Factors & Multiples
Category : 6th Class
FACTORS AND MULTIPLES
FACTOR:
Example: 16 = 1, 2, 4, 8, 16., so, 1, 2, 4, 8 and 16 are factors of 16.
25 = 1, 5, 25. so, 1,5 and 25 are factors of 25.
MULTIPLE:
or
Example: 5=5, 10, 15, 20.........and so on
\[5\times 1=5,\,\,\,5\times 2=10,5\times 3=15,5\times 4=20....................\]Thus,
5, 10, 15, 20.....................are multiple of 5.
Properties of factors and multiples
Example: \[1=1\times 1,\text{ }2=1\times 2,\text{ }5=1\times 5,\]and so on.
Example: \[2\times 1=2,\text{ }15\times 1=15,\text{ }18\times 1=18,\text{ }20\times 1=20\]and so on.
Example: \[0=1,2,3,4,..........\left( \because \,\,\,\,\frac{0}{1}=0,\,\,\frac{0}{2}=0,\,\,\frac{0}{3}=....... \right)\]
\[0=1,2,3,4,..............(\because \,\,\,0\times 1=0,\,0\times 2=0,.....)\]
Some Number related to factors and multiple
Example: 6 = 1, 2, 3, 6 the factors of 6 are 1, 2, 3 and 6.
Sum of all factors \[=1+2+3+6=12=2\times 6\]i.e., sum of all factors of \[6=2\times 6\]so, 6 is a perfect number.
Note: In 1 to 1000 there are three perfect numbers, 6, 28 and 496.
EVEN NUMBERS
Here we know that 0, 2, 4, 6, 8, 10, 12, 14,...............are multiples of 2.
Question: Why zero is an even number?
Answer: check the first definition of "even", It is an integer multiple of 2, specifically\[0\times 2\].
ODD NUMBERS
Example: 1, 3, 5, 7, 9, 11, 13,.......... are odd numbers.
PRIME AND COMPOSITE NUMBERS
Prime Numbers
2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97.
Composite Number
or
Example: 6 = 1, 2, 3, 6 (1, 2, 3 and 6 are the factors of 6) 6 have more than two factors.
\[\therefore \] 4 = 1, 2, 4 (1, 2 and 4 are the factors of 4)
Example: 4, 6, 8, 9, 10, 12, 14, 15...........are composite numbers.
Some facts about factors
Let N denote the number, and consider
\[N={{a}^{p}}{{b}^{q}}{{c}^{r}}\]............... where a, b, c,..................
are different prime numbers and p, q, r,..................
are positive integers.
then no. of factors\[=(p+1)(q++1)(r+1)\]......................
Example: How many factors 48 have.
\[48={{2}^{4}}={{3}^{1}}\]
2 
48 
2 
24 
2 
12 
2 
6 

2 
\[\therefore \] No. of factors of \[48=(4+1)\times (1+1)=5\times 2=10\]
Example: 2 Find the no. of factors of 120.
Solution: \[120={{2}^{3}}\times {{3}^{1}}\times 5\]
2 
120 
2 
60 
2 
30 
2 
15 

5 
\[\therefore \] No. of factors of \[120=(3+1)\times (1+1)\times (1+1)=4\times 2\times 2=16\]
Let N denote the number, and consider \[N={{a}^{p}}.{{b}^{p}}.{{c}^{r}}\].....................where a, b, c ................are different prime numbers and p, q, r ...are positive integers.
Then the required sum \[\frac{{{a}^{p+1}}1}{a1}.\frac{{{b}^{q+1}}1}{b1}.\frac{{{c}^{r+1}}1}{c1}............\]
Example: Find the sum of all the factors of 180.
2 
180 
2 
90 
2 
45 
2 
15 

5 
Solution: \[180={{2}^{2}}\times {{3}^{2}}\times 5\]
The no. of factors \[=(2+1)(2+1)\times (1+1)\]\[=3\times 3\times 2=18\]
\[\therefore \] Sum of the factors \[=\frac{{{2}^{2+1}}1}{21}.\frac{{{3}^{2+1}}}{31}.\frac{{{5}^{1+1}}1}{51}\]
\[={{2}^{3}}1\times \frac{{{3}^{3}}1}{2}\times \frac{{{5}^{2}}1}{4}\]
\[=\frac{7\times 26\times 24}{8}=546\]
Twin  Primes
Example: (3, 5), (5, 7) etc...
Prime Triplet;
Co  Prime:
(2, 3), (3, 4), (5, 6), (8, 13), etc... are pairs of coprimes.
Prime Factorization
Example:
2 
50 
5 
25 
5 
5 

1 
\[\therefore \,\,\,\,50=2\times 5\times 5\]
2 
180 
2 
90 
3 
45 
3 
15 
5 
5 

1 
\[\therefore \,\,\,180=2\times 2\times 3\times 3\times 5\]
Highest common factor (H. C. F)
HCF by Prime factorization method
Example: Find the HCF of 20 and 25.
\[20=2\times 2\times 5\]
\[25=5\times 5\]
Common factor of 20 and 25 is 5.
Hence HCF of 20 and 25 is 5.
HCF by long division method
To find the HCF follows these Rules.
Example: 1 Find the HCF of 120 and 150.
Solution:
\[\therefore \] HCF of 120 and 150 is 30.
Example: 2 Find the HCF of 120, 160 and 200.
\[\therefore \]HCF of 120, 160 and 200 is 40.
Note: (i) HCF of the distinct prime numbers is one.
(ii) HCF of two coprime number is one.
(iii) HCF of an even number and an odd number is one.
(iv) HCF of two consecutive even number is two.
Least Common Multiple (LCM)
LCM can be determined by three method.
1. By writing the multiple of given number.
2. By prime factorization method.
3. By division method.
LCM by writing multiples of given number
Example: Find the LCM of 5 and 15.
Solution: Multiples of 5 are 5, 10, 15, 20, 25, 30, 35, 40, 45...............
Multiples of 15 are 15, 30, 45, 60, 75, 90, 105...................
Common multiple of 5 and 15 are 30, 45.........................
Here the least common multiple of 5 and 15 is 30.
LCM by Prime factorization
Follows these steps
Example: 1 Find the LCM of 20, 24.
Solution \[20=2\times 2\times 5,\text{ }24=2\times 2\times 2\times 3\]
\[20={{2}^{2}}\times 5,\text{ }24={{2}^{3}}\times 3\]
Select the greater power of every factor.
LCM \[=23\times 3\times 5=120.\]
Example: 2 Find the LCM of 24, 36 and 40 by prime factorization method.
Solution \[24=2\times 2\times 2\times 3=23\times 3\]
\[36=2\times 2\times 3\times 3={{2}^{2}}\times {{3}^{2}}\]
\[40=2\times 2\times 2\times 5=23\times 51\]
\[\therefore \] LCM of 24, 36, 40 is\[{{2}^{3}}\times {{5}^{1}}\times {{3}^{2}}=360\].
Example: 3 Find LCM of 32, 25, 100 by prime factorization method.
Solution \[32=25,\,\,25=52,\,\,100=22\times 52\]by taking maximum powers of prime factors.
\[\therefore \] LCM\[=25\times 52=32\times 25=800\]
By Division method:
Follows these steps.
Example: 1 Find LCM of 30, 20, 60.
Solution:
2 
20, 30, 60 
2 
10, 15, 30 
5 
5, 15, 15 
3 
1, 3, 3 

1, 1, 1 
LCM of 20, 30, 60 is\[2\times 2\times 5\times 3=60.\]
Example: 2 Find LCM of 24, 20 and 30 by division method.
Solution:
2 
20, 24, 30 
2 
10, 12, 15 
3 
5, 6, 5 
5 
5, 2, 5 

1, 2, 1 
LCM of 20, 24 and 30 is\[2\times 2\times 3\times 5\times 2=120.\]
Note:
Relation between LCM and HCF
Let us take two number says 15 and 25 then,
HCF of 15 and 25 is 5.
LCM of 15 and 25 is 75.
Now, apply this relation
Product of numbers = LCM \[\times \] HCF
\[15\times 25=75\times 5\]
375 = 375
LHS = RHS
Example: The HCF and LCM of two numbers are 6 and 36 if one of the number is 12 then other number is.
Solution: We know that,
Product of numbers = LCM \[\times \] HCF
\[12\,\,\times \]second number\[=36\times 6\]
Second number\[=\frac{36\times 6}{12}=18\]
\[\therefore \] Second number = 18.
TEST OF DIVISIBILITY
Example: 2540 is divisible by 2. (since units digit is 0)
5426 is divisible by 2. (since units digit is 6)
Example: 723 is divisible by 3.
(Since sum of 723 is 12 is divisible by 3).
Example: 615 is not divisible by 4 (since 15 is not divisible by 4)
1248 is divisible by 4 (since 48 is divisible by 4)
Example: 24620 is divisible by 5.
76625 is divisible by 5.
Example: 2070 is divisible by 6 (since 2070 is divisible by 2 and 3)
71232 is divisible by 6 (since 71232 is divisible by 2 and 3)
Example: 2401 is divisible by 7.
Hence 2401 is divisible by 7.
Example: 12000, 20416. These both number are divisible by 8.
Example: 729, 65664 is divisible by 9, because the sum of the number is divisible by 9.
Example: I860, 254670 etc.
Example: Consider the number 61809.
(O denote odd place, E denote the even place)
\[\therefore \]Sum of its digit in odd place \[=6+8+9=23\]
\[\therefore \]Sum of its digit in even place\[=0+1=1\]
\[\therefore \]Difference of the two sum \[=231=22,\]
Which is divisible by 11.
Therefore, 61809 is divisible by 11.
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