Calculating Minimum Book Copies ($n$)

The problem requires finding the minimum value of $n$ where buyer A purchases $n$ book copies at a 20% discount, and buyer B purchases $n+1$ copies (one extra) at a 30% discount, spending the exact same total amount.

Cost Calculation Steps

• Let the marked price (MP) of a single book be $M$.
• Buyer A's Costs:
  • Discount = 20%
  • Cost per book = $M \times (1 - 0.20) = 0.80M$
  • Total cost for $n$ books = $n \times 0.80M = 0.8nM$
• Buyer B's Costs:
  • Discount = 30%
  • Cost per book = $M \times (1 - 0.30) = 0.70M$
  • Number of books = $n+1$
  • Total cost for $n+1$ books = $(n+1) \times 0.70M = 0.7(n+1)M$

Solving the Equation for $n$

Equating the total costs:

Total Cost (A) = Total Cost (B)

$0.8nM = 0.7(n+1)M$

Assuming the marked price $M$ is not zero, we can cancel $M$:

$0.8n = 0.7(n+1)$

Distribute the 0.7:

$0.8n = 0.7n + 0.7$

Subtract $0.7n$ from both sides:

$0.1n = 0.7$

Solve for $n$:

$n = \frac{0.7}{0.1}$

$n = 7$

Conclusion

The calculation shows that the minimum value for $n$ is 7. Therefore, Buyer B needs to buy 8 copies (7+1) to spend the same amount as Buyer A spending on 7 copies.