1) Factors

• a whole number that divides exactly into another number.
• a whole number that multiplies with another number to make a third number.

Factors Pair

• a pair of numbers multiplied together form another number called their product.




Number of factors

A number N can be written as \( N=p^xq^yr^z\), where p,q, and r are prime factors of the number N, and x,y, and z are positive integers.

Number of factors : (including 1 and the number itself)\(=(x+1)(y+1)(z+1)\);

Sum of all the factors

\(\left[ \begin{array}{cc|r} \frac{p^{x+1}-1}{p-1} \end{array} \right]\) \(\left[ \begin{array}{cc|r} \frac{q^{y+1}-1}{q-1} \end{array} \right]\) \(\left[ \begin{array}{cc|r} \frac{r^{z+1}-1}{r-1} \end{array} \right]\)


Product all the factors

\(N^{\frac{number of factors}{2}}\)

Suppose 120 is our number and the prime factorization is \(2^3*3^1*5^1\)

1) Number of factors: simply take the exponent of each factor and add 1. \((3+1)*(1+1)*(1+1)=16\)

2) Sum of all factors: \(\frac{[(2^0+2^1+2^2+2^3)(3^0+3^1)(5^0+5^1)]}{[(2-1) (3-1)(5-1)]} = 45\)

3) Product of all factors: \(120^{\frac{16}{2}}=120^8\)







2) Primes


A number greater than 1 which has no factor other than 1 and the number itself is a Prime Number
A number with two distinct factors.
We do have infinite prime numbers and apparently, to our knowledge today, there is NO clear pattern in their unfolding toward the infinite on the number line.
Prime numbers are only positive.
Two (2) is the oNLy positive Even prime numbers
The difference between any two primes greater than 2 is always even.



Verifying the primality (checking whether the number is a prime) of a given number \(n\) can be done by trial division, that is to say dividing \(n\) by all integer numbers smaller than \(\sqrt{n}\), thereby checking whether \(n\) is a multiple of \(m\leq{\sqrt{n}}\).
Example: Verifying the primality of \(161\): \(\sqrt{161}\) is little less than \(13\), from integers from \(2\) to \(13\), \(161\) is divisible by \(7\), hence \(161\) is not prime.
Note that, it is only necessary to try dividing by prime numbers up to \(\sqrt{n}\), since if n has any divisors at all (besides 1 and n), then it must have a prime divisor.

If \(n\) is a positive integer greater than 1, then there is always a prime number \(p\) with \(n < p < 2n\).







3) Multiples





HCF (GCD / GCF) & LCM OF NUMBERS


HCF : It is the greatest factor common to two or more given numbers. It is also called GCF OR GCD (greatest common factor or greatest common divisor). e.g. HCF of 10 & 15 = 5, HCF of 55 & 200 = 5, HCF of 64 & 36 = 4

To find the HCF of given numbers, resolve the numbers into their prime factors and then pick the common term(s) from them and multiply them if more than one. This is the required HCF.

LCM : Lowest common multiple of two or more numbers is the smallest number which is exactly divisible by all of them. e.g. LCM of 5, 7, 10 = 70, LCM of 2, 4, 5 = 20, LCM of 11, 10, 3 = 330

To find the LCM resolve all the numbers into their prime factors and then pick all the quantities (prime factors) but not more than once and multiply them. This is the LCM.

NOTE:
1. LCM x HCF = Product of two numbers (valid only for “two”)
2. HCF of fractions = HCF of numerators ÷ LCM of denominators
3. LCM of fractions = LCM of numerators ÷ HCF of denominators


Calculating LCM :

Method 1 : Factorization Method

Rule – After expressing the numbers in terms of prime factors, the LCM is the product of highest powers of all factors.

Q. Find the LCM of 40, 120, 380.

A. 4\(0 = 4 \times 10 = 2 \times 2 \times 2 \times 5 = 2^3 \times 5^1\)

\(120 = 4 \times 30 = 2 \times 2 \times 2 \times 5 \times 3 = 2^3 \times 5^1 \times 3^1\)

\(380 = 2 \times 190 = 2 \times 2 \times 95 = 2 \times 2 \times 5 \times 19 = 2^2 \times 5^1 \times 19^1\)

⇒ Required LCM = \(2^3 \times 5^1 \times 3^1 \times 19^1 = 2280\).

Method 2 : Division method

Q. Find the LCM of 6, 10, 15, 24, 39




Ans. LCM = 2 x 2 x 3 x 5 x 2 x 13 = 1560.


Calculating HCF


Factorization
After expressing the numbers in term of the prime factors, the HCF is product of COMMON factors.

Ex. Find HCF of 88, 24, 124

\(88 = 2 \times 44 = 2 \times 2 \times 22 = 2 \times 2 \times 2 \times 11 = 2^3 \times 11^1\)
\(24 = 2 \times 12 = 2 \times 2 \times 6 = 2 \times 2 \times 2 \times 3 = 2^3 \times 3^1\)
\(124 = 2 \times 62 = 2 \times 2^1 \times 31^1 = 2^2 \times 31^1 ⇒ HCF = 2^2\)