After taking out two-sevenths of A for the last time, there is still 155, so there is still 155*(7/5)=2 17 A before the third distribution.

Give 62 to C.

Similarly, we know that there was136 * (110) b before the third allocation.

, assigned to c136 * (110).

Before the third distribution, there were 209-62-136 * (110).