Teachers Eligibility Test - Paper 2
Class – VI Mathematics
NUMBER SYSTEM
Odd and Even Numbers
A
number is called an odd
number if it cannot be grouped
Equally in twos. 1, 3, 5, 7, …, 73, 75, …,
2009,… are
Odd numbers.
All odd numbers end with anyone of the digits 1, 3, 5, 7 or 9.
A number is called an even number if it can be
grouped
Equally in twos. 2, 4, 6, 8... 68, 70, . . ,
4592... Are
Even numbers.
All even numbers end with anyone of the digits
0, 2, 4, 6 or .
In
whole numbers, odd and even numbers come alternatively.
Prime Numbers
A natural number
greater than 1, having only two factors namely 1 and the number itself, is
called a prime
number.
For example, 2 (1 x 2) is a prime number as is 13 (1 x 13).
Composite Numbers
A natural number
having more than 2 factors is called a composite number.
For example, 15 is a composite number (15 = 1 x 3 x 5) as is 70 (1 x
2 x 5 x 7).
Expressing
a Number as the Sum of Prime Numbers
Any
number can always be expressed as the sum of two or more prime numbers.
Example
1:
Express 42 and 100 as the sum of two consecutive primes.
Solution:
42 = 19+23;
100 = 47+53
Example
2:
Express 31 and 55 as the sum of any three odd primes.
Solution:
31 = 5+7+19 (find another way, if possible!)
55 = 3 + 23+29
Twin
Primes
A pair of prime numbers, whose difference is
2, is called twin primes.
For example, (5, 7) is a twin prime pair as is (17, 19).
Rules for Test of Divisibility of Numbers
Divisibility by 2
A number is divisible by 2, if its ones place is any one of the even numbers 0, 2, 4, 6 or 8.
Examples:
1. 456368 are divisible by 2, since its ones
place is even (8).
2. 1234567 are not divisible by 2, since its
ones place is not even (7).
Divisibility by 3
Divisibility of a number by 3
is interesting! We can find that 96 are divisible by 3.
Here, note that the sum of its digits 9+6 =
15 is also divisible by 3.
Even 1+5 = 6 is also divisible by 3. This is
called as iterative or repeated addition.
So, a number is divisible by 3 if the sum of its digits is divisible by 3.
Examples:
1. 654321 are divisible by 3.
Here 6+5+4+3+2+1= 21 and 2+1=3 is divisible
by 3.
Hence, 654321 are divisible by 3.
2. The sum of any three consecutive numbers
is divisible by 3.
(For example: 33+34+35=102, is divisible by
3)
3. 107 is not divisible by 3 since 1+0+7=8,
is not divisible by 3.
Divisibility by 4
A number is divisible
by 4 if the last two digits of the given number are divisible by 4.
Note that if the last two digits of a number
are zeros, then also it is divisible by 4.
Examples:
71628, 492, 2900 are divisible by 4, because 28 and 92 are divisible by 4
and 2900 is also divisible by 4 as it has two zeros.
Divisibility by 5
Observe the multiples of 5.
They are 5, 10, 15, 20, 25,.., 95, 100, 105,
…., and keeps on going.
It is clear, that multiples of 5 end either
with 0 or 5 and so, A number is divisible by 5 if its
ones place is either 0 or 5.
Examples:
5225 and 280 are divisible by 5
Divisibility by 6
A number is divisible
by 6 if it is divisible by both 2 and 3.
Examples:
138,
3246, 6552 and 65784 are divisible by 6.
Divisibility by 8
A number is divisible by 8 if the last three digits of the given number are divisible by 8.
Note that if the last three digits of a
number are zeros, then also it is divisible by 8.
Examples:
2992 is divisible by 8 as 992 is divisible by 8 and 3000 is
divisible by 8 as its last three digits are zero.
Divisibility by 9
A number is divisible by 9 if the sum of its digits is divisible by 9. Note that the numbers
divisible by 9 are divisible by 3.
Examples:
9567 is divisible
by 9 as 9+5+6+7=27 is divisible by 9.
Divisibility by 10
A number is divisible by 10 if its ones place is only zero.
Observe that numbers divisible by 10 are also
divisible by 5.
Examples:
1. 2020 is divisible by 10 (2020÷10 = 202) whereas
2021 is not divisible by 10.
2. 26011950 is divisible by 10 and hence
divisible by 5.
Divisibility by 11
A number is divisible
by 11 if the difference between the sum of
alternative digits of the number is either 0 or divisible by 11.
Examples:
Consider the number 256795.
Here, the difference between the sum of alternative digits =(2+6+9)−(5+7+5)=17−17=0.
Hence, 256795 are divisible by 11.
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