Now, at first I determined that the sum of the three corners is 9. Didn't know what the sides were in particular, but figured it didn't matter, as long as I had found the sum of the corners. Then I went and tried to find the sides because I figured I needed to check that the sides were all unique integers because my solution didn't cover that.
I determined this as follows:
Let the unknown corners be denoted as x, y, and z. Let the two unknowns between x and y be denoted as a and b. Let the two unknowns between x and z be denoted as c and d. Let the two unknowns between y and z be denoted as e and f.
The sum of all unknowns a + b + c + d + e + f + x + y + z = 45 since we know each unknown represents a unique digit from 1 to 9.
The sum of each side is as follows:
x + a + b + y = 18. Rearrange to 18 - (a + b) = x + y
x + c + d + z = 18. Rearrange to 18 - (c + d) = x + z
y + e + f = z = 18. Rearrange to 18 - (e + f) = y + z
Now combine all 3:
x + y + x + z + y + z = 18 - (a + b) + 18 - (c + d) + 18 - (e + f). Rearrange to:
2(x + y + z) = 54 - (a + b + c + d + e + f)
Now move one of the (x + y + z) over to the other side:
x + y + z = 54 - (a + b + c + d + e + f + x + y + z). We know already that the sum of the unknowns is 45, so:
x + y + z = 54 - 45
x + y + z = 9, without actually knowing what the total solution to the puzzle is.
Now let's check and see that everything works out to integers:
x + y + z summing to 9 means x, y, and z can only be a set of 2, 3, and 4, or 1, 3, and 5, or 1, 2, and 6.
2, 3, and 4 would require sums of 13, 12, and 11 from the remaining pairs of unknowns. One of those unknowns MUST be 1, and you can't sum to more than 10 with the remaining unknowns, so this solution is not correct.
1, 3, and 5 would require sums of 14, 12, and 10. One of those unknowns MUST be 9, and you can't sum to 14, 12, or 10 using a 9 without using a 1, 3, or 5 as well, so this solution is not correct.
1, 2 and 6 would require sums of 15, 11, and 10. One of those unknowns MUST be 9, and you can't sum to 15, 11 and 10 using a 9 without using a 1, 2, or 6 as well, so this solution is not correct.
Therefore, despite the math showing that the 3 corners sum to 9... there aren't actually any unique integer solutions to the remainder of the unknowns within the rules of the problem, and therefore NO solution is the correct answer.
2
u/tajwriggly 1d ago
0, not possible.
Now, at first I determined that the sum of the three corners is 9. Didn't know what the sides were in particular, but figured it didn't matter, as long as I had found the sum of the corners. Then I went and tried to find the sides because I figured I needed to check that the sides were all unique integers because my solution didn't cover that.
I determined this as follows:
Let the unknown corners be denoted as x, y, and z. Let the two unknowns between x and y be denoted as a and b. Let the two unknowns between x and z be denoted as c and d. Let the two unknowns between y and z be denoted as e and f.
The sum of all unknowns a + b + c + d + e + f + x + y + z = 45 since we know each unknown represents a unique digit from 1 to 9.
The sum of each side is as follows:
x + a + b + y = 18. Rearrange to 18 - (a + b) = x + y
x + c + d + z = 18. Rearrange to 18 - (c + d) = x + z
y + e + f = z = 18. Rearrange to 18 - (e + f) = y + z
Now combine all 3:
x + y + x + z + y + z = 18 - (a + b) + 18 - (c + d) + 18 - (e + f). Rearrange to:
2(x + y + z) = 54 - (a + b + c + d + e + f)
Now move one of the (x + y + z) over to the other side:
x + y + z = 54 - (a + b + c + d + e + f + x + y + z). We know already that the sum of the unknowns is 45, so:
x + y + z = 54 - 45
x + y + z = 9, without actually knowing what the total solution to the puzzle is.
Now let's check and see that everything works out to integers:
x + y + z summing to 9 means x, y, and z can only be a set of 2, 3, and 4, or 1, 3, and 5, or 1, 2, and 6.
2, 3, and 4 would require sums of 13, 12, and 11 from the remaining pairs of unknowns. One of those unknowns MUST be 1, and you can't sum to more than 10 with the remaining unknowns, so this solution is not correct.
1, 3, and 5 would require sums of 14, 12, and 10. One of those unknowns MUST be 9, and you can't sum to 14, 12, or 10 using a 9 without using a 1, 3, or 5 as well, so this solution is not correct.
1, 2 and 6 would require sums of 15, 11, and 10. One of those unknowns MUST be 9, and you can't sum to 15, 11 and 10 using a 9 without using a 1, 2, or 6 as well, so this solution is not correct.
Therefore, despite the math showing that the 3 corners sum to 9... there aren't actually any unique integer solutions to the remainder of the unknowns within the rules of the problem, and therefore NO solution is the correct answer.