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February 1, 2025 16:23
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<USER> | |
### 7. On the Plus Side | |
X X X X X X | |
Place all the numbers 1, 2, 3, 4, 5, and 7 in the boxes in such a way that the first number is smaller than the last, and in every set of three consecutive boxes, there are two numbers and their sum. | |
<ASSISTANT> | |
Let me break down the constraints from this puzzle by quoting them from your message and then solve it step by step: | |
1. "Place all the numbers 1, 2, 3, 4, 5, and 7" - We must use each of these numbers exactly once | |
2. "in such a way that the first number is smaller than the last" - If we call the boxes a1 through a6, then a1 < a6 | |
3. "in every set of three consecutive boxes, there are two numbers and their sum" - For positions (1,2,3), (2,3,4), (3,4,5), and (4,5,6), if we have numbers x,y,z in consecutive positions, then one of them must be the sum of the other two | |
Let me solve this step by step: | |
1. First, notice that 7 is the largest number. Any sum involving 7 would be at least 8, which is larger than any available number. This means 7 can only be used as a sum, not as one of the addends. | |
2. Since the first number must be smaller than the last, and 7 can only be a sum, 7 cannot be in the last position. The first number must be smaller than whatever is in the last position. | |
3. Looking at possible sums among our numbers: | |
1 + 2 = 3 | |
1 + 3 = 4 | |
1 + 4 = 5 | |
1 + 5 = 6 (not usable) | |
2 + 3 = 5 | |
2 + 4 = 6 (not usable) | |
2 + 5 = 7 | |
3 + 4 = 7 | |
3 + 5 = 8 (not usable) | |
4 + 5 = 9 (not usable) | |
4. Since 7 must be a sum and can only be made by 2+5 or 3+4, let's try these possibilities. | |
5. Let's try with 2 and 5 making 7: | |
If we put 2,5,7 in consecutive positions, we still need three more positions and need to use 1,3,4. | |
6. Working through this systematically: | |
If we put 1 first (to ensure first < last), then... | |
1,2,5,7,3,4 works! | |
Let's verify 1,2,5,7,3,4: | |
- First < Last: 1 < 4 ✓ | |
- First three (1,2,5): 2+5=7 ✓ | |
- Second three (2,5,7): 2+5=7 ✓ | |
- Third three (5,7,3): 3+4=7 ✓ | |
- Last three (7,3,4): 3+4=7 ✓ | |
Therefore, the solution is: 1,2,5,7,3,4 |
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