1. BASICS NUMBERS

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1) Real_numbers (Cat syllabus)

2) Imaginary_numbers ( not in cat syllabus)

*Real numbers ranges from (-infinite to infinite). It includes all points below

I) Integers : { -infi..-3,-2,-1,0,1,2,3…infi}

2) Natural_numbers : {1,2,3,…infi}

3) Whole_numbers : {0,1,2,….infi}

4) Rational_numbers : I) All the above 3 types are Rational

ii) All the numbers whose decimal values are infinite (non terminating) but which have recurring pattern. Ex, 4.3333=13/3, 1.148148. = 93/81 are Rational numbers.

5) Irrational_numbers : Infinite non recurring decimal values ie neither they have pattern nor do they have end. Ex; rt3 rt5, Pi, e. are all irrational numbers.

=> And whenever you find there is a number less than 0 inside root, samajh jao ki imaginary hai . Ex : root (-2), root (-3)… etc

2. BASICS OF HCF AND LCM

-Number system mostly deals with natural numbers.

-Factors,Lcm,Hcf, multiples all mostly belongs to natural numbers. Though hcf, lcm can be applied on fractions too but also they are applied separately when numerator and denominator will be natural.

– #LCM (Least Common Multiple)

Common multiples of 7,2 are set of {14,28,42…..}

The least of these multiples ie 14 is the Least common Multiple. (Lcm)

– #HCF (Highest common Factor).

Factors of 80 are {1,2,4,5…80}

Factors of 144 are {1,2,3,….144}

List of common factors are: {1,2,4,8,16}

Highest of the common factors ie 16 is the Highest Common Factor (HCF).

3. PRIME NUMBERS

Prime numbers are those which have only two factors. 1 and the number itself. And hence its divisible by no other number.

While all other numbers apart from prime and 1 are composite numbers which have more than 2 factors.

=> 1 is neither prime nor composite.

=> 2 is the smallest prime

=> 2 is the only even prime

=> Lowest odd prime is 3.

=> Any prime number > 3 is of the form 6k+/-1 while converse may not be true.

=> For any prime p>3, p^2-1 is divisible by 24.

#Proof : as p^2-1= (6k+/-1)^2 -1 = 36k^2+/-12k+1-1 = 36k^2+/-12k=12k(3k+/-1)

Now if k= even, => 12*2a (odd)=24k

if k=odd, => 12k (even)=24k.

So always p^2-1 is divisible by 24.

4. METHOD TO CHECK IF THE NUMBER IS PRIME

i) Any number greater than 3 if gives a remainder of 1 or 5 when divided by 6 then it may be a prime and if not then it’s definitely not a prime.

ii) When we have to check if a number N is prime, we need to only check for its divisibility by prime factors below root(N).

Lets check 239, 15 < root(239) < 16

Prime numbers below 15 => 2,3,5,7,11,13.

As 239 is not divisible by any of the primes below 15 hence 239 is a prime number.

=> Why this works?

Whenever we have to find the factors of any number N we will get the factors in pairs (ie factor pair)

Further factor pairs will be such that in each pair,one of the factors will be lower than root(N) while other will be higher than root(N).

Extending this logic, wecan say that if we are not able to find a factor upto root (N), we will not be able to find any above root (N) and hence the number will be prime.

5. NUMBER OF FACTORS

Let N = a^x*b^y*c^z, where a,b,c are primes

=>Then number of factors = (x+1)(y+1)(z+1)

Ex: Number of factors of 360 = 2^3*3^2*5, is (3+1)(2+1)(1+1) = 4*3*2=24.

If N = 2^x*b^y*c^z, then number of Even Factors will be counted by taking a 2 common and finding factors of remaining numbers ie. 2(2^(x-1)*b^y*c^z.

and number of even factors = x(y+1)(z+1).

Ex: Number of even factors of 360=2^3*3^2*5 ie., factors of 2(2^2*3^2*5) is (2+1)(2+1)(1+1) = 3*3*2=18

And number of Odd Factors can be counted by eliminating the 2^x and find remaining factors of b^y*c^z => (y+1)(z+1)

Ex: Number of odd factors of 360=2^3*3^2*5 ie.,factors of 3^2*5 is (2+1)(1+1)=3*2= 6.

So Even factors+odd factors = Total factors

= > 18+6 = 24. #Hence_proved :p

6. FACTORS SUM

Let N = 360= 2^3*3^2*5

Then sum of all its factors is (2^0 + 2^1 + 2^2 + 2^3)(3^0 + 3^1 + 3^2)(5^0 + 5^1) = (15)(13)(6) = 1170

Sum of even factors: It can be easily calculated by just eliminating 2^0 from above and summing up remaining.

Sum of even factors of 360 = (2^1 + 2^2 + 2^3)(3^0 + 3^1 + 3^2)(5^0 + 5^1) = (14)(13)(6) = 1092

Why this works ? Coz even*odd is always even and after removing 2^0=1 we left with all evens in the first term which on multiplication with any other will give an even factor.

Sum of odd factors: For this remove all the even terms so that we only left with odd and odd*odd is always odd.

So Sum of odd factors of 360 = (2^0)(3^0 + 3^1 + 3^2)(5^0 + 5^1) = 1(13)(6) = 78

So Even sum + odd sum = total sum

=> 1092+78 = 1170. #Hence_Proved

7. FACTOR PRODUCT

The Product of factors of a number N is N^(m/2), where m is the number of factors of N.

Ex; The product of factors of 360 = 360^(m/2)

and number of factors,m = 24, so its 360^(24/2) = 360^12.

Why this works? coz in a number product of each factor pair will give the Number N.

1*360=360

2*180=360

3*120=360

4*90=360

5*72=360

6*60=360

8*45=360

9*40=360

10*36=360

12*30=360

15*24=360

18*20= 360

So observe carefully above, as there are 24 factors and hence there are 12 “Factor Pairs”

So multiplying all 24 factors = multiplying 12 factor pairs = multiplying 360 12 times, hence PRODUCT = 360^12

easynet1

NUMBERS WHOLE BASICS

How to determine if the number is prime

Why this works?

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