### Today’s Puzzle:

Joseph Nebus is nearly finished with all the posts in his Little 2021 Mathematics A to Z series. Every year he requests that his readers give him mathematical subjects to write about. At my suggestion, he recently wrote about subtraction, and how it is a subject that isn’t always as elementary as you might expect. With a touch of humor, we learn that subtraction opens up whole new topics in mathematics.

I wanted to make a puzzle to commemorate his post. I gave it some thought and remembered a tweet from Sarah Carter @mathequalslove:

NEW PUZZLE!

Triangle Sums – Arrange the numbers 1 to 9 in the squares so that the sum of the numbers in each pair of squares is the same as the number in the triangle that touches both squares in the pair. https://t.co/YQknFMTYiO #mtbos #iteachmath #puzzlingclassroom pic.twitter.com/AjRgVc1YFG

— Sarah Carter (@mathequalslove) September 23, 2021

That puzzle originated from *The Little Giant Encylopedia of Puzzles* by the Diagram Group. I wondered how the puzzle would work if it were a subtraction puzzle instead of an addition puzzle, and here’s how I tweaked it:

There is only one solution. I hope you will try to find it! If you would like a hint, I’ll share one at the end of this post.

### Factors of 1702:

- 1702 is a composite number.
- Prime factorization: 1702 = 2 × 23 × 37.
- 1702 has no exponents greater than 1 in its prime factorization, so √1702 cannot be simplified.
- The exponents in the prime factorization are 1, 1, and 1. Adding one to each exponent and multiplying we get (1 + 1)(1 + 1)(1 + 1) = 2 × 2 × 2 = 8. Therefore 1702 has exactly 8 factors.
- The factors of 1702 are outlined with their factor pair partners in the graphic below.

### More About the Number 1702:

1702 is the hypotenuse of a Pythagorean triple:

552-1610-1702, which is (12-35-**37**) times **46**.

1702² = 2896804, and

2197² = 4826809.

Do you notice what OEIS.org noticed about those two square numbers?

### Puzzle Hint:

Here’s how I solved the puzzle: I let the rightmost box be x. Then using the values in the adjacent triangles and working from right to left, I wrote the values of the other boxes in terms of x.

x – 5 went in the box that is second to the right,

x – 5 + 2 = x – 3 went in the next box,

x – 3 + 5 = x + 2,

x + 2 – 6 = x – 4,

x – 4 + 5 = x + 1, and so on until I had assigned a value in terms of x for every box.

Think about it, and this hint should be enough for you to figure out where the numbers from 1 to 9 need to go.