Free Maths Puzzles - MathSphere - Free Printable

To solve this puzzle, we need to place the numbers 1 through 7 into the circles so that every straight line of three numbers adds up to the same total.

Step 1: Understand the Structure
The puzzle has a central circle and three lines passing through it. Each line consists of:
- One outer number
- The center number
- Another outer number on the opposite side

So, there are 3 lines in total, and they all share the center number.

Step 2: Find the Center Number
Let’s call the center number C.
The sum of all numbers from 1 to 7 is:
1 + 2 + 3 + 4 + 5 + 6 + 7 = 28

Each line adds up to the same total, let’s call it S.
Since there are 3 lines, if we add up all three line totals, we get:
3 × S

But notice: when we add the three lines together, the center number C is counted three times (once for each line), while all the other six numbers are counted only once.

So:
3 × S = (Sum of all 7 numbers) + 2 × C
Because the center is extra-counted twice.

We know the sum of all 7 numbers is 28, so:
3S = 28 + 2C

We want S to be a whole number, and C must be one of the numbers 1–7.

Let’s try possible values for C:

Try C = 4:
3S = 28 + 2×4 = 28 + 8 = 36 → S = 12 ✔ Works!

Try C = 1:
3S = 28 + 2 = 30 → S = 10 → Let’s check if we can make pairs that add to 9 (since 10 - 1 = 9). Pairs from remaining numbers {2,3,4,5,6,7} that add to 9: (2,7), (3,6), (4,5) → Yes! So C=1 also works? Wait — but let’s test both.

Actually, multiple solutions may exist, but typically these puzzles expect the most balanced or symmetric solution. Let’s verify both.

But wait — let’s think again.

If C = 4, then each line must sum to 12. That means each pair of outer numbers on a line must add to 12 - 4 = 8.

Remaining numbers: 1, 2, 3, 5, 6, 7

Can we pair them to sum to 8?
- 1 + 7 = 8
- 2 + 6 = 8
- 3 + 5 = 8 → Perfect!

So with center = 4, we can arrange:
Line 1: 1, 4, 7
Line 2: 2, 4, 6
Line 3: 3, 4, 5

All lines sum to 12.

Now check C = 1:
Then each line sums to 10 → outer pairs must sum to 9.
Pairs from {2,3,4,5,6,7}:
- 2+7=9
- 3+6=9
- 4+5=9 → Also works!

So two possible centers: 1 and 4? But wait — what about C=7?

Try C=7:
3S = 28 + 14 = 42 → S=14 → outer pairs sum to 7.
Remaining: 1,2,3,4,5,6
Pairs summing to 7: (1,6), (2,5), (3,4) → Also works!

So actually, any odd number as center might work? Let’s check C=3:

C=3 → 3S = 28 + 6 = 34 → S = 34/3 → not integer ✘

C=5 → 3S = 28 + 10 = 38 → 38/3 ✘

C=2 → 3S = 28 + 4 = 32 → 32/3 ✘

C=6 → 3S = 28 + 12 = 40 → 40/3 ✘

So only C = 1, 4, 7 give integer S.

Thus, there are three valid solutions, depending on which number you put in the center.

But since the problem doesn’t specify uniqueness, any of these is acceptable. However, often in such puzzles, the “middle” number (4) is expected because it balances the set.

Let’s go with center = 4, as it’s the median and gives equal spacing.

Final arrangement:

Center: 4

Lines:
- 1 — 4 — 7
- 2 — 4 — 6
- 3 — 4 — 5

Each line sums to 12.

You can assign the outer pairs to any of the three lines — the diagram doesn’t label positions, so any rotation or reflection is fine.

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Final Answer:
Place 4 in the center. Then pair the outer numbers as follows: (1 and 7), (2 and 6), (3 and 5) — each pair goes on opposite ends of a line through the center. Every line will sum to 12.