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 the two brothers so each brother received equal number of packets. How much money should one brother give to the other to make the division financially equitable?

• A. 1
• B. 2
• C. 4
• D. 5
• E. 7

Hint:

When money equals \(10k+b\) with packets split equally, \(k\) must be odd, so use congruences modulo \(20\) (not just modulo \(10\)) to pin down \(b\).
Last digits of perfect squares are only \(0,1,4,5,6,9\); combining with the parity condition quickly yields \(b=6\).

Answer 1

The Correct Option is B

Step 1: Let the number of pots sold be \(n\). Then the price per pot is also \(n\) (\(₹ n\) each), so the total money is \(n^2\). Suppose they buy \(k\) packets of potato chips at \(₹ 10\) each and one packet of banana chips at \(₹ b\) with \(b<10\). Then \[ n^2 \;=\; 10k + b. \]

Step 2: The total number of packets is \(k+1\), which is split equally between the two brothers. Hence \(k\) must be odd so that \(k+1\) is even. Thus \(\frac{n^2 - b}{10}=k\) is odd, giving \[ n^2 - b \equiv 10 \pmod{20}. \] Squares modulo \(20\) are \(\{0,1,4,5,9,16\}\). Therefore \(b \equiv n^2 - 10 \pmod{20}\) forces \(b\) to be the only value \(<10\) that fits, namely \[ b=6. \] (Indeed, \(16-10\equiv 6 \pmod{20}\) works, and no other square class yields a number \(<10\).) 

Step 3: One brother gets the banana packet plus \(\frac{k-1}{2}\) potato packets; the other gets \(\frac{k+1}{2}\) potato packets. Their values are \[ V_1 = b + 10\cdot\frac{k-1}{2}, \qquad V_2 = 10\cdot\frac{k+1}{2}. \] Hence the difference is \[ V_2 - V_1 = 10 - b = 10 - 6 = 4. \] To make the division equitable, the richer brother gives half of this difference, \[ \frac{4}{2} = 2. \] \[\boxed{2}\]