Comment by mattnewton

Claude 3.5 seems to get it, (thought the method is not what I would use?) -

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Let's think through this step-by-step:

  We're looking for three numbers a, b, and c where:
    (a / b) / c = a + b + c

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Certainly, let's approach this problem step-by-step:

1) First, let's write out our equation: (a / b) / c = a + b + c

2) To simplify the left side, we can rewrite it as: a / (b * c) = a + b + c

3) Now, to solve this, we need to find values for a, b, and c that satisfy this equation. Let's start by considering some properties: - All numbers must be positive (because we're dividing by b and c) - The left side is a fraction, while the right side is a sum

4) One way to approach this is to try to make the left side equal to 1, which would mean: a = b * c

5) If we substitute this into our original equation: 1 = b * c + b + c

6) Now we're looking for two numbers b and c where their product plus their sum equals 1.

7) Let's try some small numbers: If b = 1/2 and c = 1/2: (1/2 * 1/2) + 1/2 + 1/2 = 1/4 + 1 = 5/4 ≠ 1

8) Let's try smaller numbers: If b = 1/3 and c = 1/3: (1/3 * 1/3) + 1/3 + 1/3 = 1/9 + 2/3 = 7/9 ≠ 1

9) Even smaller: If b = 1/4 and c = 1/4: (1/4 * 1/4) + 1/4 + 1/4 = 1/16 + 1/2 = 9/16 ≠ 1

10) We're getting closer. Let's try b = 1/4 and c = 1/3: (1/4 * 1/3) + 1/4 + 1/3 = 1/12 + 1/4 + 1/3 = 1/12 + 3/12 + 4/12 = 8/12 = 2/3 ≠ 1

11) Let's adjust further. If b = 1/3 and c = 1/4: (1/3 * 1/4) + 1/3 + 1/4 = 1/12 + 1/3 + 1/4 = 1/12 + 4/12 + 3/12 = 8/12 = 2/3 ≠ 1

12) One more try. Let's use b = 1/2 and c = 1/3: (1/2 * 1/3) + 1/2 + 1/3 = 1/6 + 1/2 + 1/3 = 1/6 + 3/6 + 2/6 = 6/6 = 1

13) We found it! When b = 1/2 and c = 1/3, the equation holds.

14) Remember, we set a = b * c, so a = 1/2 * 1/3 = 1/6

Therefore, one solution to the equation (a / b) / c = a + b + c is: a = 1/6, b = 1/2, and c = 1/3

We can verify: (1/6 / 1/2) / 1/3 = (1/3) / 1/3 = 1 1/6 + 1/2 + 1/3 = 1/6 + 3/6 + 2/6 = 1

This solution satisfies the equation.

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In general I have been seeing much better results across the board with Claude, and I wonder if open ai has been more focused on scale while Claude is doing larger more expensive models to get ahead.