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
Wilson, Mathan and Guna can complete 1 round of a circular track in 10,15 and 20 minutes respectively.
Starting point at 7 a.m.

LCM = \(5 \times 2 \times 3 \times 2\) = 60 min = 1 hour = 1a.m.
⇒ 7a.m. + 1a.m. = 8a.m.
Therefore after 60 minutes at 8 a.m. will they meet together again at the starting point.

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.

i) The greatest 2 digit number 9A is divisible by 2.
Greatest two digit number = 99.
But, given 9A is divisible by 2,
So, two digit number = 98
Therefore A = 8
ii) The least number 567A is divisible by 3.
The least number = 5670
Sum of the digits = 5 + 6 + 7 + 0 = 18 is divisible by 3.
Therefore A = 0
iii) The greatest 3 digit number 9A6 is divisible by 6
The greatest 3 digit number = 999.
But, given 9A6 is divisible by 6.
Therefore 9A6 = 996.
Therefore A = 9
iv) The number A08 is divisible by 4 & 9
Given number =  A08
A08 is divisible by 4 means 108.
A08 is divisible by 9 means 108.
Therefore 108 is divisible by both 4 & 9.
Therefore A = 1
v) The number 225A85 is divisible by H.
Given number = 225A85
Also given 225A85 is divisible by 11.
Difference between the sum of alternative digits = (2 + 5 + 8) - (2 + A + 5) = 15 - (2 + 8 + 5) = 15 - 15 = 0
Therefore 225A85 = 225885
Therefore A = 8

All common multiples of 254 and 508 will be divisible by both the numbers. Let us find the LCM of 254 and 508 (by division method).
LCM of 254, 508 = 2 \(\times\) 2 \(\times\) 127 = 508
Thus, 508 is the smallest common number that is divisible by 254 and 508. Now, as we need remainder 4 while dividing, the required number is 4 more than the LCM and so, the required number is 508 + 4 =512.

This is a HCF related problem. So, we need to find the HCF of 175, 245 and 385.

175 = 5 \(\times\) 5 \(\times\) 7; 245 = 5 \(\times\) 7 \(\times\) 7; 385 = 5 \(\times\) 7 \(\times\) 11
The HCF of 175, 245 and 385 is the product of the common factors 5 and 7 i.e, 5 \(\times\) 7=35
Since each box contains equal number of books, the greatest possible number of boxes = 35
The number of English books in each box = 175 \(\div \) 35 = 5
The number of Science books in each box = 245 \(\div \) 35 = 7
The number of Maths books in each box = 385 \(\div \) 35 = 11
Hence, the total number of books in each box is 5+7+11 = 23.

                   
The product of common factors of 40 and 56
= 2 \(\times\) 2 \(\times\) 2 = 8 and so, HCF (40, 56) = 8
Dividing by the common factor 2, (in 3 steps)
HCF = Product of common factors
 = 2 \(\times\) 2 \(\times\) 2 = 8

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.