizen, whose number is y, is receiving the donations from all the workers whose number is a factor of y. i.e. is receiving as many donations as the number of factors of y.

So, we just have to find out, from 1 to 70, how many numbers have odd number of factors.

And from the previous discussion, we know that only perfect squares have the odd number of factors.

And from 1 to 70, there are 8 perfect squares, which are 1, 4, 9, 16, 25, 36, 49 and 64.

So, the answer is, only 8 senior citizens received donations from an odd number of workers.

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