A Magic DiamondD 1Mm Rr HB d WJj ZzgoGa JWw m k La Ns TCc EeKBP q
Place 16 different positive integers on the vertices of this graph so that the ten products of three numbers in a straight line are all equal.
-
3$\\begingroup$ All ones works, but pretty clearly isn't what you are looking for. Might want to specify that they are distinct. $\\endgroup$ – hdsdv 8 hours ago
-
1$\\begingroup$ Indeed, distinct. $\\endgroup$ – Bernardo Recamán Santos 8 hours ago
-
1$\\begingroup$ Don't you mean 15 different integers? $\\endgroup$ – hexomino 31 mins ago
1 Answer
There are a number of ways to do this. An easy strategy:
To turn this into a puzzle about addition instead of multiplication, we can solve the addition version of the puzzle and take (for example) 4 to the power of everything.
So here, I put a magic square in the middle 3x3, and then just pick numbers for the left and right corners so that the remaining four are all different. (This is why I chose 4 instead of 2 as a base: this way, I can use halves, ensuring no conflict between the wings and the center 3x3.)
The additive solution I found:
And to turn this into a multiplicative solution, all you have to do is raise 4 to the power of each of the circles.
-
2$\\begingroup$ What if the smallest possible product is required? $\\endgroup$ – Bernardo Recamán Santos 7 hours ago