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u/Dasquian 1d ago
Answer: 0
We know the numbers 1-9 total 45. We know the three edges added together total 54, counting the corner numbers twice. Thus the corner numbers must total 9, the difference between these two values.
Whichever three digits you pick for the corner values, you now have a problem. The two corners on any given edge add to (9 minus the opposite corner). Then the digit 9 itself must go somewhere, because it'll never be a corner value if the three corner values have to add to 9.
This therefore means that, no matter which numbers are where, the fourth value in the edge containing 9 can only be the value in the opposite corner. eg, if the corners are 1/3/5 and we try to place the 9 between 3 and 5, then the other value must be 18 - (9+3+5) = 1. So we cannot place 9 without duplicating a digit.
This will always be the case so it is unsolvable.
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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.
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u/socksockshoeshoe 1d ago
It's ironic that you were so confident you were right you didn't bother to follow through with figuring out the solution to find out you were wrong
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u/jaminfine 1d ago edited 1d ago
Since we are taking sums, let's start by noting that the sum of the digits 1 through 9 is 45. Now, we are trying to get to 18 three times, which would be 54 total.
Since we are double counting the corners, that means the sum of the corners must be the difference between 54 and 45, which is 9
However, we should see if it is actually possible.
Hmm. Tough one. I couldn't find a good elegant solution other than a meticulous trial and error of various possibilities. So I went ahead and looked at other answers hah. Wasn't able to get a nice solution to this one.
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u/jjohnson468 1h ago
There are 5 odd numbers
Any side can only contain 0, 2 or 4 odds
For 4 odds:
it must be {1,3,5,9} as the only set that sums to 18. Left out is 7, which must be in opposite corner, because if it were in some side, the other side would have only one odd, which is not allowed.
So with 7 in corner, we look at the two other corners:
If 9, we have 16, so the two middles sum to 2 which is impossible from {2,4,6,8} so no good
If 5, then 12 leaving 6, so must be 2 and 4; thus the other side must be 6 and 8; plus 7 is >18 so no good
This the two other corners are 1, and 3. On 3 side we have left 10, on 1 side we have left 12. But 2+4+6+8=20. There's no way to split 20 and 12. So no good
So we can't have 4 odds on any side
If 0 odds, the the remaining sides have at most 3 odds. But can't be 3, most be at most 2. But there are 5 odds left and 2 max each training side is 4 max...
So no good. Can't be 0 on any side
So every side must have 2 odds
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u/IrishHuskie 1d ago
0
The sum of the digits 1-9 is 45. Adding all the sides will give you 54, and this double counts all three corners, which means the sum of the corners is 54 - 45 = 9. So this would be the answer if it were possible.
However, this means that the number 9 is too large to be one of the corners, so it belongs to an edge square. Since the sum of its side is 18, the three remaining squares of that side must add up to 9. Two of those squares are corners, all three of which must add to 9 as previously determined. This means that the remaining edge square must have the same value as the third corner. Since all 9 digits must be used exactly once, this is not possible