[Daily puzzle] Another MSM puzzle \ stacker news ~science

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696 sats \ 10 comments \ @south_korea_ln 18 Oct science

Too little time last night (ok, I admit, one too many beers) to post the daily puzzle at the usual time. So here it is.

The four squares below form a Venn diagram with ten regions, labeled A to J. The letters A to J each stand for a number between 1 and 10, such that no two regions share the same number and all numbers from 1 to 10 are used. The regions in each square add up to the same number. What value is D, the intersection of all four squares?

Previous iteration: #728017

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500 sats \ 1 reply \ @Murch 19 Oct

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.

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.
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.
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.

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.
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.

42 sats \ 0 replies \ @south_korea_ln OP 19 Oct

Give this man a medal.

142 sats \ 2 replies \ @WeAreAllSatoshi 18 Oct

What does MSM mean?

0 sats \ 1 reply \ @south_korea_ln OP 18 Oct

Mainstream media. Today's and yesterday's puzzle are from the same article in an MSM journal. I'll link to it after this one finishes, so as not to spoil the post with the answers provided in the article.

0 sats \ 0 replies \ @WeAreAllSatoshi 18 Oct

Ohh ok! I know that acronym but I didn’t associate it with this context.

112 sats \ 2 replies \ @SimpleStacker 18 Oct

I brute forced it

import numpy as np
import itertools

A,B,C,D,E,F,G,H,I,J = 0,1,2,3,4,5,6,7,8,9
x = np.arange(1,11)

for p in itertools.permutations(x, 10):
    red = p[A]+p[B]+p[C]+p[D]+p[E]
    yellow = p[C]+p[D]+p[F]+p[G]+p[H]
    green = p[B]+p[C]+p[D]+p[E]+p[G]+p[H]+p[I]
    blue = p[D]+p[E]+p[H]+p[I]+p[J]
    if (red==yellow) and (red==green) and (red==blue):
        break

p

D = 7

168 sats \ 1 reply \ @Undisciplined 19 Oct

I think I get the logic of it but I don't have a proof.

A = G+H+I F = B+E+I J = B+C+G

That means they are each at least 6 and none of B, C, E, G, H, or I can be larger than 6, which means they make up 1-6 in some order.

Then, it must be impossible to arrange them such that any of A, F, or G are equal to seven without one of them being greater than 10.