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 number is 564872.
i) If the last 3 digits of a number are zeros, then also it is divisible by 8.
564872
872 is divisible by 8.
Therefore 564872 is divisible by 8.
ii) Difference between the sum of alternative digits.
= (5 + 4 + 7) - (6 + 8 + 2) ,
= 16 - 16 = 0
Therefore 564872 is divisible by 11.
88 = 8 \(\times\)11
Therefore 564872 is divisible by 88.

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.

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.