# Products or sums

Arrange the integers between 1 to 20 on twenty of the cells of this board, precisely two on each row and each column. The sum or product of the two numbers in each row must be the number on its right, while the sum or product of the two numbers in each column must be the number below it. • Darn it! I have a solution if that 25 was a 24 :p – El-Guest Sep 20 '18 at 18:19

Here is the final grid with all rows and columns filled:

Not a complete write-up, just some steps and thoughts:

The highest possible sum is $$39$$ ($$20 + 19$$), so all numbers higher than that must be products.

The rows were pretty straight forward:

• $$44$$ must be $$4 * 11$$. ($$2 * 22$$ does not work)
• $$50$$ must be $$5 * 10$$. ($$2 * 25$$ does not work)
• Now $$72$$ must be $$9 * 8$$. ($$4 * 18$$ does not work anymore)
• Now $$48$$ must be $$3 * 16$$. ($$6 * 8$$ does not work anymore)
• Now $$42$$ must be $$6 * 7$$. ($$3 * 14$$ does not work anymore)
• Now $$40$$ must be $$2 * 20$$. ($$4 * 10$$ and $$8 * 5$$ do not work anymore)
• Now $$26$$ must be $$12 + 14$$. (No other combination of the remaining numbers fits)
• Now $$13$$ must be $$1 * 13$$.
• The remaining sums must be $$17 + 19 = 36$$ and $$15 + 18 = 33$$

Two columns were immediately clear: $$57 = 19 * 3$$ and $$65 = 5 * 13$$. I couldn't find a similarily clear reasoning as I did for the rows, so eventually it was some more or less clever try-and-error. The "clever" part was to focus on columns where only two combinations were possible. This way, if a combination turned out to be wrong, the other one had to be right.