Number 9 and age of mother when child is born. If a mother's age is divisible by 9 when a child is born then once you go to the next decade,n every 11 years the child's age and mother's age are always the same two numbers in reverse order.
For instance mother is 18 when she gives birth.  When she is 20 then the child is 02, when she is 31 the child is 13, when she is 42, the child is 24.  The same works any time the mother's age is divisible by 9.  For instance, if the mother is 36 when she give birth, then when she is 40, the child is 04, when she is 51, the child is 15 etc.  I am just interested in why this works for only those mother's whose age at birth of the child is divisible by nine?
 A: Let us say the mother's age is 9x. Now, note that the next decade will start at 10x. So, the mother's age will be 10x and the child's age will be x.
See, the digits are reversed. (Mother's age: x0 and Child's age: 0x)
Now you keep increasing by 11, each digit will increase by 1 and the symmetry will remain as it is (that is, the ages will be in reverse order)
A: The short answer is that it happens because our number system is base $10$, and $9$ is one less than $10$.
Numbers whose digits add up to the same number are equivalent to each other modulo $9$; that is, they leave the same remainder when divided by $9$.  This is the property that depends on the number system being base $10$.  Thus, in order for the property you describe to obtain, the mother and child (or any two people) must have ages separated by a multiple of $9$.  Therefore, the older person's age must be a multiple of $9$ when the younger person is born.
To see that it always obtains (for two-digit mother ages), observe that we can always write $9k$ (the mother's age at childbirth) as $(10k+0)-(10\cdot0+k)$ (e.g., $9\cdot4 = 36$ can be written as $(10\cdot4+0)-(10\cdot0+4) = 40-04$).  Eleven years later, this becomes $(10(k+1)+1)-(10\cdot1+(k+1))$ (e.g., $51-15 = 36$).  And so on.
