Table of contents  
Introduction  
Understanding Numbers  
Classification of Numbers  
Divisibility Test  
Properties of Number  
Number System Questions  
Absolute Value of a Number 
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All numbers divisible by 2 are called even numbers.
Example: 2, 4, 6, 8, 10.…
Even numbers can be expressed in the form 2n, where n is an integer. Thus 0,  2, − 6, etc. are also even numbers.
Even and odd numbers
All numbers not divisible by 2 are called odd numbers.
Odd numbers can be expressed in the form (2n + 1) where n is an integer. Thus  1, − 3, − 9 etc. are all odd numbers.
Example: 1, 3, 5, 7, 9…
A natural number that has no other factors besides itself and unity is a prime number.
Examples: 2, 3, 5, 7, 11, 13, 17, 19 …
Important Observation About Prime Numbers
Tip: If a number has no prime factor equal to or less than its square root, then the number is a prime number.
Must Know
1 to 25 ⇒ 9 prime
1 to 50 ⇒ 15 prime
1 to 100 ⇒ 25 prime
1 to 200 ⇒ 45 prime
Example: If a, a + 2, a + 4 are consecutive prime numbers. Then how many solutions ‘a’ can have?
(a) One
(b) Two
(c) Three
(d) More than three
Correct Answer is Option (a)
 No even value of ‘a’ satisfies this. So ‘a’ should be odd.
 But out of three consecutive odd numbers, at least one number is a multiple of 3.
 So, the only possibility is a = 3 and the numbers are 3, 5, 7.
A composite number has other factors besides itself and unity.
Example: 8, 72, 39, etc.
On the basis of this fact that a number with more than two factors is a composite, we have only 34 composites from 1 to 50 and 40 composites from 51 to 100.
Tip: 1 is neither a prime number nor a composite number.
Example: 6 is a perfect number because the factors of 6, i.e., 1, 2 and 3, add up to the number 6 itself.
Also, 1/6 + 1/3 + 1/2 = (1+ 2+3/6) = 6/8 = 1 (Unity)
Other examples of perfect numbers are 28, 496, 8128, etc.
There are 27 perfect numbers discovered so far.
Two numbers are (prime or composite) said to be coprime to one another, if they do not have any common factor other than 1.
Example: 35 & 12, since they both don’t have a common factor among them other than 1.
Tip: The HCF of Coprime number is always 1.
A fraction denotes the part or parts of a unit. Several types are:
Examples: Finite decimal numbers, whole numbers, integers, fractions
i.e. 3/5, 16/9, 0.666...∞, 2/3, 7, 0 etc.
Examples: π, √5, √7, e
A non  terminating number on the basis of occurrence of digits after decimal can be classified as follows:
(a) Pure Recurring to Fractions
Funda 1: If a number is of the form of 0. ababab……. then divide the repeating digits with as many 9’s as we have repeated digits.
Example: 0.363636...= 36/99 = 4/11
(b) Mixed Recurring to Fractions
Funda 2: If N = 0. abcbcbc…. Then N = abc  a/990 = Repeated & non Repeated digits  Non Repeated digits/ As many 9's as repeated digits followed by as many zero as non  repeated digits
Example: 0.25757...= 257  2/990 = 255/990 = 17/66
(c) Funda 3. If N = a. bcbc…. Then
Write N = a + 0. bcbc….Proceed as Funda 1
5.3636… = 5 + 0.3636… = 5 + 36/99 = 59/11
As the name suggests, divisibility tests or division rules in Maths help one to check whether a number is divisible by another number without the actual method of division. If a number is completely divisible by another number then the quotient will be a whole number and the remainder will be zero.
The rule for divisibility by 7 is a bit complicated which can be understood by the steps given below:
Example: Is 1073 divisible by 7?
 From the rule stated remove 3 from the number and double it, which becomes 6.
 Remaining number becomes 107, so 1076 = 101.
 Repeating the process one more time, we have 1 x 2 = 2.
 Remaining number 10 – 2 = 8.
 As 8 is not divisible by 7, hence the number 1073 is not divisible by 7.
If the difference of the sum of alternative digits of a number is divisible by 11, then that number is divisible by 11 completely.
In order to check whether a number like 2143 is divisible by 11, below is the following procedure:
Example 1: 28 + 82 = 110, this is divisible by 11. At the same time, the difference between those numbers will be divisible by 9.
Example 2: 82 – 28 = 54, this is divisible by 9.
Example: 3^{3} + 2^{3} = 35 and is divisible by 5, which is (3 + 2).
The given expression can be broken as: 18n^{2}/n + 9n/n + 8/n. This gives us: 18n + 9 + 8/n.
 Now we can see that whatever the value of ‘n’, 18n + 9 will always give an integral value. Therefore, it now depends upon 8/n the only ⇒ n can have any integral value, which is a factor of 8.
 The integers, which will satisfy this condition are ±1, ±2, ±8, ±4. Thus, in total, n can take 8 values.
Example 2. N = 960. Find the total number of factors of N.
 The number of divisors of a composite number: If D is a composite number in form D = a^{p} × b^{q} × c^{r}, where a, b, c are primes, then the no. of divisors of D, represented by n is given by n = (p+1)(q+1)(r +1).
 Following the same, after dividing 960 into prime factors: 2^{6} × 3^{1} × 5^{1}, we can calculate the total number of factors as (6+1). (1+1).(1+1) = 28.
Example 3. Find the unit digit of the following expression: (123)^{34} × (876)^{456} × (45)^{86}.
 Whenever an even unit digit and a 5 at the unit digit are present, they will always give a 0 at the unit digit, no matter if any other number is present or not.
Hence, with this approach, this question proves to be a sitter.
Example 4. What will be the number of zeroes at the end of the product of the first 100 natural numbers?
 In this kind of questions, you need to find the greatest power of 5, which can divide the product of the first 100 natural numbers.
 Remember, a multiple of 5 multiplied by any even number, gives you a zero. Now divide 100 by 5 and take 20 as the quotient. Then divide 20 (the quotient) by 5 and get the new quotient 4, which further cannot be divided by 5.
 The sum of all such quotient gives you the greatest power of 5, which can divide that number. The sum is 24 and this is the number of zeroes at the end of the product of the first 100 natural numbers.
Example 5. Which letter should replace the $ in the number 2347$98, so that it becomes a multiple of 9?
 As you know that if the sum of all the digits is divisible by 9, then the number is divisible by 9.
 Now sum of the given digits is 2 + 3 + 4 + 7 + 9 + 8 = 33 + $. Now think the next multiple of 9 after 33 i.e. 36. This means you add 3 in this. The value of $ is 3.
Example 6. In a party, there are 20 persons present. If each of them shakes hand with all the other persons, in total, how many handshakes will take place?
 Out of 20 persons, the first person will shake hand with 19 persons.
 The second person will shake hand with 18 persons (because he has already shaken hand with first person).
 The third person will shake hand with 17 persons and so on.
 The second last person shakes hand with only one person.
 And last will shake hand with none (because he has already shaken hand with all persons).
 In order to find the total number of handshakes, you have to add all the natural numbers from 1 to 19, i.e. ∑ 19. ∑19 = 19 x 20/2 = 190 handshakes.
Example 7. What is the remainder when 2354789341 is divided by 11?
 Odd place digit sum (O) = 1 + 3 + 8 + 4 + 3 = 19.
 Even place digits sum (E) = 4 + 9 + 7 + 5 + 2 = 27.
 Difference (D) = 19  27 =  8
 Remainder = 11  8 = 3.
Tip: ⇒ When any number with even number of digits is added to its reverse, the sum is always divisible by 11. e.g. 2341 + 1432 = 3773, which is divisible by 11.
⇒ Any number written 6 times consecutively will be divisible by 7 and 13.
Example 8. If 567P55Q is divisible by 88; Find the value of P + Q.
(a) 11
(b) 12
(c) 5
(d) 6
(e) 10
Correct Answer is Option (e)
 The number is divisible by 8 means; the number formed by the last 3 digits should be divisible by 8, which are 55 Q. Only Q = 2 satisfies this.
 From the divisibility rule of 11, (2 + 5 + 7 + 5)  (5 + P + 6) is divisible by 11. So 8P is divisible by 11.
 If P= 8, then only it is possible. So P = 8 and Q = 2.
 So, P + Q = 10.
Example 9. If the first 100 natural numbers are written side by side to form a big number and it is divided by 8. What will be the remainder?
(a) 1
(b) 2
(c) 4
(d) 7
(e) Cannot be determined
Correct Answer is Option (c)
 The number is 1234…..9899100.
 According to the divisibility rule of 8, we will check only the last 3 digits.
 If 100 is divided by 8, the remainder is 4.
Example 10. What will be the remainder when 4444……..44 times is divided by 7?
(a) 1
(b) 2
(c) 5
(d) 6
(e) 0
Correct Answer is Option (b)
 If 4 is divided by 7, the remainder is 4.
 If 44 is divided by 7, the remainder is 2.
 If 444 is divided by 7, the remainder is 3.
 By checking like this, we come to know that 444444 is exactly divisible by 7.
 So if we take six 4’s, it is exactly divisible by 7.
 Similarly, twelve 4’s is also exactly divisible by 4 and 42 4’s will be exactly divisible by 7. So out of 44, the remaining two 4,s will give a remainder of 2.
Example 1. If ‘X’ is an even number; Y is an odd number, then which of the following is even?
(a) X^{2} + Y
(b) X + Y^{2}
(c) X^{2} + Y^{2}
(d) X^{2}Y^{2}
(e) None of these
Correct Answer is Option (d)
 Since X is even, X^{2} is even.
 Y is odd, Y^{2} is odd.
 So options (1), (2), (3) are even + odd = odd.
 Option (4) is (even) (odd) = Even.
Example 2. What is the difference between 0.343434....…and 0.2343434…… in fraction form?
(a) 6/55
(b) 6/11
(c) 9/55
(d) 9/13
(e) 5/11
Correct Answer is Option (a)
0.343434.....= 34/99 and 0.23434.... = 234  2/990 = 232/990
∴ Difference = 34/99  232/990 = 108/990 = 6/55
Example 3. How many of the following numbers are divisible by at least 3 distinct prime numbers 231, 750, 288 and 1372?
(a) 0
(b) 1
(c) 2
(d) 3
(e) 4
Correct Answer is Option (c)
 231 = 3 × 7 × 11 (3 prime factors)
 750 = 2 × 3 × 53 (3 prime factors)
 288 = 25 × 32 (only 2 prime factors)
 1372 = 22 × 73 (only 2 prime factors)
So, only 231 & 750 has 3 prime factors
Example 4. n^{3} + 6n^{2} + 11n + 6 (where n is a whole no) is always divisible by
(a) 4
(b) 5
(c) 6
(d) 8
(e) 12
Correct Answer is Option (c)
 n^{3} + 6n^{2} + 11n + 6 = (n + 1) (n + 2) (n + 3).
 Product of 3 consecutive numbers is always divisible by 3! = 6. (or)Take n = 0, 1, 2, 3 and check it is always divisible by 6.
Example 5. What is the remainder, if 351 × 352 × 353 ×        × 356 is divided by 360?
(a) 0
(b) 1
(c) 2
(d) 3
(e) 359
Correct Answer is Option (a)
 Since the given is the product of 6 consecutive numbers, it is always divisible by 6! = 720.
 It is divisible by 360 also. So, the remainder will be 0.
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