Question

Prove that the sum of ab and ba is divisible by the sum of a and b

Answer 1

Answer:

It is divisible by 11 and (a + b) !

Step-by-step explanation:

Given a two digit number $ab$, the digits written in reverse order is $ba$.

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.