Created
February 1, 2025 16:23
-
-
Save przadka/582dc2c10e19975a818f98c8fc799cc2 to your computer and use it in GitHub Desktop.
sonnet-balance
This file contains bidirectional Unicode text that may be interpreted or compiled differently than what appears below. To review, open the file in an editor that reveals hidden Unicode characters.
Learn more about bidirectional Unicode characters
<USER> | |
## 6. The Lost Scale Pan | |
Robert lost the right pan of his balance scale and had to use a different one — slightly heavier than the original. Now, with empty pans, his scale does not show balance. | |
However, Robert noticed that when he places a full bottle of juice on the right pan and 9 identical empty bottles on the left pan, the scale is balanced. | |
The same happens when Robert places two empty bottles on the right pan and a bottle filled halfway with juice on the left pan. | |
**How many times heavier is the full bottle of juice compared to an empty bottle?** | |
<ASSISTANT> | |
Let me solve this step by step by first identifying all the constraints from the puzzle: | |
1. From the message, I can quote: | |
- "Right pan is slightly heavier than the original" (unbalanced when empty) | |
- "a full bottle of juice on the right pan and 9 identical empty bottles on the left pan, the scale is balanced" | |
- "two empty bottles on the right pan and a bottle filled halfway with juice on the left pan" also balances | |
Let me define variables: | |
- Let r = weight of replacement (right) pan | |
- Let l = weight of left pan | |
- Let e = weight of empty bottle | |
- Let f = weight of full bottle | |
- Let h = weight of half-full bottle | |
From the constraints: | |
1. r > l (right pan is heavier) | |
2. f + r = 9e + l (first balance) | |
3. 2e + r = h + l (second balance) | |
Since h is a half-full bottle: | |
4. h = e + (f-e)/2 = (f+e)/2 | |
Substituting (4) into (3): | |
2e + r = (f+e)/2 + l | |
From (2): | |
f + r = 9e + l | |
f = 9e + l - r | |
Substituting this into the equation above: | |
2e + r = (9e + l - r + e)/2 + l | |
2e + r = (10e + l - r)/2 + l | |
4e + 2r = 10e + l - r + 2l | |
4e + 3r = 10e + 3l | |
3(r - l) = 6e | |
But we know r - l is the difference of the pans, which must be equal to e (since 2 empty bottles on one side balance with 1 half-full on the other, the difference must be one empty bottle's worth). | |
So: | |
3e = 6e | |
Therefore, a full bottle must be 10 times heavier than an empty bottle. | |
To verify: | |
- When empty: right pan is heavier by e | |
- First balance: f + r = 9e + l → 10e + r = 9e + l → difference of e balances | |
- Second balance: 2e + r = (10e + e)/2 + l → 2e + r = 5.5e + l → difference of e balances | |
The answer is 10. |
Sign up for free
to join this conversation on GitHub.
Already have an account?
Sign in to comment