# Three-digit multiplication puzzle

Place different three-digit numbers (000-999) on each of the seven nodes of the following diagram:

There are six lines in the diagram, on each of which are three of the nodes. On each of those lines, the result of multiplying the numbers on those nodes should end in 001. The same thing for the circle, on which are also three of the nodes.

This works:

Proof:

$$249 \cdot 251 \cdot 499 = 31187001$$ $$249 \cdot 501 \cdot 749 = 93437001$$ $$249 \cdot 999 \cdot 751 = 186812001$$ $$499 \cdot 501 \cdot 999 = 249749001$$ $$499 \cdot 749 \cdot 751 = 280687001$$ $$251 \cdot 501 \cdot 751 = 94439001$$ $$251 \cdot 749 \cdot 999 = 187811001$$

I found this

using a brute-force computer program; there aren't that many 'degrees of freedom'. Once you have chosen a value for three points not on a line (e.g. 249, 251 and 501), the rest are uniquely determined and you can just check whether all products end in 001.

There's obviously a pattern; all numbers are ±1 a multiple of 250. When you write out the product $$(250a±1)(250b±1)(250c±1)$$ you can probably solve it by hand, but you first have to come up with the idea. I'm really curious how the OP invented this puzzle...

• Yes. In fact this is unique up to permutation. – deep thought Dec 2 '18 at 19:30
• May I ask how you did this? Obviously each line must contain either one or three numbers ending in 1, with the others ending in 9. But after that... computer, or additional cleverness? – Floris Dec 4 '18 at 0:33
• I hate to tell but I brute-forced it ... – Glorfindel Dec 4 '18 at 11:13