number system is very important chapter for all competitive exam . everything is included in it like all numbers,some special number,divisibility concept,unit digit,divisibility remainder concept,decimal and number of factor,key to remainder concept,lcm and hcf etc.
UNIT DIGIT
Unit digit is used to represent the "ones" place value of any number or product.
Example:- 12 x 10 = 120, '0' is the unit's digit in the given product.
In competitive examination we get different types of questions of Unit digits like
* find the unit digit of (562)^44
* find the unit digit of (4569)^927 x (7596)^44..... and soon.
Find the unit digit of an exponential number
First , divide the power of that number by 4
1. Condition (a):- if you get the remainder zero then change the power of that number by 4.
Example: we have to find the unit digit of this number (3846)84,
Step 1: At first we have to divide 84 by 4
Step 2nd: After dividing you will get 0 as remainder
Step 3rd: change the powers of given number with 4 and then find the unit digit by multiplying the unit digit four times
In the above condition (3846)^84 = (3846)^4
So, the unit digit will be the unit digit of 6 x 6 x 6 x 6. That is 6.
II. Condition (b):- if you get the remainder other than zero i.e (1, 2, 3)
Step 1st:- change the power of given number with the remainder that you get in the above division
Step 2nd: and then multiply the unit digit of that number in the times equal to the remainder.
Example:- if we have to find the unit digit of (342)^43
After dividing 43 by 4, we will get 3 as remainder
So, we can write (342)^43 as (342)^3 to find the unit digit
Now after multiplying 2 three times we will get the unit digit of (342)^43 number.
Number of zeros at the end of the number can be found by finding the number of pairs of 2 and 5.
DIVISIBILITY AND REMAINDER CONCEPT
Dividend = Divisor x Quotient + Remainder
key to Remember
When any positive number A is divided by any other positive number B and if B > A, then the remainder will be A itself. In other words, if the numerator is smaller than the denominator, then numerator will be the remainder.
Example:- Remainder of 5/12 = 5, Remainder of 21/45 = 21 etc.
Common Mistake
The remainder will calculate in its actual from i.e. you will not reduce the fraction to its lower ratio.
Example:- The remainder of 1/2 = 1, The remainder of 2/4 = 2, The remainder of 3/6 = 3 etc. It will observe that despite all the fraction being equal, remainders are different in each case.
Theorem Base
The product of any two or more than two natural numbers have the same remainder when divided by any natural number, as the product of their remainders.
Example:- Remainder of (12x13)/7= Remainder of 156/7 = 2
Normal way : Product > Remainder
Theorem Base: Remainder > Product > Remainder
therefore, first of all, we shall find out the remainder of each individual number and then we will multiply these individual remainders to find out final remainders. Since, Remainder of 12/7 = 5 and Remainder of 13/7 = 6 So, Remainder of (12 x 13)/7= Remainder of (5 x 6)/7= Remainder of 30/7 = 2
* If a^x/a-1 then remainder will always be 1, whether x is even or odd.
* if a^(p-1)/p then remainder will be 1,where P is a prime number.
* if (ma-1)^x/a then remainder will be (a-1), where x is an odd number.
* If (ma+1)^x/a then remainder will be 1,where x is an even number.
* If (ma+1)^x/a then remainder will be 1,whether x is even or odd.
DECIMAL AND NUMBER OF FACTOR
* The decimal representation of a rational number is either finite or infinite recurring.
Example:- 2/5= 0.4 (finite), 5/3 = 1.6666..... (infinite recurring)
* If decimal number 0.a and 0.ab are given, then they can be expressed in the form of p/q.
Example:- 0.a = a/10 and 0.ab =ab/100
* If decimal recurring numbers 0.a and 0.ab are given, then they can be expressed in the form of p/q As 0.a recurring = a/9 and 0.ab( recurring of seperate a and b) = ab/99.
Example:- 0.4(recurring only 4) = 4/9, 0.86( recurring of 86) = 86/99
* The recurring decimal numbers of type 0.ab( recurring of b) or 0.abc( recurring of seperate b and c)may be converted to rational form as p/q follows: (ab-a)/90 and (abc-a)/990.
Example:- 0.57 = (57-5)/90
NUMBER FACTOR
To find the factors of any number, number of factors, sum of all factors, number of even factors, sum of all even factors, number of odd factors and the sum of odd factors, we can follow these rules and formulae:
(Let the number be 60)
Rule: First of all find all the prime factor of the given number by factorization method
Factor of 60 = 2^2 multiply 3 multiply 5 .
KEYS TO REMAINDER CONCEPT
* Number of factors = multiplication of (powers of each prime factor + 1)
So, Number of factors= (6+1) x (2+1) × (3+1)=84
* Sum of all factors = (x^0 + x^1+ +x^m) x (y^0 + y^1 + ... + y^n) x (z^0 + z^1 + ... + z^q), where x, y and z are the factors and m, n and q are the powers of the prime factor respectively.
So, Sum of all factors (2^0 + 2^1 + ... + 2^5) x (3^0+3^1) x (5^0 + 5^1) = 1512
* Number of even factors = multiplication of (powers of each prime factor + 1, except even number)
So, Number of even factors = 7 x (1+1) x (1 + 1) =28
* Sum of even factors = (a^1 + ....+ a^m) x (b^0 + b^1+...+ b^n) x (c^0 + c^1+ ....+ c^q), where a (even prime factor), b and c are prime factors and m, n and q are the powers of the prime factor respectively.
So, Sum of even factors = (2^1+. ....+2^5) x (3^0 + 3^1) x (5^0 + 5^1) = 1488
* Number of odd factors = multiplication of (powers +1) of odd prime factors
So, Number of odd factors = (3 + 1) x (1+1)=8
* Sum of odd factors = (b^0 + b^1+...+ b^n) x (c^0+ c^1+...+c^q).......... Where b and c are odd prime factors and n and q are the powers of the odd prime factor respectively .
So, Sum of odd factors = (3^0+ 3^1) x (5^0 + 5^1)=24
* Number of prime factors = Sum of powers of prime factors
So, Number of prime factors = 7+5+1=13
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