Newton's Crosses Puzzle 3: Solve two math puzzles by placing numbers in a cross shape to achieve target sums of 15 and 17.

Let’s solve each puzzle step by step.

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Puzzle 1:

We need to place the numbers 2, 3, 5, 6, and 7 into the empty squares of the cross so that each line (horizontal and vertical) adds up to 15.

The cross has:
- A top square already filled with 4.
- One center square (shared by both lines).
- Two side squares (left and right) for the horizontal line.
- Two bottom squares (below center) for the vertical line.

So the vertical line is:
Top (4) + Center + Bottom1 + Bottom2 = 15

The horizontal line is:
Left + Center + Right = 15

We have to use exactly these five numbers: 2, 3, 5, 6, 7 — no repeats, all must be used.

Let’s think about the total sum of all numbers we’re placing:
2 + 3 + 5 + 6 + 7 = 23

But notice: the center square is counted in BOTH lines. So when we add the two lines together:

Vertical line sum + Horizontal line sum = 15 + 15 = 30

But this counts the center number twice, and the other four numbers once.

So:
Sum of all five numbers + center number = 30
→ 23 + center = 30
→ center = 7

Great! So the center must be 7.

Now plug that in.

Vertical line: 4 (top) + 7 (center) + ? + ? = 15
→ 4 + 7 = 11 → remaining two bottom squares must add to 4

Which two numbers from {2, 3, 5, 6} add to 4? Only 2 and 2 — but we don’t have two 2s. Wait — we already used 7, so available numbers are 2, 3, 5, 6.

Wait — 2 + 3 = 5 → too big? No, 4+7=11, 15-11=4 → we need two numbers that add to 4. But smallest two are 2 and 3 → 5. That’s too much.

Hmm… did I make a mistake?

Wait — let me double-check.

Total of given numbers: 2+3+5+6+7 = 23 ✔️

Two lines: 15 + 15 = 30 ✔️

Center is counted twice → so 23 + center = 30 → center = 7 ✔️

Then vertical line: 4 + 7 + A + B = 15 → A + B = 4

But from remaining numbers {2,3,5,6}, no pair adds to 4. Minimum is 2+3=5.

That means something’s wrong.

Wait — maybe I miscounted the structure?

Looking back at the image description: The cross has 6 squares? Let me visualize:

It says “write the numbers 2,3,5,6,7” — that’s 5 numbers. Plus the top is already 4. So total 6 squares? But only 5 numbers to place? That makes sense — one square is pre-filled (the 4), and we fill 5 others.

But how many squares are in the cross?

Typical Newton’s Cross for this level: It’s a plus sign with 3 squares vertically and 3 horizontally, sharing the center → total 5 squares? But here top is already filled with 4, and we have to place 5 more numbers? That would be 6 squares total.

Wait — looking again: In puzzle 1, it shows a cross with:

- Top square: 4 (given)
- Then below it: center square
- Then below that: another square
- Left and right of center: two squares

So that’s 5 squares to fill? But we have 5 numbers: 2,3,5,6,7 — yes.

But then vertical line is: top (4) + center + bottom = 3 squares? Or 4?

Wait — in the diagram, it looks like vertical line has 4 squares: top (4), then three below? But that can’t be because we only have 5 numbers to place.

Actually, re-examining standard Newton’s Cross puzzles — usually it’s a 3x3 grid minus corners, making a cross of 5 squares: center, up, down, left, right.

But here, in puzzle 1, the top square is labeled “4”, and there are 4 empty squares? But we have 5 numbers to place? That doesn’t match.

Wait — look at the instruction: “Write the numbers 2,3,5,6 and 7” — that’s 5 numbers. And the cross has 5 empty squares? But in the diagram, it shows:

- One square on top (with 4)
- Then a row of three: left, center, right
- Then two squares below center? That would be 6 squares total.

I think I see the issue. Let me count the squares in the cross as drawn:

From the description: It's a cross shape with:

- Top: 1 square (has 4)
- Middle row: 3 squares (left, center, right)
- Bottom: 2 squares stacked under center? That would be unusual.

Actually, looking at common versions, Puzzle 1 likely has:

Vertical line: 4 squares — top (4), then three below? But we only have 5 numbers to place for 5 empty squares? Confusing.

Perhaps it's a standard 5-square cross: positions are:

U
L C R
D

But in this case, U is given as 4, and we need to fill L, C, R, D — that’s only 4 squares, but we have 5 numbers? Doesn't fit.

Wait — the problem says: "Write the numbers 2,3,5,6 and 7" — 5 numbers. And the cross must have 5 empty squares. So probably the cross has 6 squares total: one is pre-filled (4), and 5 to fill.

And the lines are:

- Vertical: includes top (4), center, and two below? So 4 squares in vertical line?
- Horizontal: left, center, right — 3 squares.

That makes sense with the diagram described.

So let's assume:

Vertical line: 4 squares — positions: V1 (top, given as 4), V2 (center), V3, V4 (bottom two)

Horizontal line: 3 squares — H1 (left), H2 (center, same as V2), H3 (right)

So center is shared.

We need to assign 2,3,5,6,7 to the 5 empty squares: V2, V3, V4, H1, H3.

Constraints:

Vertical sum: V1 + V2 + V3 + V4 = 4 + V2 + V3 + V4 = 15 → so V2 + V3 + V4 = 11

Horizontal sum: H1 + V2 + H3 = 15

Also, all five numbers 2,3,5,6,7 used exactly once in V2,V3,V4,H1,H3.

Let S = sum of all five numbers = 2+3+5+6+7 = 23

Note that V2 is in both sums.

So if we add the two required sums:

(V2 + V3 + V4) + (H1 + V2 + H3) = 11 + 15 = 26

But this equals (V2 + V3 + V4 + H1 + H3) + V2 = 23 + V2

So 23 + V2 = 26 → V2 = 3

Ah! So center is 3.

Then vertical: V2 + V3 + V4 = 11 → 3 + V3 + V4 = 11 → V3 + V4 = 8

Horizontal: H1 + 3 + H3 = 15 → H1 + H3 = 12

Remaining numbers: 2,5,6,7 (since 3 is used)

We need two numbers that add to 8 for V3,V4: possible pairs from {2,5,6,7}: 2+6=8, or 5+3 but 3 used, so only 2 and 6.

Then H1 and H3 must be 5 and 7, which add to 12 — perfect.

So:

Center (V2) = 3

Bottom two (V3,V4) = 2 and 6 (order doesn't matter for sum, but typically we can put smaller first or as per diagram — since no specification, either way)

Left and right (H1,H3) = 5 and 7

Check vertical: 4 (top) + 3 (center) + 2 + 6 = 4+3+2+6=15 ✔️

Horizontal: 5 + 3 + 7 = 15 ✔️

Perfect.

So for Puzzle 1:

- Center: 3
- Left: 5 or 7
- Right: 7 or 5
- Below center: 2 and 6 (say, first below center is 2, then 6, or vice versa — but since the diagram has two separate boxes below, order might not matter as long as they are in the vertical line)

In the diagram, it's likely:

Top: 4

Middle row: left, center, right

Then below center: two boxes stacked — so vertical line is top, center, then two below.

So positions:

Vertical: 4, center, boxA, boxB

Horizontal: left, center, right

We have:

center = 3

boxA and boxB = 2 and 6

left and right = 5 and 7

Any assignment works as long as sums are correct. For definiteness, let's say:

- Left: 5
- Right: 7
- First below center: 2
- Second below center: 6

Or swap left/right or the two bottoms — all fine.

But to write the answer clearly, we can specify the values in the cross.

Since the problem doesn't specify positions beyond the lines, we can present one valid solution.

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Puzzle 2:

Write numbers 2,3,4,5,6,7 in the squares so that each line adds to 17.

This cross has no pre-filled numbers — all 6 squares are empty? But we have 6 numbers to place.

Structure: likely same as before — vertical line has 4 squares, horizontal has 3, sharing center.

So total squares: 6 (since 4 + 3 - 1 shared = 6)

Yes.

Let the squares be:

Vertical: V1, V2, V3, V4 (from top to bottom)

Horizontal: H1, H2, H3 (left to right), with H2 = V2 (center)

We need to assign 2,3,4,5,6,7 to these six positions.

Constraints:

Vertical sum: V1 + V2 + V3 + V4 = 17

Horizontal sum: H1 + V2 + H3 = 17

Sum of all six numbers: 2+3+4+5+6+7 = 27

When we add the two line sums: 17 + 17 = 34

This counts V2 twice, and all others once.

So: sum of all six numbers + V2 = 34

→ 27 + V2 = 34 → V2 = 7

So center is 7.

Then vertical: V1 + 7 + V3 + V4 = 17 → V1 + V3 + V4 = 10

Horizontal: H1 + 7 + H3 = 17 → H1 + H3 = 10

Remaining numbers: 2,3,4,5,6

We need three numbers that add to 10 for V1,V3,V4, and two numbers that add to 10 for H1,H3.

Possible pairs for H1,H3 that add to 10: from {2,3,4,5,6}: 4+6=10, or 5+5 invalid, so only 4 and 6.

Then remaining numbers for V1,V3,V4: 2,3,5 — sum is 2+3+5=10 — perfect.

So:

Center = 7

Left and right: 4 and 6 (in any order)

Top, and two below: 2,3,5 in some order.

Check:

Vertical: say V1=2, V3=3, V4=5 → 2+7+3+5=17 ✔️

Horizontal: H1=4, H3=6 → 4+7+6=17 ✔️

Perfect.

So for Puzzle 2:

- Center: 7
- Left: 4 or 6
- Right: 6 or 4
- Top: one of 2,3,5
- Below center: the other two of 2,3,5

Again, multiple solutions, but we can pick one.

For example:

- Top: 2
- Center: 7
- First below: 3
- Second below: 5
- Left: 4
- Right: 6

Sums: vertical 2+7+3+5=17, horizontal 4+7+6=17.

Good.

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Final Answer:

For Puzzle 1:
- Place 3 in the center.
- Place 5 and 7 in the left and right squares (either way).
- Place 2 and 6 in the two bottom squares of the vertical line (either order).

For Puzzle 2:
- Place 7 in the center.
- Place 4 and 6 in the left and right squares (either way).
- Place 2, 3, and 5 in the top and two bottom squares of the vertical line (any order, as long as all are used).

To write it neatly:

Puzzle 1 Solution:
- Top: 4 (given)
- Center: 3
- Left: 5
- Right: 7
- Below center (first): 2
- Below center (second): 6

Puzzle 2 Solution:
- Top: 2
- Center: 7
- Left: 4
- Right: 6
- Below center (first): 3
- Below center (second): 5

(Note: Other arrangements are possible as long as the sums are correct.)