Given Sum of my digits is 10.
Factors of 57 are 1, 3, 19, 57
Here, 19 and 57 are two digit prime numbers.
Therefore 19 = 1 + 9 = 10.
Therefore I am 19.

One month have 30 or 31 days.
2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30 are divisible by 2.
3, 6, 9, 12, 15, 18, 21, 24, 27, 30 are divisible by 3.
Therefore 6, 12, 18, 24, 30 (excluding February) are divisible by both 2 and 3.

Answer : Given statement is True.
Example : Let the composite number be 4.
4 has 3 factors namely, 1, 2 and 4.
Now, let 6 be the another composite.number,
6 has ⇾ 1, 2, 3, 6 4 factors.
Therefore Each of the composite numbers has atleast three factors.

Answer : Given statement is false.
Example : Let the odd number be 9.
9 is an odd number, but not prime.
Therefore, every odd number is not prime.

The given prime number 13 can be reversed
to get another prime number 31.
Similarly, other any such pairs are (17, 71), (37, 73) and (79, 97)

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

The Smallest two digit prime number = 11
The Biggest two digit prime number = 97

31 = 5+7+19 (find another way, if possible)
55 = 3 + 23+29

42 = 19+23;
100 = 47+53

First let us find the LCM of 72 and 108 (by division method).
LCM of 72 and 108 = 2 \(\times\) 2 \(\times\) 2 \(\times\) 3 \(\times\) 3 \(\times\) 3 = 216
Now, all multiples of 216 will also be common multiples of 72 and 108.
The smallest 5 digit number = 10,000.
Now, 10,000 \(\div \) 216 gives quotient as 46 and remainder as 164.
Hence the next multiple of 216 i.e., 216 \(\times\) 47 = 10,152 is the required smallest 5 digit number that is exactly divisible by 72 and 108.

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.

Now, 18 = 2 \(\times\) 3 \(\times\) 3 and 30 = 2 \(\times\) 3 \(\times\) 5
and their HCF is 2 \(\times\) 3 = 6 and LCM is 2 \(\times\) 3 \(\times\) 3 \(\times\) 5 = 90
Hence, HCF : LCM = 6 : 90 = 1 : 15

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.

                   
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

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.