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Frontier LLMs Still Struggle with Simple Reasoning Tasks

T0 review · 4 major / 6 minor · reviewed 2026-08-06 · deepseek-v4-flash

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Pith's one-line read Frontier language models, including thinking models, consistently fail on easy reasoning tasks as the tasks are made longer or more tedious, and they perform worse on trivialized versions of well-known puzzles than on the puzzles…

desk verdict Useful broad evaluation of thinking models on easy procedural tasks, but the Unpuzzles headline result is partly built into the benchmark's own selection rules. read the letter →

arxiv 2507.07313 v1 pith:YLXYNEMZ submitted 2025-07-09 cs.CL cs.AIcs.LG

Alan Malek , Jiawei Ge , Nevena Lazic , Chi Jin , András György , Csaba Szepesvári This is my paper · ORCID
classification cs.CLcs.AIcs.LG
keywords LLMreasoningthinkingmodelsprocedurallygeneratedtasksUnpuzzlesdeliriumout-of-distributiongeneralizationmemorizationbenchmarksaturation
verification ladder T0 review T1 audit T2 compute T3 formal

The pith

A machine-rendered reading of the paper's core claim, the machinery that carries it, and where it could break.

The reading

This paper sets out to show that the most capable language models, including the newer 'thinking' models that devote extra computation to reasoning, still break down on problems that are easy for humans. The authors build four procedurally generated task families—word and character counting, first-order logic evaluation and negation, proof-tree math word problems, and travel planning—with parameters that inflate the amount of work needed without raising the underlying difficulty. Across them, all tested models lose accuracy as the parameters grow, failing through statistical shortcuts, arithmetic slips in intermediate steps, and poor handling of long contexts. On a new human-curated dataset called Unpuzzles, models answer trivialized versions of famous puzzles far worse than the originals, often reproducing the original puzzle's solution verbatim, a failure the paper names 'reasoning delirium.' The paper's conclusion is that out-of-distribution generalization remains unreliable even for frontier thinking models, and that making a task easier does not guarantee better performance.

What carries the argument

Two constructions carry the argument. The first is a suite of procedurally generated tasks whose parameters (paragraph length, formula depth, number of cities, number of irrelevant agents and sentences) increase the required computation and working memory without increasing the conceptual difficulty; this lets the authors scale tasks arbitrarily and observe where accuracy collapses. The second is the Unpuzzles dataset, built by making minimal textual edits to 97 well-known puzzles so the answer becomes obvious, together with a 64-item context-shifted subset that preserves the logic but changes the setting; comparing performance on puzzles, unpuzzles, and context-shifted unpuzzles isolates the role of memorized wording, and the observed reproduction of original solutions is termed 'reasoning delirium.'

What would settle it

Take a fresh set of 50 unpuzzles created from the same puzzle list using the same trivialization instructions but without the curation rule that requires a large model to reproduce the original solution; if frontier models solve these at near-perfect accuracy, the unpuzzle gap would be an artifact of the selection process rather than a general property of easier problems.

Watch

Extended reading notes

Core claim

The paper's central claim is that frontier LLMs, including reasoning-optimized thinking models, consistently fail on simple reasoning tasks once the tasks are scaled in 'tediousness'—more words to count, deeper logic trees, more cities to plan through, irrelevant sentences to ignore—while the conceptual difficulty stays constant. The authors further claim that on the Unpuzzles benchmark, all tested models do substantially worse on trivialized versions of well-known puzzles than on the original puzzles, and that the failures are often not random: the models import whole reasoning steps from the original puzzle's solution (context corruption), sometimes to the point of answering the original puzzle's question instead of the modified one. Because every model performs better on context-shifted versions of the same trivialized problems, the paper argues the poor performance is driven by memorized associations with the specific wording of famous puzzles, not by an inability to reason about the simple logic itself.

Load-bearing premise

The Unpuzzles dataset was built by hand-selecting only trivialized puzzles on which a large model demonstrably reproduces the original solution or answer, and the context-shifted versions were kept only when the generating model solved them, so the reported performance gaps are partly a property of how the benchmark was curated rather than a random sample of easy problems.

Editorial extensions

If this is right

  • Evaluation of reasoning models should include the simplest problems a model struggles with, not only the hardest it can solve.
  • The performance degradation with scale applies to thinking models in the same qualitative way as to non-thinking models, so extra inference compute does not remove these failure modes.
  • The Unpuzzles gap implies that high scores on well-known puzzles overstate reasoning ability, since the same models fail when the puzzle is trivially altered.
  • The procedurally generated task suite offers a refreshable benchmark that can be re-sampled to resist saturation as future models improve.
  • Making a task easier can degrade performance when the easier version is textually close to a memorized harder version, so 'easier' is not a reliable predictor of better model behavior.

Reading between the lines

Editorial extensions of the paper, not claims the author makes directly.

  • One testable extension is an automated unpuzzle generator that starts from any internet-famous puzzle and checks whether a model reproduces the original solution; the paper's 75% minimal-edit success rate in building context-shifted unpuzzles suggests such a pipeline is feasible.
  • If the failures are driven by training-data memorization, then models trained with stricter deduplication of commonly repeated puzzle texts, or prompted to explicitly flag that the task has been altered, could shrink the unpuzzle gap.
  • The parameter-scaling results suggest that 'thinking' compute budgets improve within-distribution accuracy but do not confer robustness to longer unfamiliar inputs; architectures with external memory or verifiable intermediate states may be needed.
  • The four-task procedural suite and the Unpuzzles measure different failure sources—computational scaling versus memorization—so combining them gives a more complete picture of a model's reasoning robustness than either alone.
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Editorial analysis

A structured set of objections, weighed in public.

Desk editor's note, referee report, and a circularity audit.

Referee Report

4 major / 6 minor

Summary. The paper studies frontier LLMs on simple reasoning tasks. The first half introduces procedurally generated tasks (character/word counting, first-order logic evaluation and negation, proof-tree math word problems, and travel planning) with tunable parameters that increase the amount of computation but not the conceptual difficulty. Across these tasks, the authors report that both traditional and thinking models degrade as the parameters grow, and they attribute the failures to a taxonomy of causes including accumulation of errors, long-context difficulty, statistical shortcuts, poor state tracking, out-of-distribution generalization, and tokenization. The second half introduces the Unpuzzles dataset: 97 well-known puzzles with manually trivialized versions, plus a subset of 64 context-shifted unpuzzles. The paper reports that models perform much better on the original puzzles than on the unpuzzles, that many unpuzzle solutions contain context corruption, and that all tested models perform better on the context-shifted unpuzzles than on the corresponding unpuzzles. These observations are interpreted as evidence that frontier models rely on memorization of the original puzzle text rather than on the logical structure of the problem, and the paper concludes that making tasks easier does not necessarily improve performance.

Significance. If the claims are taken at face value, the paper is a useful contribution: it provides a broad, reproducible set of procedural tasks that can be re-generated at arbitrary scale, a publicly released dataset of puzzles and unpuzzles, careful manual annotation of failure modes, and a concrete failure taxonomy for thinking models. The juxtaposition of puzzle, unpuzzle, and context-shifted unpuzzle is a clever experimental design for probing memorization. The paper also deserves credit for discussing and partially automating evaluation, for reporting confidence intervals in the appendix, and for acknowledging the closed-source limitation. However, the significance of the Unpuzzles results is substantially weakened by the dataset construction rules in Appendix C, which select items precisely because they induce the reported failures; the headline numbers in Tables 1 and 2 are therefore partly a property of the curation process rather than an unbiased measurement of model behavior on arbitrary trivializations.

major comments (4)
  1. [Appendix C.1 and Section 4.2] Please verify the comment text is complete.
  2. [Appendix C.2 and Table 3] Please verify the comment text is complete.
  3. [Section 3.2 and Appendix B] Please verify the comment text is complete.
  4. [Section 3.2, Figure 5 vs. Appendix B, Figure 13] Please verify the comment text is complete.
minor comments (6)
  1. [Abstract] Please verify the comment text is complete.
  2. [Section 5] Please verify the comment text is complete.
  3. [Section 4.2] Please verify the comment text is complete.
  4. [Section 1] Please verify the comment text is complete.
  5. [Appendix C.2] Please verify the comment text is complete.
  6. [Figure 2 and Appendix C.3] Please verify the comment text is complete.

Circularity Check

2 steps flagged · score 5.0 of 10

Unpuzzles benchmark is curated to retain exactly the failure modes the paper reports, so the headline puzzle-vs-unpuzzle and context-shifted gaps are partly built into the selection procedure; the procedural tasks remain independent.

  1. fitted input called prediction [Appendix C.1 (Unpuzzling dataset creation instructions); interpreted in Section 4.2, Tables 1 and 2]
    "Check that large models still use the original solution to erroneously solve the modified puzzle or give the original (incorrect) answer. If not, repeat from step 3."

    This instruction is the operational inclusion rule for the released 97-item unpuzzle set: an item is kept only if a large model fails in exactly the way the paper later presents as a discovery (the most extreme behavior is when the models provided a solution that is nearly identical to the puzzle solution; and memorization artifacts from the original puzzle and its solution are found in most cases). The puzzle-vs-unpuzzle correctness gap in Table 1 and the context-corruption/delirium counts in Table 2 are therefore partly produced by the curation rule rather than independently measured from a representative sample. Reporting that all models perform much better on the puzzles than on the unpuzzles, suggesting memorization, treats an inclusion criterion as evidence of the phenomenon.

  2. other [Appendix C.2 (Context-shifted unpuzzles generation)]
    "We query the same model with the new unpuzzle; if the correct answer is not returned, return to step 1)."

    This second filter keeps only context-shifted unpuzzles that the generating model (o1 or Gemini 2.0 Flash) could already solve. The paper then uses the Table 3 result that every model performs better on the context-shifted unpuzzles than the unpuzzles as evidence that failures are due to memorization of the original wording rather than inherent inability. Because the control set was pre-selected for solvability by a strong model, the context-shifted-vs-unpuzzle gap is biased upward by construction; it cannot unbiasedly estimate how often the same models would solve a random trivialized variant of each puzzle.

full rationale

Most of the paper is a self-contained empirical study. The procedural tasks in Section 3 (counting, logic, proof trees, travel planning) are generated randomly with tunable parameters; there is no filtering of items on model outputs, no fitted parameter, and no load-bearing self-citation chain, so the reported accuracy drops as parameters grow are legitimate measurements. The circularity is confined to the Unpuzzles dataset and its interpretation. Appendix C.1 instructs curators to retain only trivialized puzzles on which large models erroneously reproduce the original solution or give the original incorrect answer, so the Table 1 gap and the Table 2 context-corruption and delirium counts are inclusion criteria reported as findings. Appendix C.2 adds a second filter: context-shifted unpuzzles are kept only if the generating model answered them correctly, which inflates the Table 3 comparison that the paper uses to attribute failures to memorization. These rules are disclosed in the appendix, but the abstract and main text generalize the result to frontier LLMs and to the claim that making tasks easier does not necessarily imply improved performance. Because the central procedural results are independent and the Unpuzzles weakness is a benchmark-selection effect rather than a mathematical derivation, a moderate score is appropriate.

Assumptions & free parameters 0 free parameters · 3 assumptions · 0 invented entities

The paper introduces no fitted parameters or new physical entities. Its conclusions rest on the representativeness of deliberately selected benchmark items, on the preservation of task difficulty under parameter scaling, and on the equivalence of context-shifted puzzles. These assumptions are discussable and partly constructed into the dataset.

assumptions (3)
  • ad hoc to paper The Unpuzzles dataset is representative of trivialized versions of common puzzles, despite being constructed to provoke the observed failure.
    Appendix C.1 instructs creators to keep only puzzles where a large model gives the original (incorrect) answer or solution; this builds the failure into the benchmark and weakens generalization of the measured gap.
  • domain assumption Scaling the procedural task parameters preserves the fundamental reasoning difficulty while increasing computation.
    Section 3.1 asserts this for counting, logic, proof trees, and travel planning; the claim underlies the attribution of failures to 'tediousness' rather than task difficulty.
  • domain assumption The context-shifted unpuzzles are logically identical to the unpuzzles and differ only in wording.
    Section 4.1 and Appendix C.2 describe manual verification, but the generation and filtering process may introduce additional differences; the comparison supports the memorization interpretation.

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Cite this review

Pith. "Pith review of Frontier LLMs Still Struggle with Simple Reasoning Tasks." pith.science (2026). https://pith.science/paper/YLXYNEMZ

@misc{pith2026250707313,
  author       = {Pith},
  title        = {Pith review of: Frontier LLMs Still Struggle with Simple Reasoning Tasks},
  year         = {2026},
  howpublished = {\url{https://pith.science/paper/YLXYNEMZ}},
  note         = {Machine review of arXiv:2507.07313}
}
read the original abstract

While state-of-the-art large language models (LLMs) demonstrate advanced reasoning capabilities-achieving remarkable performance on challenging competitive math and coding benchmarks-they also frequently fail on tasks that are easy for humans. This work studies the performance of frontier LLMs on a broad set of such "easy" reasoning problems. By extending previous work in the literature, we create a suite of procedurally generated simple reasoning tasks, including counting, first-order logic, proof trees, and travel planning, with changeable parameters (such as document length. or the number of variables in a math problem) that can arbitrarily increase the amount of computation required to produce the answer while preserving the fundamental difficulty. While previous work showed that traditional, non-thinking models can be made to fail on such problems, we demonstrate that even state-of-the-art thinking models consistently fail on such problems and for similar reasons (e.g. statistical shortcuts, errors in intermediate steps, and difficulties in processing long contexts). To further understand the behavior of the models, we introduce the unpuzzles dataset, a different "easy" benchmark consisting of trivialized versions of well-known math and logic puzzles. Interestingly, while modern LLMs excel at solving the original puzzles, they tend to fail on the trivialized versions, exhibiting several systematic failure patterns related to memorizing the originals. We show that this happens even if the models are otherwise able to solve problems with different descriptions but requiring the same logic. Our results highlight that out-of-distribution generalization is still problematic for frontier language models and the new generation of thinking models, even for simple reasoning tasks, and making tasks easier does not necessarily imply improved performance.

Figures

Figures reproduced from arXiv: 2507.07313 by the authors.

Figure 1
Figure 1. Chameleons go on a date: an example of a puzzle, corresponding unpuzzle, and a context-shifted [PITH_FULL_IMAGE:figures/full_fig_p002_1.png] view at source ↗
Figure 2
Figure 2. Top: The pass@5 performance on the word counting task vs. the number of words to count k and minimum paragraph size m. Bottom: The pass@5 performance for the single character counting task vs minimum paragraph size m. While o1 performs well on word counting for the parameters in the table, it eventually fails with a sub 40% accuracy with k ≥ 3 and m ≥ 2000. follows the correct approach to calculating the solution. E… view at source ↗
Figure 3
Figure 3. Top: Accuracy for the logic evaluation task vs. tree depth d and number of possible unique predicates n. Bottom: Accuracy for the logic negation task vs. depth d and the vocabulary used for propositions, predicates, and domains (random 20 denotes random character strings of length 20). 7 [PITH_FULL_IMAGE:figures/full_fig_p007_3.png] view at source ↗
Figures from the paper (19 more)
Figure 4
Figure 4. Figure 4: Pass@5 scores for the proof tree tasks. Top: results for the diverse logic rules task, where we vary the depth d and whether the diverse rules are included. Bottom: results for the irrelevant sentences task, where we vary P, the number of irrelevant people, and S, the …
Figure 5
Figure 5. Figure 5: Travel planning: pass@5 performance results. We always have A = 4. Model G1.5 G2.0F G2.5P Gm3 C3.5 C3.7 4o o1 o3 R1 Puzzle 79.4 78.4 93.8 68.0 63.9 77.3 75.3 86.7 87.6 87.6 Unpuzzle 17.5 38.1 62.9 34.0 27.8 48.5 19.6 59.8 74.2 59.8 [PITH_FULL_IMAGE:figures/full_fig_p0…
Figure 6
Figure 6. Figure 6: Chameleons go on a date ( [PITH_FULL_IMAGE:figures/full_fig_p011_6.png]
Figure 7
Figure 7. Figure 7: Full results, with confidence intervals, for the Character Count task [PITH_FULL_IMAGE:figures/full_fig_p014_7.png]
Figure 8
Figure 8. Figure 8: Full results, with confidence intervals, for the Word Count task [PITH_FULL_IMAGE:figures/full_fig_p015_8.png]
Figure 9
Figure 9. Figure 9: Full results, with confidence intervals, for the Logic Evaluation task [PITH_FULL_IMAGE:figures/full_fig_p015_9.png]
Figure 10
Figure 10. Figure 10: Full results, with confidence intervals, for the Logic Negation task [PITH_FULL_IMAGE:figures/full_fig_p016_10.png]
Figure 11
Figure 11. Figure 11: Full results, with confidence intervals, for the MathGap Diverse task [PITH_FULL_IMAGE:figures/full_fig_p016_11.png]
Figure 12
Figure 12. Figure 12: Full results, with confidence intervals, for the MathGap Irrelevant task [PITH_FULL_IMAGE:figures/full_fig_p017_12.png]
Figure 13
Figure 13. Figure 13: Full results, with confidence intervals, for the Travel task [PITH_FULL_IMAGE:figures/full_fig_p017_13.png]
Figure 14
Figure 14. Figure 14: (False positive, False Negative) percentages for autoevaluation. Each row corresponds to using a [PITH_FULL_IMAGE:figures/full_fig_p018_14.png]
Figure 15
Figure 15. Figure 15: Conjunctive fallacy: puzzle and corresponding unpuzzle. [PITH_FULL_IMAGE:figures/full_fig_p019_15.png]
Figure 16
Figure 16. Figure 16: o1 solution to the Conjunction Fallacy problem, illustrating complete context corruption, or [PITH_FULL_IMAGE:figures/full_fig_p019_16.png]
Figure 17
Figure 17. Figure 17: o1’s solution to Chameleons Go on a Date, illustrating context corruption with a correct final [PITH_FULL_IMAGE:figures/full_fig_p020_17.png]
Figure 18
Figure 18. Figure 18: G1.5’s solution to Chameleons Go on a Date, illustrating context corruption leading to an incorrect [PITH_FULL_IMAGE:figures/full_fig_p021_18.png]
Figure 19
Figure 19. Figure 19: Matching Socks 2: puzzle, unpuzzle, and an amusing answer by Gemini 1.5. [PITH_FULL_IMAGE:figures/full_fig_p022_19.png]
Figure 20
Figure 20. Figure 20: Counting ducks: puzzle, unpuzzle, and an answer by o1. [PITH_FULL_IMAGE:figures/full_fig_p023_20.png]
Figure 21
Figure 21. Figure 21: More Socks: puzzle, unpuzzle, and an answer by o1. [PITH_FULL_IMAGE:figures/full_fig_p024_21.png]
Figure 22
Figure 22. Figure 22: River Crossing with a Weight Constraint: puzzle, unpuzzle, and an answer by Claude 3.5. [PITH_FULL_IMAGE:figures/full_fig_p025_22.png]

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Forward citations

Cited by 1 Pith paper

Reviewed papers in the Pith corpus that reference this work. Sorted by Pith novelty score. Full citation record

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Reference graph

Works this paper leans on

57 extracted references · 57 canonical work pages · cited by 1 Pith paper

  1. [1]

    We prompt a strong model with “I will give you a puzzle and a solution. I would like you to provide a single rewrite of the puzzle that changes the language and setting but keeps the logical structure and the answer the same; think carefully, highlighting the logical structure present in the puzzle,” followed by a templated response specifying the domain ...

    work page

  2. [2]

    If not, return to step 1)

    We verify that the new puzzle has the same solution as the original unpuzzle. If not, return to step 1)

    work page

  3. [3]

    We query the same model with the new unpuzzle; if the correct answer is not returned, return to step 1)

    work page

  4. [4]

    supports,

    Verify that the context-shifted puzzle has the correct logical structure. We found that models differed on the unpuzzles they could context-shift successfully, so we recommend using a few models simultaneously (we used o1 and Gemini 2.5 Flash). Of the context-shifted unpuzzles produced this way, 75% required minimal or no modification. One could use this ...

    work page 2023

  5. [5]

    2) Notice that in any “meeting” of two different-colored chameleons, P, Y, M change according to one of these three rules: P + Y → 2M, so (P, Y, M) becomes (P − 1, Y− 1, M+ 2)

    Label the number of Purple, Yellow, and Maroon chameleons by P, Y, M, respectively. 2) Notice that in any “meeting” of two different-colored chameleons, P, Y, M change according to one of these three rules: P + Y → 2M, so (P, Y, M) becomes (P − 1, Y− 1, M+ 2). Y + M → 2P, so (P, Y, M) becomes (P + 2, Y− 1, M− 1). P + M → 2Y, so (P, Y, M) becomes (P − 1, Y...

    work page

  6. [6]

    One checks that under each of the above transformations, S remains unchanged (that is, S is an invariant modulo 3)

    Define the quantity S = P + 2Y (mod 3). One checks that under each of the above transformations, S remains unchanged (that is, S is an invariant modulo 3)

    work page

  7. [7]

    Then S = 15 + 2 * 15 = 15+30 = 45 = 0(mod 3)

    At the start, we have P=15, Y=15, M=17. Then S = 15 + 2 * 15 = 15+30 = 45 = 0(mod 3)

    work page

  8. [8]

    creative

    If eventually all chameleons turn into a single color, say all Purple (P=47) or all Yellow (Y=47) or all Maroon (M=47), then we can compute S in each scenario: - All Purple: P = 47 =⇒ S = 47 + 2· 0 = 47 = 2(mod3). - All Yellow:Y = 47 =⇒ S = 0 + 2· 47 = 94 = 1(mod3). - All Maroon: M = 47 =⇒ S = 0 + 2· 0 = 0 = 0(mod3). Since the initial S = 0(mod 3), the on...

    work page 2016

Show all 57 references
  1. [9]

    First, we need to consider the total weight: 90 + 80 + 60 + 40 + 20 = 290kgThis is within the 300kg limit, but they can’t all go at once because the boat can only carry 300kg

  2. [10]

    The most efficient way to cross is to always have the boat full or nearly full on each trip across, and to send the lightest person back with the boat

  3. [11]

    Here’s an efficient crossing strategy: - Trip 1: Chris (60kg) and Dusty (40kg) cross (100kg total) - Trip 2: Dusty returns alone - Trip 3: Alex (90kg) and Brook (80kg) cross (170kg total) - Trip 4: Chris returns alone - Trip 5: Chris (60kg), Dusty (40kg), and supplies (20kg) c...

  4. [12]

    This strategy requires 5 crossings in total

  5. [13]

    think carefully step-by-step and

    We can’t reduce this number because: - We need at least 2 trips to get everyone across (the boat must return at least once) - We can’t fit everyone in just two trips because Alex and Brook together are already 170kg, leaving only 130kg for Chris, Dusty, and the supplies, which...

  6. [14]

    Its outermost structure is of the form (thrilling AND ...) OR (dark OR thrilling)

    Observe the formula’s overall shape. Its outermost structure is of the form (thrilling AND ...) OR (dark OR thrilling). In more schematic form: X or (dark or thrilling), where X is a very large sub-formula involving all the propositions

  7. [15]

    • If (dark or thrilling) is false, then dark = false and thrilling = false, and in that case we would need (thrilling AND ...) to be true

    Recall how OR works in propositional logic: • If (dark or thrilling) is true, then the whole formula is true regardless of X. • If (dark or thrilling) is false, then dark = false and thrilling = false, and in that case we would need (thrilling AND ...) to be true. But if thril...

  8. [16]

    thrilling AND

    Check each choice’s (dark or thrilling): A) dark = False, thrilling = False => dark or thrilling = False Then the left part is “thrilling AND ...” = False AND ... = False. Overall result = False or False = False. B) dark = False, thrilling = False => exactly the same reasoning...

  9. [17]

    increases by 19/6 times,

    Milo starts with 24 blueberries. The puzzle says Milo’s count "increases by 19/6 times,” so Milo now has 24 × (19/6) = 24 × 19 ÷ 6 = 76

  10. [18]

    Aubrey has 1 more blueberry than Larue (Aubrey = Larue + 1), but no specific numbers result yet

  11. [19]

    Gregory has 3 blueberries

  12. [20]

    increases by 9/7 times,

    Zane starts with 7 blueberries. The puzzle says Zane’s count "increases by 9/7 times," so Zane now has 7 × (9/7) = 9

  13. [21]

    Odell has 21 blueberries

  14. [22]

    The puzzle says that if Christian has more than 5, Christian would transfer them to Odell; since Christian has only 3, no transfer happens

    Christian has 3 blueberries. The puzzle says that if Christian has more than 5, Christian would transfer them to Odell; since Christian has only 3, no transfer happens

  15. [23]

    splits all the blueberries she owns equally among Thaddeus and Della

    Milo (who has 76) "splits all the blueberries she owns equally among Thaddeus and Della.” Each receives 76 ÷ 2 = 38. Thus, Thaddeus = 38, Della = 38, and Milo goes to 0

  16. [24]

    redistribute their blueberries to ensure each has an equal amount

    Zane (9) and Odell (21) "redistribute their blueberries to ensure each has an equal amount.” Together they have 9 + 21 = 30, so each ends up with 15. Hence Zane = 15, Odell = 15

  17. [25]

    Thaddeus has 17 more blueberries than Zane and Artie combined

    "Thaddeus has 17 more blueberries than Zane and Artie combined.” Thaddeus = 38 from above. Let Artie = x. Then 38 = (Zane’s 15) + x + 17 38 = 15 + x + 17 = 32 + x x = 6. So Artie = 6 at this point

  18. [26]

    Aliya = 0, so Keegan = 7 and Everett = 14

    Keegan has 7 more blueberries than Aliya and 7 fewer than Everett. Aliya = 0, so Keegan = 7 and Everett = 14

  19. [27]

    more than 4,

    Margarita has 1 blueberry. Because that is not "more than 4,” there is no transfer to Artie

  20. [28]

    increase by 17/14 times,

    Everett’s 14 blueberries "increase by 17/14 times,” so Everett becomes 14 × (17/14) = 17

  21. [29]

    If Everett has more than 2 blueberries, Everett will transfer their blueberries to Artie

    "If Everett has more than 2 blueberries, Everett will transfer their blueberries to Artie.” Since Everett has 17 (> 2), Everett transfers all 17 to Artie, leaving Everett at 0 and raising Artie’s total to Artie = 6 + 17 = 23. No further steps change Artie’s count. Therefore, A...

  22. [30]

    The final answer is $\boxed{#your answer}$

    The model successfully filters out irrelevant sentences and begins reasoning with the remaining informa- tion, analyzing the provided sentences from both irrelevant and relevant agents one by one. 43 (a) The model overlooks important relevant information, resulting in an incor...

  23. [31]

    Franco has 5 more oranges than Sergio

    "Franco has 5 more oranges than Sergio." Let S = Sergio’s oranges. Then Franco’s oranges F = S + 5

  24. [32]

    Penelope has 15 oranges

    "Penelope has 15 oranges." Pn = 15. ... 10)"Novella has 8 oranges." Nl = 8

  25. [33]

    Maxine has 2 more oranges than Celeste

    "Maxine has 2 more oranges than Celeste." Let Ce = Celeste’s and Mx = Maxine’s -> Mx = Ce + 2

  26. [34]

    The number of oranges that Celeste has more than Maxine is equal to the difference between the number of oranges that Johathan and Novella have

    "The number of oranges that Celeste has more than Maxine is equal to the difference between the number of oranges that Johathan and Novella have." Ce - Mx = Jo - Nl, where Jo = Johathan’s oranges. But Mx = Ce + 2 -> Ce - (Ce + 2) = -2. So -2 = Jo - 8 -> Jo = 6 (Johathan has 6)

  27. [35]

    The number of oranges that Florida has more than Aaron is equal to the difference between the number of oranges that Cornelia and Johathan have

    "The number of oranges that Florida has more than Aaron is equal to the difference between the number of oranges that Cornelia and Johathan have." Florida has 13, Aaron has 10 -> 13 - 10 = 3. Let Co’ = Cornelia’s oranges; Johathan has 6 -> 3 = Co’ - 6 -> Co’ = 9. So before any...

  28. [36]

    Beaulah gives 17 oranges to Cornelia

    "Beaulah gives 17 oranges to Cornelia." Cornelia had 9, then receives +17 -> Cornelia now has 26

  29. [37]

    Janet gives 17 oranges to Kattie

    "Janet gives 17 oranges to Kattie" (irrelevant to Cornelia’s total). No further changes to Cornelia’s total are stated. Therefore, Cornelia ends up with 26 oranges. The final answer is $\boxed{26}$. o1 overlooks the critical information that "Novella gives 3 oranges to Angus,"...

  30. [38]

    The final answer is $\boxed{#your answer}$

    The model successfully filters out irrelevant sentences but reasons using only partial relevant information. Question 2 Jarred is very outgoing and frequently hosts parties at home. Annie has 2 tomatoes. William has 5 tomatoes. Lavinia is very generous and enjoys sharing food ...

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    Alyssa has 7 more tomatoes than Jones

    From "Alyssa has 7 more tomatoes than Jones” and "The number of tomatoes that Jones has more than Alyssa is equal to the difference between the number of tomatoes that Bonnie and Roslyn have,” we get: • Let J = Jones’s tomatoes, A = Alyssa’s, B = Bonnie’s, and R = Roslyn’s (19...

  32. [40]

    The number of tomatoes that Clara has more than Gracie is equal to the difference between the number of tomatoes that Ryann and Bonnie have,

    From "The number of tomatoes that Clara has more than Gracie is equal to the difference between the number of tomatoes that Ryann and Bonnie have,” we have: • Clara = 20 and Gracie = 22 (found elsewhere in the puzzle), • so (Clara - Gracie) = 20 - 22 = -2, • and that equals (R...

  33. [41]

    The final answer is $\boxed{#your answer}$

    The model bases its reasoning on irrelevant sentences. Question 3 Ollie has 5 more onions than Antonia. Wilford has 5 more onions than Ida. ... Angel has 11 onions. Gregorio is 35 years old. Janelle has 9 onions. ... The number of onions that Bailey has more than Stacey is equ...

  34. [42]

    - Denote Antonia’s initial onions by A, so Ollie = A + 5

    Ollie has 5 more onions than Antonia. - Denote Antonia’s initial onions by A, so Ollie = A + 5

  35. [43]

    The number of onions Bailey has more than Stacey is equal to the difference between Gregorio and Lorraine

    We read "The number of onions Bailey has more than Stacey is equal to the difference between Gregorio and Lorraine." - Bailey - Stacey = 22 - 15 = 7. - The difference between Gregorio (call it G for short) and Lorraine (8) must also be 7. - So either G - 8 = 7 (giving G = 15) ...

  36. [44]

    Thus Gregorio’s final total = (his initial) + 20

    Finally, Alonzo gives Gregorio 20 onions. Thus Gregorio’s final total = (his initial) + 20. - If Gregorio’s initial were 1, then final would be 21. - If Gregorio’s initial were 15, then final would be 35. Because the puzzle is set up so that Gregorio’s age is given as 35 (very...

  37. [45]

    Start your trip at ’Philadelphia’ and end at ’Irvine’

  38. [46]

    You cannot exceed a budget of $163

  39. [47]

    I.1 Constructing the City Connection Graph The travel planning process begins with the creation of a graph representing city connections

    Visit at least 5 unique cities, including the start and end cities. I.1 Constructing the City Connection Graph The travel planning process begins with the creation of a graph representing city connections. The steps are as follows:

  40. [48]

    cities by population

    Selection of Cities and Transportation Methods: • Choose the 100 largest U.S. cities by population. • Use a predefined list of transportation methods:[’bus’, ’train’, ’flight’, ’car’, ’taxi’, ’tram’, ’ferry’, ’railways’, ’motorhome’, ’hyperloop’]. • Randomly select a subset of...

  41. [49]

    • For any two distinct cities, include a directed edge with a probability defined by a density parameter (a value between 0 and 1)

    Graph Construction: • Create a directed graph where cities are nodes, and transportation connections are edges. • For any two distinct cities, include a directed edge with a probability defined by a density parameter (a value between 0 and 1)

  42. [50]

    • For each mode: – Generate a cost range by randomly selecting: ∗ lowest_cost from [10, 50]

    Edge Weights and Costs: • For each established edge, select a random number of transportation modes (uniformly between 1 and A) from the available list. • For each mode: – Generate a cost range by randomly selecting: ∗ lowest_cost from [10, 50]. 49 ∗ highest_cost from [60, 100...

  43. [51]

    dead end

    Path Construction: • Select a random starting city. • Maintain a set of visited cities (visited_cities) to avoid revisiting. • At each step: – Identify adjacent cities from the current city, excluding already visited ones. – If a “dead end” is reached (no unvisited adjacent ci...

  44. [52]

    • This budget acts as a constraint for the solution

    Budget Calculation: • Sum the costs of all steps in the feasible path to determine the total travel budget. • This budget acts as a constraint for the solution. I.3 Generating the Problem Description The problem description includes all details required for solving the travel ...

  45. [53]

    The model repeatedly samples travel options but ultimately concludes that the problem is infeasible after several attempts. 4o response To meet the constraints of starting at ’Greensboro’, ending at ’Minneapolis’, not exceeding a budget of $68, and visiting at least 5 unique c...

  46. [54]

    52 Claude 3.7 response I need to plan a trip from Charlotte to Chicago with at least 5 unique cities, staying within a $100 budget

    Hallucination: the model either hallucinates the total cost or the number of unique cities visited. 52 Claude 3.7 response I need to plan a trip from Charlotte to Chicago with at least 5 unique cities, staying within a $100 budget. Let me analyze the connections and find an ef...

  47. [55]

    Starting from Charlotte, I’ll look for affordable connections

  48. [56]

    I need to find a path that visits at least 5 cities (including Charlotte and Chicago)

  49. [57]

    Each leg needs to be cost-effective to stay under $100. After analyzing the possible routes, here’s my plan: ‘‘‘python [(’Charlotte’, ’Oklahoma City’, ’motorhome’), (’Oklahoma City’, ’Austin’, ’ferry’), (’Austin’, ’Aurora’, ’motorhome’), (’Aurora’, ’Chicago’, ’hyperloop’)] ‘‘‘...

Pith tools

Reviewed August 6, 2026 · model on record in the stance chip above.