Introduction / Context:
This is a classic number puzzle modeled as an algebraic equation. It uses staged consumption (half of the arrows, then fixed amounts, then an amount depending on the square root of the original total) to form a solvable equation in the original number of arrows, n.
Given Data / Assumptions:
- Half of n are used first: n/2.
- Six more arrows are used next: +6.
- Then 1 arrow each for chariot, flag, bow: +3 total.
- Finally, 4*sqrt(n) + 1 arrows are used.
- All arrows are exhausted by these uses.
Concept / Approach:
Translate the staged usage into the equation: (n/2) + 6 + 3 + (4*sqrt(n) + 1) = n. Solve by isolating sqrt(n), converting to a quadratic in t = sqrt(n), and choosing the positive root only (since counts are nonnegative).
Step-by-Step Solution:
Total used = n/2 + 6 + 3 + (4*sqrt(n) + 1) = n/2 + 10 + 4*sqrt(n).Set equal to n: n/2 + 10 + 4*sqrt(n) = n.Rearrange: 10 + 4*sqrt(n) = n/2 ⇒ multiply by 2: 20 + 8*sqrt(n) = n.Let t = sqrt(n). Then n = t^2, so t^2 − 8t − 20 = 0.Solve: t = [8 ± sqrt(64 + 80)]/2 = [8 ± 12]/2 ⇒ t = 10 (reject −2).Hence n = t^2 = 100.
Verification / Alternative check:
Check usage: half = 50; then 6 + 3 = 9; total 59. Remaining 41. Final use = 4*10 + 1 = 41. All used.
Why Other Options Are Wrong:
- 90, 101, 104: Substituting these into the equation fails to balance total used with the original stock.
Common Pitfalls:
- Missing the extra “+1” in the final usage term.
- Forgetting that the square root applies to the original count n, not the remaining arrows.
Final Answer:
100
Discussion & Comments