Prove A+B+B+B+80+90=100

Question

There are 100 coins placed on a table in a dark room, out of which 20 coins are showing tails and 80 coins are showing heads. The task is to divide this set of 100 coins into 2 groups (not necessarily same size) such that both groups have the same number of coins showing the tails.

Answer 1

(A)kkad (B)akkad (B)ambey (B)o (80) (90) = (100)

Answer 2

Its a old Hindi local saying
Akkad Bakkad Bambey Bo 80 90 Pure 100
Which implies

A+B+B+B+80+90=100