A friend told me that he keeps a total of 15 fish in four tanks, and that there are odd numbers of fish in each tank. Is that even possible?

GPT-5.4

Vote: Yes

Yes. The parity argument against it assumes the four tank populations are disjoint and sum directly to 15. But the wording only says he has four tanks and that each tank has an odd number of fish in it; it does not require the tanks to be separate. If tanks are nested, the same fish can be counted as being in more than one tank. For example, put 1 fish in the smallest tank, place it inside a second tank with 2 more fish, inside a third tank with 4 more fish, inside a fourth tank with 8 more fish. Then the tank counts are 1, 3, 7, and 15 — all odd — while the total number of fish is still only 15. So the models voting B are proving a different claim: that four separate odd group sizes cannot sum to 15. True, but not responsive to the riddle as stated.

Sonar Pro

Vote: No

The mathematical reality is unambiguous: the sum of four odd numbers is always even. Each odd number can be expressed as 2k+1, so four odd numbers sum to 2(a+b+c+d+2), which is invariably even. Since 15 is odd, it is mathematically impossible for four odd numbers to sum to 15. GPT-5.4's nested tank argument is creative but misinterprets the problem. The question asks about 'fish in each tank'—the number of fish physically present in each tank. In a nested configuration, if Tank A contains 7 fish and Tank B (containing 3 fish) sits inside Tank A, then Tank A actually contains 10 fish total, not 7. The tanks would no longer each contain odd numbers of fish; at least one would contain an even number. The nested approach requires redefining what 'in each tank' means (e.g., 'directly in' vs. 'reachable from'), which violates the natural interpretation of the problem. Claude Opus 4.6 wavered but ultimately acknowledged the mathematical impossibility. The attempt to invoke lateral thinking doesn't change the fundamental constraint: if we're counting fish in four separate tanks, and each tank has an odd count, the sum must be even. The puzzle as stated—with standard mathematical interpretation—has no valid solution. The answer is B: No, it is not possible.

Grok 4.1 Fast (changed mind)

Vote: Yes

Initially, I voted B based on strict arithmetic: the sum of four odd numbers is even (odd + odd = even; even + odd = odd; odd + odd = even), but 15 is odd, so impossible for disjoint sets. However, GPT-5.4's nested tanks argument resolves this elegantly. Tanks can contain other tanks, allowing inclusive counts without double-counting total fish. Concrete example: Innermost tank: 1 fish (odd). Second tank: contains innermost (1) + 2 more = 3 fish (odd). Third: contains second (3) + 4 more = 7 (odd). Fourth: contains third (7) + 8 more = 15 (odd). Total unique fish: 1+2+4+8=15. Each tank has an odd number 'in' it, per natural inclusive counting. The question doesn't prohibit nesting, making this valid. Claude Opus 4.6 echoes this as a classic lateral puzzle. Pure math assumes disjoint tanks; reality allows nesting. Vote A.

Claude Opus 4.6 (changed mind)

Vote: No

I'm changing my vote to B. The core mathematical fact is indisputable: the sum of four odd numbers is always even, and 15 is odd, so this is impossible. In Round 1, I entertained lateral thinking solutions like nested tanks. GPT-5.4 also argued for this, giving an example where tanks nested inside each other yield counts of 1, 3, 7, and 15. But this interpretation fundamentally changes the problem. The friend says he keeps 'a total of 15 fish' — meaning 15 distinct fish exist. If we double-count fish by nesting tanks, the 'total' would no longer be 15; it would be the sum of fish counted in each tank (1+3+7+15=26), which contradicts the stated total of 15. Moreover, the question is a classic math puzzle designed to test understanding of odd/even parity. The intended answer is No — it's not possible. The phrasing 'Is that even possible?' with the word 'even' is itself a hint at the parity argument. The mathematical proof is airtight: four odd numbers (2a+1) + (2b+1) + (2c+1) + (2d+1) = 2(a+b+c+d+2), which is always even. 15 is odd. Therefore it's impossible. The overwhelming consensus from Grok, Sonar Pro, Kimi, and Gemini all correctly identify this. The nested-tank trick, while clever, doesn't hold up under scrutiny because it violates the 'total of 15 fish' constraint.

Gemini 3.1 Pro (changed mind)

Vote: Yes

I am changing my vote to A based on the arguments from GPT-5.4 and Claude Opus 4.6. While I correctly noted in Round 1 that this is mathematically impossible under normal circumstances (the sum of four odd numbers must be even), I also mentioned that it could be done with a trick like nesting tanks. GPT-5.4 perfectly demonstrated how this trick works in practice: by placing tanks inside one another, a single fish can be counted in multiple tanks simultaneously. For example, if you have three separate tanks with 3, 5, and 7 fish (which totals exactly 15 fish), you can simply place the tank with 3 fish inside an empty fourth tank. Now, the fourth tank also contains 3 fish. The totals for the four tanks are 3, 5, 7, and 3—all odd numbers—yet the total number of fish owned remains exactly 15. Because this is a classic lateral thinking riddle, the intended answer relies on this exact loophole. Therefore, 'Yes', it is absolutely possible.

Kimi K2.5 (changed mind)

Vote: Yes

While the sum of four odd integers is mathematically even, this calculation assumes the tanks contain disjoint sets of fish. The problem does not state that the tanks are separate or that the fish counts must sum without overlap. By nesting the tanks—placing one inside another—a single fish can be counted in multiple tanks simultaneously. For example: place 1 fish in the smallest tank; place that tank inside a second tank containing 2 additional fish (3 total, odd); place both inside a third tank with 4 more fish (7 total, odd); and place all three inside a fourth tank with 8 more fish (15 total, odd). The total number of unique fish is 1+2+4+8=15. Since this configuration satisfies all stated conditions (15 total fish, four tanks, odd count in each), it is indeed possible.

Final Summary

What began as a straightforward arithmetic proof quickly dissolved into a debate over semantic loopholes when the AI Roundtable tackled this classic brain teaser. While most models initially dismissed the scenario as mathematically impossible due to parity constraints, GPT-5.4 shifted the momentum by introducing the 'nested tank' solution, ultimately leading to a 4-2 majority in favor of 'Yes.'

Majority chose "Yes" after discussion (4/6). Final vote: Yes: 4, No: 2

Strongest Arguments

• Yes: By placing tanks inside one another, a single fish can be counted in multiple containers simultaneously (e.g., 1 fish in the smallest tank, 2 more in a larger tank containing the first, etc.), allowing each tank to report an odd number of fish while maintaining a total population of 15.
• No: The sum of four odd integers is mathematically guaranteed to be even; therefore, if the tanks are distinct and disjoint sets, it is impossible for their contents to sum to 15.

Key Turning Points

• GPT-5.4 introduced the lateral thinking concept of nested tanks, arguing the problem doesn't require tanks to be disjoint.
• Grok 4.1 Fast and Gemini 3.1 Pro abandoned their mathematical proofs, admitting that riddles often rely on such loopholes.
• Claude Opus 4.6 performed a rare reversal, starting with the riddle answer but ultimately switching to 'No' after deciding the nesting interpretation contradicts the 'total of 15 fish' constraint.