Potter's Two Sons Pots and Chips

Question

A potter asked his two sons to sell some pots in the market. The amount received for each pot was same as the number of pots sold. The two brothers spent the entire amount on some packets of potato chips and one packet of banana chips. One brother had the packet of banana chips along with some packets of potato chips, while the other brother just had potato chips. Each packet of potato chips costs 10/- and the packet of banana chips costs less than 10/-. The packets of chips were divided between

Accepted Answer

B. Let total pots sold = $N$, total revenue = $N^2$. Total packets = $n$ potato chips + 1 banana chips = $n+1$. Each gets $\frac{n+1}{2}$ packets $\implies n$ is odd. Total cost = $10n + x = N^2$, where $x < 10$. Since $n$ is odd, the tens digit of $10n + x$ is odd. Perfect squares with an odd tens digit always end in 6 $\implies x = 6$. So one brother gets packets worth $10k + 6$ (potato + banana), and the other gets $10(k+1)$ (all potato). Difference = $10(k+1) - (10k + 6) = 4$. To equalize, one gives $\frac{4}{2} = 2$ rupees to the other.

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Potter's Two Sons Pots and Chips

Explanation

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