I have a calendar with daily puzzles, and I want to verify something, so here is today's puzzle. I answered it myself and have the "official" answer as well. Please be sure to clearly state your reasoning for your answer.

Determine the number from 1 to 10 that goes at each point on this star according to the rules below.

Going clockwise from the top of the star, the points are A, C, E, D, B. Apologies for the manual illustration instead of a nice digital version.

  1. The sum of all five numbers is 28, and no number is duplicated.

  2. Consecutive numbers may not appear (a) at adjacent points nor (b) across a straight line from each other.

  3. Point A is (a) greater than the sum of B and C and (b) less than the sum of D and E.

  4. Points C and D sum to 11, and neither point is 10.


If I've read/interpreted correctly, the following should work

A = 6
B = 1
C = 3
D = 8
E = 10

My reasoning was more or less:

Lemma 1: we can't have consecutive numbers at all with rule 2,

Possible pairs for C and D are $(1,10)$, $(2,9)$, $(3,8)$, $(4,7)$, and $(5,6)$, but $(1,10)$ isn't possible because of rule 4, $(5,6)$ is impossible by lemma 1, and $(2,9)$ and $(4,7)$ aren't possible because then we can no longer choose 5 numbers with no two pairwise points being consecutive numbers, leaving us with C and D being 3 and 8.

Now we need a sum of 17 with 3 numbers from the set $\{1,5,6,10\}$, which obviously leaves us with 1, 6, and 10 for A, B, and E.

The arrangement was then straightforward, though as Wesley pointed out, swapping A and E works as well

  • $\begingroup$ I agree that it should work, but I'd like to see exactly how you got there. I believe this to be one of two answers. $\endgroup$
    – Allevu
    Feb 3 '16 at 18:32
  • 2
    $\begingroup$ I agree with this answer, but I think you could switch the values for A and E and it would still work. $\endgroup$ Feb 3 '16 at 18:35
  • $\begingroup$ I just ran a brute force script, switching A and E gives the only other possible configuration. $\endgroup$
    – DiscOH
    Feb 3 '16 at 23:06

I came up with:

A = 10
B = 1
C = 3
D = 8
E = 6

Basically figuring:

There can't be any consecutive numbers
Can't be 1 3 5 7 9 So need 10
And also 8
Then 6 3 1
Made the points fit the other criteria

  • $\begingroup$ Thanks! This was the "official" answer. Good to see both up here as confirmation. $\endgroup$
    – Allevu
    Feb 3 '16 at 18:51
  • $\begingroup$ @Allevu Yes, now see Will's answer works fine as well. $\endgroup$
    – Paul Evans
    Feb 3 '16 at 18:53

There can be no consecutive numbers, so we could take the first odd numbers and add 3 to them in total, obtaining 28. Even then, two of the numbers would end up being 1 and 3. If $a + b + c + d + e = 28$, $c + d = 11$ and $d + e > a > b + c$, then $11-c + e > a > b + c$. The expression $11-c + e$ must be greater than 10, and since $e-c > 1$, it's at least 13. $a, c, d$ and $e$ > 1, so $b=1$. $a, e$ and at least one of $c$ or $d$ are greater than 3, therefore the one left out equals 11-3=8. According to the "no consecutives" rule, $a+e$ can only be written as 10+6. Since $b + c$ must be less than 9...

The solution:

$b=1, c=3, d=8$. $a$ and $e$ can either be 6 or 10.


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