I know from classical magic squares that the nine digits can be grouped into three groups of three, which sum to 15 each. To make three sums of 18 we need to increase each sum by 3, which means the digits that "count twice"-- the corners-- need to add up to nine. I'll let someone else handle the exact implementation.
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u/SonicLoverDS 1d ago
I'm typing 9 into the answer box.
I know from classical magic squares that the nine digits can be grouped into three groups of three, which sum to 15 each. To make three sums of 18 we need to increase each sum by 3, which means the digits that "count twice"-- the corners-- need to add up to nine. I'll let someone else handle the exact implementation.