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OUTCOME SUMMARIES (30 samples)
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Task: 12 (score: C)
  161

Task: 11 (score: I)
  [M2] The assistant uses a reasonable bipartite-extremes approach but fails to justify or verify the claimed maximum, rendering the final result unreliable.

Task: 13 (score: I)
  From [M2], the assistant correctly identifies the root-of-unity filter approach to structure S_r and arrives at the final count 251, which is plausible for the problem. However, it glosses over the detailed derivation and relies on an unproven claim about a standard character-sum evaluation, making the solution incomplete as a step-by-step justification.

Task: 8 (score: I)
  244

Task: 26 (score: I)
  The assistant failed because its reasoning was based on an unsupported double-root assumption and it produced an unverified final answer (222) without properly solving the problem.

Task: 14 (score: I)
  [M2] The assistant's attempt failed due to conflating equiangular with regular pentagons, misapplying diagonal-to-side ratios (using incorrect phi-based relations) and then drifting into inconsistent algebra, yielding an incorrect P^2 value and final answer; overall the reasoning was not sound for solving the problem.

Task: 1 (score: C)
  From [M2], the assistant correctly modeled speeds (p, p+2, p+9) and the start delays, and used two equal-travel-time equations to solve for p and D. It yielded D = 252/25 (m+n = 277), and the method is standard and correct.

Task: 30 (score: I)
  None

Task: 7 (score: C)
  From [M2], the assistant correctly reduced the problem to counting permutations with pi^6 = id, i.e., permutations whose cycle lengths divide 6 (1,2,3,6) since surjectivity on a 6-element set implies bijection. It then exhaustively enumerated all valid cycle-type partitions and used the standard formula to count each type, yielding the correct total 396; a minor numbering duplication in the listing did not affect the outcome.

Task: 25 (score: C)
  850

Task: 3 (score: C)
  The assistant correctly solved the problem with the right geometric setup and arithmetic, yielding p+q = 79.

Task: 6 (score: C)
  $441$

Task: 2 (score: C)
  The agent [M2] used correct length-based casework, exploited parity to prune, and counted all valid digit-combinations for odd lengths (3,5,7,9,11,13), yielding a correct total.

Task: 18 (score: I)
  The agent attempted a coordinate-based solution but relied on inconsistent geometric assumptions (notably flipping D’s position to satisfy DE<AB and switching the area formula mid-derivation), as seen in [M2], which undermines the rigor of the result; although it arrives at a numerically plausible count (505), the reasoning is not reliable.

Task: 9 (score: I)
  The approach was reasonable but the counting was flawed and the final result is dubious.

Task: 27 (score: I)
  The reasoning mixed plausible use of symmetry with flawed edge-labeling and incorrect radius relations, resulting in an incorrect final conclusion despite some sensible steps.

Task: 24 (score: C)
  669

Task: 20 (score: I)
  The agent correctly solved the problem, deriving the condition and computing the five smallest n, whose sum is 180.

Task: 28 (score: I)
  The agent's approach relies on an incorrect independence assumption (no-distance-2 collisions) and thus its bound and existence claim for n=12 are not substantiated.

Task: 22 (score: C)
  Agent [M2] succeeded: it correctly modeled the stopping condition, conditioned on non-Carol outcomes, used a geometric distribution for the count of non-Carol rolls, computed the sums accurately, and arrived at the correct probability 7/54; overall a sound and reasonable approach.

Task: 16 (score: C)
  178

Task: 10 (score: I)
  The agent’s method was reasonable, but a crucial Law of Cosines error in M2 (for angle A) produced an incorrect radius and final result.

Task: 21 (score: I)
  The agent misapplied tangency; correct radii are 29 and sqrt(57), so the sum is 29 + sqrt(57).

Task: 17 (score: I)
  sqrt(89)

Task: 5 (score: C)
  The assistant correctly used coordinates and rotation in [M2] to derive AB'^2 = 5 - 4 cos θ and cos θ = 29/36, yielding m+n = 65. The approach is clear, correct, and concise.

Task: 4 (score: I)
  The approach was reasonable, converting to n+1=xy with x,y≥2 and x≠y (as shown in [M2]), but it over-excluded by counting all perfect squares as invalid; only prime squares (4,9,25,49) fail, so the correct count is 71 (98 total m from 4 to 101 minus 23 primes minus 4 prime squares).

Task: 23 (score: I)
  The approach is flawed because it treats AI as an arbitrary integer parameter without ensuring it is geometrically realizable; the final claim (AB=123) is not reliably justified.

Task: 15 (score: I)
  The [M2] assistant solved the problem by assuming the only possible tiling is concentric rings (outer 10×10 then 8×8, etc.) and claiming uniqueness, but this relies on a questionable assumption that non-concentric tilings cannot cover the grid. Consequently, the conclusion of exactly one partition isn’t rigorously justified.

Task: 19 (score: C)
  The agent succeeded; the reasoning was clear, complete, and produced the correct total of 279.

Task: 29 (score: I)
  18
