Given: numbers are 1 to 9.

Therefore the smallest number
= \(2 \times 2 \times 3 \times 3 \times 5 \times 7 \times 2\)
= 2520

All the prime numbers between 10 and 20 are 11, 13, 17 and 19
sum = 11 + 13 + 17 + 19
= 60 - single digits ( 1 to 9 )
i)  60 is divisible by 1.
ii)  60 is divisible by 2.
iii)  60 is divisible by 3.
iv)  60 is divisible by 4.
v)  60 is divisible by 5.
vi)  60 is divisible by 6.
vii)  60 is not divisible by 7, 8 and 9.
Therefore the sum 60 is divisible by 1, 2, 3, 4,  5 and 6.

Given: the statement is Every even number greater than 2 can be expressed as the sum of two prime numbers.
To verify : This statement for every even number upto 16.
Even numbers are 4, 6, 8, 10, 12, 14, 16
Now 4 = 2 + 2, 6 = 3 + 3,
8 = 3 + 5,
10 = 3 + 7 (or) 5 + 5
12 = 5 + 7
14 = 7 + 7 (or) 3 + 11
16 = 5 + 11 (or) 3 + 13
\(\therefore\) "Every even number greater than 2 can be expressed as the sum of two prime number" is true.

Total no. of math books = 143
Let the number of books be 'x' and Let the number of stacks be 'y'
put x = 11, ⇒ 143/ 11 = 13
⇒ y = 13.
when y = 13
⇒ x = 143/ 11 = 13.
Therefore number of books in each stack and also the number of stacks
⇒ (11, 13) or (13, 11) .

Let 3 odd natural numbers are 3, 7 and 9.
Sum = 3 + 7 + 9 = 19 is odd.
This statement is true.

The smallest 3 digit composite number = 100
The biggest three digit composite number = 999

This is a LCM related sum. So, we need to find the LCM of 15, 20, 25 and 30.

The LCM of 15, 20, 25 and 30 is 2 \(\times\) 2 \(\times\) 3 \(\times\) 5 \(\times\) 5
= 300 minutes = 5 \(\times\) 60 minutes = 5 \(\times\) 1 hour = 5 hours
Thus, the four Mobile Phones will ring together again at 10.00 a.m.

Solution: By Division method
Step 1: Start with the smallest prime factor and go on dividing till all the numbers are divided as given below.
Step 2: LCM = product of all prime factors
= 2 \(\times\) 2 \(\times\) 3 \(\times\) 13 \(\times\) 31 = 4836
Thus, the LCM of 156 and 124 is 4836.
By Prime Factorisation method
Step 1: We write the prime factors of 156 and 124 as given below (use of divisibility test rules will also help).
156 = 2 \(\times\) 78 = 2 \(\times\) 2 \(\times\) 39 = 2 \(\times\) 2 \(\times\) 3 \(\times\) 13
124 = 2 \(\times\) 62 = 2 \(\times\) 2 \(\times\) 31
Step 2: The product of common factors is 2 \(\times\)2 and also the product of the factors that are not common is 3 \(\times\) 13 \(\times\) 31.
Step 3: Now, LCM = product of common factors x product of factors that are not common
= (2 \(\times\) 2) \(\times\) (3 \(\times\) 13 \(\times\) 31) = 4 \(\times\) 1209 = 4836
Thus, LCM of 156 and 124 is 4836.
                          (or)
156 = 2 \(\times\) 78 = 2 \(\times\) 2 \(\times\) 39 = 2 \(\times\) 2 \(\times\) 3 \(\times\) 13;
124 = 2 \(\times\) 62 = 2 \(\times\) 2 \(\times\) 31
The prime factor 2 appears a maximum of 2 times in the prime factorization of 156 and 124, the prime factor 3 appears only 1 time in the prime factorization of 156, the prime factor 13 appears only 1 time in the prime factorization of 156 and the prime factor 31 appears only 1 time in the prime factorization of 124.
Hence, the required LCM = (2 \(\times\) 2) \(\times\) 3 \(\times\) 13 \(\times\) 31 = 4836.

Let us find the factors of 18, 24 and 30 (use of divisibility test rules will also help).
The factors of 18 are  9 and 18.
The factors of 24  8, 12 and 24.
The factors of 30 are  10, 15 and 30.
The factors that are common to all the three given numbers are 1, 2, 3 and 6 of which 6 is the highest.
Hence, HCF (18, 24, 30) = 6.
Note that 1 is a trivial factor of all numbers.
Let us find the factors of 24 by tree method.

Here, 24 = 2 \(\times\) 2 \(\times\) 2 \(\times\) 3
Similarly, we can find the factors of 18 and 30.

Given statement
The sum of any two successive odd numbers is always divisible by 4.
Example:
Let any two successive odd numbers are
(i) 13, 15 (ii) 17, 19
(i) Sum = 13 + 15 = 28 is divisible by 4.
(ii) Sum = 17 + 19 = 36 is divisible by 4.
Therefore given statment is true.

Given statement
The numbers divisible by 4 and 6 are divisible. by 24.
Example:
Let the number be 12.
12 is divisible by both 4 and 6 but not divisible by 24.
Therefore statement is False.

Given
The product of 2 two digit numbers = 300, HCF = 5
300 = \(15 \times 20=60 \times 5=100 \times 3=2 \times 150=75 \times 4\)
But, given that, two digit numbers.
Therefore 60 \(\times\) 5, 100 \(\times\) 3, 2 \(\times\) 150, 75 \(\times\) 4 is not possible.
300 = 15 \(\times\) 20

HCF = 5
Therefore numbers are 15, 20.

We know that, the product of the two numbers = LCM \(\times\) HCF
As the HCF of co-primes is 1,
65 \(\times\) (the other number) = 5005 \(\times\) 1
The other number = 5005 \(\div \) 65 = 77

We know that, the product of the two numbers = LCM \(\times\) HCF
108 \(\times\) (the other number) = 432 \(\times\) 36
The other number = (432 \(\times\) 36) \(\div \) 108 = 144

Get all the common factors of 62 − 2, 78 − 3 and 109 − 4, i.e., 60, 75 and 105 and see that the common factors will divide them all. The greatest number is the H.C.F of 60, 75 and 105.
60 = 2 \(\times\) 2 \(\times\) 3 \(\times\) 5         75 = 3 \(\times\) 5 \(\times\) 5        105 = 3 \(\times\) 5 \(\times\) 7
Hence, the HCF is 3 \(\times\) 5=15, which is the greatest number that will divide 62, 78, 109 leaving remainders 2, 3 and 4 respectively.