500 sats \ 1 reply \ @Murch 19 Oct \ on: [Daily puzzle] Another MSM puzzle science
red: A+B+C+D+E = x green: B+C+D+E+G+H+I = x yellow: C+D+G+H+F = x blue: D+E+H+I+J = x
red - green: A - (G+H+I) = 0 A = G+H+I
green - yellow: B+E+I - F = 0 F = B+E+I
green - blue: B+C+G - J = 0 J = B+C+G
We notice that there are three symmetric groups of letters. As starting points for any solution, A, F, and J are interchangeable, B, G, and I are interchangeable, and C, E, and H are interchangeable.
Since A…J are each unique values and integers from [1..10], the minimum value for three letters to be summed is 1+2+3 = 6. A, F, J must be 6 or bigger.
None of B, C, E, G, H, I can be bigger than 7, as 7 + 2 + 1 = 10
We suspect that D is 7. Proof by contradiction:
Let’s assume A = 6. If A is six, G, H, and I must be 1, 2, and 3 in some configuration. If that’s the case, for F = B+E+I to be valid, I must be 1 and B and E must be 4 and 5. Otherwise, F would be bigger than 10. However, for the same reason, G would have to be one and B+C would be 9 which is in contradiction to the requirement that each number only appears once. Due to the symmetry, it follows that none of A, F, and J can be 6.
Let’s assume A = 7. This means that G, H, and I must be 1, 2, 4 in some order. It would follow that I is either 1 or 2.
  1. If I were 4, F would necessarily be greater than 10 as the next two lowest numbers we can use for B and E are 3 and 5.
  2. If I is 2, B and E must be 3 and 5, which makes F 10. This leaves C to be at least 6, in which case J would be at least 10 as well 3+6+1 = 10. F and J cannot both be 10.
  3. If I is 1, B and E could be 3 and 5 or 3 and 6. 3a. Let’s assume they are 3 and 6, which makes F = 10. Then the lowest remaining number for C is 5. This would make J at least ten via J = 3+C+2. Again, both J and F would be 10. We follow that due to the symmetry, neither A, F, or J can be 7. A, F, and J must be 8, 9, and 10.
Among B, C, E, G, H, and I three characters appear twice in our formulas for A, F, and J and three appear only once.
Let’s assume one of B, G, and I were 7, for example B. This leads to F = 7+E+I and J = 7+C+G. Let’s assume E and I are 1 and 2, which makes F 10. J ends up being bigger than 10. It follows that B, G, and I cannot be 7.
Let’s assume one of C, E, or H is 7, e.g. H. If H is 7, G and I have to be 1 and 2 for A to only be 10, let’s say G is 1 and I is 2 (the other pick is symmetric). It follows that F and J have to be either 8 or 9 each.
  1. Let’s say F is 8, it follows that B + E = F - I = 6. This would only be possible if B and E could either 1 and 5, 2 and 4, or 3 and 3. 1 and 2 are already assigned, and 3 and 3 is not permitted.
  2. Let’s say F is 9. Because I is already 2, B and E must be 3 and 4. This leaves J to be 8. G is 1 and B is either 3 or 4. C would have to be 4 or 3 respectively, but both are taken.
Because neither the three letters A, F, and J, nor the three letters B, G, and I, nor the three letters C, E, or H can be 7, given all the restrictions, only D can be 7.
Per a close look, we guess that the three letters that only appear once should be 4, 5, and 6, and see that the following configuration would be viable:
A = G+H+I = 1 + 6 + 3 = 10 F = B+E+I = B + E + 3 = 2 + 4 + 3 = 9 J = B+C+G = 2 + 5 + 1 = 8
I.e. A = 10, B = 2, C = 5, D = 7, E = 4, F = 9, G = 1, H = 6, I = 3, J = 8
Checking back in the original formulation of the problem: red: A+B+C+D+E = 10 + 2 + 5 + 7 + 4 = 28 green: B+C+D+E+G+H+I = 2 + 5 + 7 + 4 + 1 + 6 + 3 = 28 yellow: C+D+G+H+F = 5 + 7 + 1 + 6 + 9 = 28 blue: D+E+H+I+J = 7 + 4 + 6 + 3 + 8 = 28
Q.E.D.
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42 sats \ 0 replies \ @south_korea_ln OP 19 Oct
Give this man a medal.
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