Answer:
It is divisible by 11 and (a + b) !
Step-by-step explanation:
Given a two digit number [tex]$ ab $[/tex], the digits written in reverse order is [tex]$ba $[/tex].
Note that a two digit number ab = 10a + b.
For example: 24 = 10(2) + 4
Similarly, ba = 10(b) + a
Now, the sum of the numbers ab and ba = 10a + b + 10b + a
= 11a + 11b
= 11(a + b)
Hence, the sum of any two digit number ab and the reverse of the number ba, is divisible by 11 and (a + b).
Hence, proved.
