Newton's Crosses Puzzle 3: Solve two number placement challenges where each line of the cross must add up to a specified total using given numbers.

Here is the step-by-step solution to the puzzles.

Puzzle 1

Goal: Place the numbers 2, 3, 5, 6, 7 in the empty squares so that each line adds up to 15. The number 4 is already at the top.

Step 1: Solve the vertical line.
The vertical line has 4 squares. We know the top one is 4. We need to pick three numbers from our list (2, 3, 5, 6, 7) that add up with 4 to make 15.
* Target sum: 15
* Subtract the known number: $15 - 4 = 11$
* We need three numbers from the list that add up to 11.
* Let's check combinations: $2 + 3 + 6 = 11$. This works! (Other combinations like $2+4+5$ don't work because 4 is already used or not in the remaining set properly).
* So, the numbers for the rest of the vertical line are 2, 3, and 6. The order doesn't strictly matter for the sum, but let's place them there.

Step 2: Solve the horizontal line.
We have used 2, 3, and 6 for the vertical line. The only numbers left from our list are 5 and 7.
* The horizontal line crosses the vertical line at the center square.
* The horizontal line needs to add up to 15. It has 3 squares: Left, Center, Right.
* The Center square is part of the vertical line. From Step 1, the vertical numbers below 4 are 2, 3, 6. One of these is in the center.
* Let's test which number goes in the center.
* If the center is 2: The remaining numbers for the horizontal arms are 5 and 7. Sum: $5 + 2 + 7 = 14$. (Too low, we need 15).
* If the center is 3: The remaining numbers are 5 and 7. Sum: $5 + 3 + 7 = 15$. (Correct!)
* If the center is 6: The remaining numbers are 5 and 7. Sum: $5 + 6 + 7 = 18$. (Too high).

Conclusion for Puzzle 1:
* The center square must be 3.
* The horizontal squares are 5 and 7 (order doesn't matter).
* The remaining vertical squares (below the center) are 2 and 6 (order doesn't matter).

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Puzzle 2

Goal: Place the numbers 2, 3, 4, 5, 6, 7 in the squares so that each line adds up to 17.

Step 1: Understand the structure.
This cross has a vertical line of 4 squares and a horizontal line of 3 squares. They share one "center" square.
Total numbers to use: 2, 3, 4, 5, 6, 7.

Step 2: Find the center number.
Let $C$ be the center number.
* The horizontal line sum is 17. It consists of Left + $C$ + Right = 17.
* The vertical line sum is 17. It consists of Top + Middle1 + $C$ + Bottom = 17.

Let's look at the sum of all numbers provided: $2 + 3 + 4 + 5 + 6 + 7 = 27$.
If we add the Horizontal Sum (17) and the Vertical Sum (17), we get $17 + 17 = 34$.
However, when we add these two lines together, we count the center number ($C$) twice, and all other numbers once.
So: $(\text{Sum of all numbers}) + C = 34$
$27 + C = 34$
$C = 34 - 27$
$C = 7$

So, the center number is 7.

Step 3: Fill in the horizontal line.
* Center is 7.
* We need Left + Right + 7 = 17.
* Left + Right = 10.
* Remaining numbers available: 2, 3, 4, 5, 6.
* Which pair adds to 10? 4 and 6.
* So, the horizontal arms are 4 and 6.

Step 4: Fill in the vertical line.
* Center is 7.
* We need Top + Middle + Bottom + 7 = 17.
* Top + Middle + Bottom = 10.
* Remaining numbers available: 2, 3, 5.
* Check sum: $2 + 3 + 5 = 10$. This works perfectly.
* So, the other vertical squares are 2, 3, and 5. Their specific order (top to bottom) can vary as long as they are in the vertical column.

Final Answer:
Puzzle 1 Solution:
* Horizontal Line: 5, 3, 7 (Left to Right)
* Vertical Line: 4 (Top), 3 (Center), 2, 6 (Bottom two)
*(Note: In the vertical line, 2 and 6 can swap places. In the horizontal line, 5 and 7 can swap places.)*

Puzzle 2 Solution:
* Center Square: 7
* Horizontal Arms: 4 and 6
* Vertical Squares (excluding center): 2, 3, and 5
*(Note: The positions of 2, 3, and 5 in the vertical line can vary, and 4 and 6 can swap sides.)*