# NUMBER SYSTEM

## Number System

Number System is divided into 4 parts we are providing short tricks to remember all the parts of number system:

• Classification
• Divisibility Test
• Division& Remainder Rules
• Sum Rules

TRICKS TO REMEMBER NUMBER SYSTEM :

1. Classification

 Types Description Natural Numbers: all counting numbers ( 1,2,3,4,5….∞) Whole Numbers: natural number + zero( 0,1,2,3,4,5…∞) Integers: All whole numbers including Negative number + Positive number(∞……-4,-3,-2,-1,0,1,2,3,4,5….∞) Even & Odd Numbers : All whole number divisible by 2 is Even (0,2,4,6,8,10,12…..∞) and which does not divide by 2 are Odd (1,3,5,7,9,11,13,15,17,19….∞) Prime Numbers: It can be positive or negative except 1, if the number is not divisible by any number except the number itself.(2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61….∞) Composite Numbers: Natural numbers which are not prime Co-Prime: Two natural number a and b are said to be co-prime if their HCF is 1.

2. Divisibility

 Numbers IF A Number Examples Divisible by 2 End with 0,2,4,6,8 are divisible by 2 254,326,3546,4718 all are divisible by 2 Divisible by 3 Sum of its digits is divisible by 3 375,4251,78123 all are divisible by 3. [549=5+4+9][5+4+9=18]18 is divisible by 3 hence 549 is divisible by 3. Divisible by 4 Last two digit divisible by 4 5648 here last 2 digits are 48 which is divisible by 4 hence 5648 is also divisible by 4. Divisible by 5 Ends with 0 or 5 225 or 330 here last digit digit is 0 or 5 that mean both the numbers are divisible by 5. Divisible by 6 Divides by Both 2 & 3 4536 here last digit is 6 so it divisible by 2 & sum of its digit (like 4+5+3+6=18) is 18 which is divisible by 3.Hence 4536 is divisible by 6. Divisible by 8 Last 3 digit divide by 8 746848 here last 3 digit 848 is divisible by 8 hence 746848 is also divisible by 8. Divisible by 10 End with 0 220,450,1450,8450 all numbers has a last digit zero it means all are divisible by 10. Divisible by 11 [Sum of its digit in odd places-Sum of its digits in even places]= 0 or multiple of 11 Consider the number 39798847(Sum of its digits at odd places)-(Sum of its digits at even places)(7+8+9+9)-(4+8+7+3)(23-12)23-12=11, which is divisible by 11. So 39798847 is divisible by 11. Divisible by 12 [The number must be divisible by 3 and 4] Consider the number 462157692 is divisible by 12 { last 2 digits 92, so divisible by 4, and sum 4+6+2+1+5+7+6+9+2 = 42 is divisible by 3} , So 462157692 is divisible by 12. Divisible by 13 [Multiply last digit with 4 and add it to remaining number in given number, result must be divisible by 13] Consider the number 4568 is not divisible by 13 { 456 + (4*8) = 488 –> 48 + (4*8) = 80, 80 is not divisible by 13.} , So 4568 is not divisible by 13. Divisible by 14 [The number must be divisible by 2 and 7. Because 2 and 7 are prime factors of 14.] Divisible by 15 [The number should be divisible by 3 and 5. Because 3 and 5 are prime factors of 15.] Divisible by 16 [The number formed by last four digits in given number must be divisible by 16.] 7852176 is divisible by 16 –> 2176 is divisible by 16. Divisible by 17 [Multiply last digit with 5 and subtract it from remaining number in given number, result must be divisible by 17] Divisible by 18 [The number should be divisible by 2 and 9] Divisible by 19 [Multiply last digit with 2 and add it to remaining number in given number, result must be divisible by 19] Divisible by 20 [The number formed by last two digits in given number must be divisible by 20.]

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TRICKS

Generalized Divisibility rules :

To check divisibility by a number You should check divisibility by highest power of each of its prime factors.
Remember that the divisibility by any two of the factors of a number is not sufficient to judge its divisibility.

TRICKS 1. Suppose you are checking for divisibility by 12. 2,3,4,6 are factors of 12. You cannot say that the number is divisible by 12 by checking divisibility by only 2 and 4 or 4 and 6. You should check divisibility by 3 and 4 to tell the divisibility by 12. Because 3 and 4 are prime factors of 12.

TRICKS 2. Now, we are checking divisibility by 18. Prime factors of 18 are {2 and 3²}. So, you should check divisibility by 2 and 9 . don’t check for just 2 and 3 and also don’t check for 3 and 6.

TRICKS 3 List of Prime Factors of numbers upto 100

Basically prime numbers have factors 1 and the number itself. So, we can’t this rule apply for prime numbers. This is only for composite numbers.

Divisibility by numbers which ends in 1,3,7,9
So, this rule counts for prime numbers which have been missed in previous rule.
TRICKS 4
To test divisibility by a number N which ends in 1,3,7,9 this method can be used.

Multiply N with any number to get 9 in the end. Add 1 to the result and divide it by 10.

Store the above result as R.

We are checking whether a number X is divisible by N or not.
Split X as X = 10 y + z ;
X is divisible by N, only if Rz + y is divisible by N.

EXAMPLE (A):
Find whether 645 is divisible by 23 or not
.
N =23 ; 23 * 3 = 69 ; so now N has 9 in the end.
R = (69 + 1) / 10 = 7 ;
X = 645 ; split X as X = 10 y + z ;
645 = (10 * 64) + 5 ;
Y = 64 ; z = 5 ;
Rz + y = (7 * 5) + 64 = 35 + 64 = 99 ;
99 is not divisible by 23. So, 645 is also not divisible by 23.

EXAMPLE (B) :
Let us find 585 is divisible by 39 or not.

N =39 ; so now N has 9 in the end.
R = (39 + 1) / 10 = 4 ;
X = 585 ; split X as X = 10 y + z ;
645 = (10 * 58) + 5 ;
Y = 58 ; z = 5 ;
Rz + y = (4 * 5) + 58 = 20 + 58 = 78 ;
78 is divisible by 39. So, 585 is also divisible by 39.
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3. Division & Remainder Rules

Suppose we divide 45 by 6 hence ,represent it as:

dividend = ( divisorquotient ) + remainder

or

divisior= [(dividend)-(remainder] / quotient

could be write it as

x = kq + r where (x = dividend,k = divisor,q = quotient,r = remainder)

Example:

On dividing a certain number by 342, we get 47 as remainder. If the same number is divided by 18, what will be the remainder ?

Number = 342k + 47

( 18 ✘19k ) + ( 18 ✘2 ) + 11

18 ✘( 19k + 2 ) +11.

Remainder = 11

4. Sum Rules:

(1+2+3+………+n) = 1/2 n(n+1)

(12+22+32+………+n2) = 1/6 n (n+1) (2n+1)

(13+23+33+………+n3) = 1/4 n2 (n+1)2

Arithmetic Progression (A.P.)

a, a + d, a + 2d, a + 3d, ….are said to be in A.P. in which

a = First Term and d = Common Difference .

Let the nth term be tn and last term = l, then

(a) nth term = a + ( n – 1 ) d
(b) Sum of n terms =n/2[2a + (n-1)d]
(c) Sum of n terms = n/2 (a+l) where l is the last term