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Three sides x 17 = 51.

The sum of the numbers 1 to 9 = 45.

So we need an extra '6' to make 45 up to 51.

The numbers at the corners of a triangle are counted twice.

So we need three numbers at the corners which total 6, because they will be counted twice.

The only three numbers which total six are 1,2,3.

So the sides are

1**2

2**3

3**1

The side which starts with 1 and ends with 2 already totals 3, so we need two numbers that total 14 to complete the sum of 17. Two unused numbers which total 14 are 9 & 5.

So side 1 is 1, 9 & 5, 2

The side which starts with 2 and ends with 3 already totals 5, so we need two numbers that total 12 to complete the sum of 17. Two unused numbers which total 12 are 8 & 4.

So side 2 is 2, 8 & 4, 3

The side which starts with 3 and ends with 1 already totals 4, so we need two numbers that total 13 to complete the sum of 17. Two unused numbers which remain are 7 & 6, which, fortunately, total 13!

So side 3 is 3, 7 & 6, 1

-------------------

The only other way is

1, 8 & 6, 2

2, 7 & 5, 3

3, 9 & 4, 1

-------------------

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Q: There are 9 circles using 1-9 what numbers can you use so the sum will be 17 on each side of the triangle?
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