**Natural Numbers:-** These are the numbers (1,2,3 etc) that are used for counting. In other words, all positive integers are natural numbers.

**Prime Numbers:-** A natural number larger than unity is a prime number if it does not have other divisors except for itself & unity.

**Whole Numbers:-** The set of numbers that includes all natural numbers and the number zero are called whole numbers.

**Composite Numbers:-** It is a natural number that has at least one divisor different from unity and itself.

**Real Numbers:-** All numbers between negative infinity & positive infinity that can be represented on a number line are called real numbers.

Rational Numbers:- A rational numbers is defined as a number of the form (a/b) where a and b are integers and b ≠ 0.

**Rational numbers that are not integers have decimal values, can be of two types:**

**1.Terminating Decimal Fraction**

**2.Non- Terminating Decimal Fractions**

Terminating decimal fractions :- For example 17/4 = 4.24 , 21/5 = 4.2

Non-Terminating decimal fractions: Nonterminating decimal fractions are of two types:

Non-Terminating Repeating Fractions: These are a non terminating repeating fraction. For Example 5.333….., 6.161616…..

Neither terminating nor repeating fractions: For example 5.273612…..

**Concept of GCD (Greatest Common Divisor or Height Common Factor)**

Consider two natural numbers n1 and n2.

If the numbers n1 and n2 are exactly divisible by same number x, then x is a common divisor of n1 and n2.

The highest of all the common divisors of n1 and n2 is called as the GCD or The H.C.F:-

Suppose we have to find H.C.F of 150,210 and 375.

**Step 1:- **Writing down the standard form of numbers

150=5×5×3×2

210=5×2×7×3

375=5×5×5×3

** **

**Step 2:- W**riting prime factors common to all the three numbers are taking their product is 5^{1} × 3^{1}

**Step 3:-** H.C.F = 5×3= 15

** Concept of LCM (Least common multiple)**

Let n1 and n2 be two natural numbers distinct from each other. The smallest natural number n that is exactly divisible by n1 and n2 is called the least common multiple (LCM)

**Steps for calculation of LCM:-**

Suppose we have to find LCM of 150,210,375.

**Step 1:-** Write down the standard form of numbers

150=5×5×3×2

210=5×2×7×3

375=5×5×5×3

**Step 2:- ** Write down all the prime factors that appear at least once in any of the numbers: 5,3,2,7.

**Step3: **-Raise each of the prime factor to their highest available power

The LCM= 2 × 3 × 5^{3} × 7 = 5270

**HCF OF TWO OR MORE FRACTION IS GIVEN BY :**

=HCF of Numerators Divided By LCM Of Denomination

**LCM OF TWO OR MORE FRACTIONS IS GIVEN BY: **

=LCM of Numerators Divided By HCF of Denominators

**QUES:- **Three traffic signals change after 36,42 and 72 seconds respectively.If the lights are first switched on at 9:00 sharp. At What time will they change simultaneously?

**Ans:-In order** to distribute $4000 and 180 pencils evenly, the numbers of employees must be a factor of each of these two numbers. We have to find the greatest number of employees possible therefore, we have to find HCF of $4000 and 180.

4000=2×2×2×5×2×5×2×5 =2^{5} × 5^{3}

180= 2×9×2×5 = 2^{2} × 3^{2} × 5

HCF = 2^{2} × 5 = 20

** Correct Answer = 20**

**Ques:-** The number of students who attended a school could be divided among 10,12 or 16 buses, such that each bus transports an equal number of students that could attend the school?

**Ans:-** No, of students must be divisible by 10,12 and 16. We have to find a minimum number of students that could attend the school or we have to find a minimum number that is divisible by 10,12 and 16.

LCM of 10,12,16 10= 2×5

=2^{4 }× 3×5 12=2×2×3

=16×15 16=2×2×2×2

ANS:- (240)

**Ques**:- ** **There are 576 boys and 448 girls in a school that are to be divided into sections containing equal no. of students of either boys or girls alone. Find the minimum total number of sections thus formed

a)24 b)32

c)16 d)20

**Ans:- **We have to find HCF of 576 and 448

576= 3^{2} × 2^{6}

448= 2^{6} × 7

HCF= 2^{6 = }64

Number of sections = 576/64 + 448/64 = (9+7) = 16 – Ans

**Divisibility:- **Any integer I (Commonly called dividend)

When divided by a natural number N (Called Divisor)

There exist a unique pair of number Q AND R ,Which are called the quotient and remainder respectively.

Dividend=Divisor × Quotient + Remainder

Suppose We divide 42 by 5. The result has a quotient of 8 and remainder of 2.

**Ques:-** Find the sum of factors and the number of factor of 240= 2^{4} × 4^{1} × 5^{1}

Total Number of divisor = (1+4) (1+1) (1+1)

=5×2×2= 20

Sum of these divisors

= (2^{0} + 2^{1} + 2^{2} + 2^{4} ) (3^{0} + 3^{1}) (S^{0} + S^{1})

= (1+2+4+8+16) (4) (6)

=24 (31) = (744)

**Ques:-** **Quantity A ** **Quantity B**

The number of distinct The number of distinct

Positive factor of 32 positive factors of 20

Ans:- 32 = 2^{5 } , 20= 2^{2 } 5^{1}

Number of distinct positive factors of 32 = (6)

Number of distinct positive factors of 20 = 3*2 = 6

**The two quantities are equal.**

The highest power of a prime number p, which divides n exactly is given by

= [n/p] + [n/p^{2}] +[n/p^{3}]……..

Where [x] denotes the greatest integer less than or equal to x

**Remainder Theorem:-**

The remainder of the expression [A_{1 }× B_{1} × C_{1} ………………+ A_{2 }× B_{2} × C_{2}] / m

Will be the same as the remainder of the expression

** **[A_{1R }× B_{1R} × C_{1R} ………………+ A_{2R }× B_{2R} × C_{2R}] / m

Where A_{1R } is the remainder in each of the following cases:

- 17 ×23×126×38 divided by 8

**Ans:- ( **17 ×23×126×38 / 8 **)**

** = (1**×7×6×6 / 8 **) = ( 42**×6 / 8**) = (2*6 / 8 ) = ( 12 / 8 )**

** **

**Remainder= ( 4 )**

** **

** **

** **

**Ques:-** If a=16 b and b is a prime number greater than 2, How many positive distinct factors does A have?

**Ans:- a **=16b = 2^{4}. b^{1}

No. of positive distinct factors = ( 4 + 1 ) ( 1 + 1 ) = ( 10 )

**Ques:- **If x is a positive integer, which one of the following could be the remainder. When 73^{x } is divisible by

a)0 (b)1 (c)2 (d) 3 (e)4 (f)5 (g) 6 (h) 7 (i) 8 (j) 9

Indicate all such remainders

** **

**Ans:-** We have to find the unit digit of 73^{x}

As with multiplication, when an integer is raised to a power, the unit digit mis determined solely by the product of units digits.

Those products will form a repeating pattern.

3^{1}= 3 3^{2} =9 3^{3 }=27 3^{4 }= 81 and 3^{5 }=243

Here the pattern returns to its original value of 3 and we have got a repeating pattern

3,9,7,1,3,7,1,3 —–

Thus the unit digit of 73^{x } must be 1,3,7,9

**Divisibility Rules:-**

Divisibility by 2-A number is divisible by 2 if the last digit is divisible by 2.

Divisibility by 5 – A number is divisible by 5 if the last digit is 5 or 0

Divisibility by 4 – A number is divisible by 4 if the last two digit are divisible by 4

Divisibility by 8 – A number is divisible by 8 if the last 3 digit are divisible by 8

Divisibility by 11 – A number is divisible by 11 if the difference of the sum of the digits in the odd places and the sum of the digits in the even places is zero or divisible by 11.