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Explanation:

Let the potter’s sons have x pots.

Hence, they received Rs. x² after selling these pots.

As the price of one banana wafer packet is less than Rs. 10 hence, x² will not be a multiple of 100.

Assume that they bought n packets of potato wafers.

Hence, total number of wafers packet = n + 1         

Hence, each son gets (n + 1)/2 packets.

Hence, n is odd.

Let b be the price of banana wafers.

Hence, they have Rs. (n × 10 + b)

As n is odd, tens place of (n × 10 + b) is odd.

Now, each brother can have equal money if total amount earned by them is even.

Hence, b must be even.

Hence, we have the following condition,

x² = 10 × n + b such that b is even and n is odd.

Hence, x is an even number.

Now, if unit’s digit is of x is 2 or 8, then tens place of x² will be even.

This is violates our condition that tens digit of x² is odd and hence it is not possible.

Hence, unit’s place digit of x is 4 or 6.

In either case, unit digit of x² is 6.

Hence, b = 6.

Hence, the son having banana wafers owes Rs. ((n – 1) × 10/2 + 6) and the other son owes ((n + 1) × 10/2)

Hence, one of the son has Rs. ((n + 1) × 10/2) – ((n – 1) × 10/2 + 6) = Rs. 4 more than the other.

Hence, he must give Rs. 2 to the other to have financially equitable division.

Hence, option (b).