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31

First, we want to take a look at the snake itself. Then, let's take a look at the grid, and see if there are some obvious features. So, let's fill those in: Then, because the snake has more white near the tail end, I thought it would be a good idea to go from the 93 to the top side as soon as possible; the grid seems to have more white along the right and ...

20

The answer is PROBABLY I don't have a proof yet, but experimenting with it will yield something like My strategy is Now my reasoning is that One idea to rigorously prove it is to use some result similar to and reason that EDIT: There was an error in my picture above: I put $18$ to the right of $15$. But this is very easy to correct. In the beginning ...

19

Using 2,3,7,11: $2 = 2$ $3 = 3$ $5 = 11 + 3 - 7 -2$ $7 = 7$ $11 = 11$ $13 = 2 + 11$ $17 = 3! + 11$ $19 = 2^3 + 11$ $23 = 3 \cdot 7 + 2$ $29 = \frac{(7-2)!}{3} - 11$ $31 = 3 \cdot 11- 2$ $37 = (11-3!) \cdot 7 + 2$ $41 = 7^2 +3 - 11$ $43 = 2 \cdot 11 + 3 \cdot 7$ $47 = 3 \cdot 11 + 2 \cdot 7$ $53 = 2^ {3!} - 11$ $59 = 3! \cdot 11 - 7$ $61 = (11-2) \cdot 3! + ... 18 Solution Your sister is and the age on the cake is Proof of uniqueness We're given several pieces of information: "transposed the digits of her age" - her age is 2 digits. "She'll thank you for the compliment" - the age on the cake makes her seem younger, so the second digit is smaller than the first. "her age is a prime number" - her age is a 2-digit ... 15 It is possible to solve this without a computer search. The proof of$\min|O|$is below. 13 Given the first 11 primes: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31 We can observe that: We'll always get 2, 3 and 7 for "free" (via other numbers, such as 23 and 17, etc), so they can be ignored leaving only 5 to be explicitly included Of the double digit numbers it is possible to get either 13 and 31 for free (using 11 + 31 or [1/2]3 + 1[1/7/9]), but ... 12 One solution to part A: 12 The full list of possibilities that each prime can be adjacent to is as follows: The key ones to notice are Now there are only two possibilities for the other number next to$5$, namely$11$or$41$. So it must be$11$next to$5$, then next to that must be either$2$or$53\$. So altogether we have I've tried to illustrate this as follows:

11

(The earlier version of this answer contained more praise for this puzzle, sometimes comparing it to the earlier one. I only now noticed that this snake was indeed not posted by the OP of the original one, so I have now toned it down a bit.) Here's the solution I found: Some things I particularly liked: All in all, this snake strikes the exact sweet spot ...

11

It might be more fun to fill the whole table. I've started with a few examples: $$\require{begingroup}\begingroup \def\*{\times } \begin{array}{|c|c|c|c|c|} \hline n& 2,3,5 & 2,3,7 & 2,5,7 & 3,5,7 \\ \hline 1& 2\*3-5 & 7-2\*3 & \frac{2+5}{7} & 3+5-7 \\ 2&... 11 I think this works: The adjacent number pairs are restricted to where each pair sums to Note that it is necessary to because 10 They were born in the year In fact, note that 10 First: Now: Next: Now We're nearly there. Let's draw up a table. Entries are scores for the team on the left when they played against the team on the top. And So we're done, and here is the final table which you can readily check has all the required properties: 10 Here is the answer: 2=2 3=3 5=2+3 7=\frac{43-3}{2}-13 11=43-2\cdot(13+3) 13=13 17=43-13\cdot2 19=13+3\cdot2 23=43-2\cdot(13-3) 29=43-13-3+2 31=43-13+3-2 37=43-3\cdot2 41=\frac{43+3\cdot13}{2} 43=43 47=2\cdot43-3\cdot13 53=43+13-3 59=43+13+3 ... 10 Solution for the 5x5 grid Other solutions for the same grid are For the 6x6 grid I noted that This leads to this conclusion for the 6x6 grid 9 83 is probably not allowed - I swear I'll find a legitimate solution soon... arrgh... Using 2,3,5,7: 2, 3, 5, 7 = 2, 3, 5, 7 11 = 7 + 5 + 2 - 3 13 = 7 + 5 + 3 - 2 17 = 7 \cdot 2 + 3 19 = 7 \cdot 2 + 5 23 = 7 \cdot 3 + 2 29 = 7 \cdot 5 - 3! 31 = 7 \cdot 5 - 3! + 2 37 = 7 \cdot 5 + 2 41 = 7 \cdot 5 + 3! 43 = 7 \cdot 5 + 3! + 2 47 = 7^2 +... 9 I tried writing a generator. I can get everything from 1 to 24 inclusive, for all 4 combinations. I can get 0 to 33 for one combination. The first ungeneratable counting number is 68.$$\begin{array}{|c|c|c|c|c|} \hline n& 2,3,5 & 2,3,7 & 2,5,7 & 3,5,7 \\ \hline 0 & (5-(2+3)) & (3-\sqrt{(2+7)}) & (7-(2+5)) & - \\ 1 & (\...

9

Improving upon Gareth's answer (the details of which I will not repeat) with consideration of ThomasL's suggestion, a solution with 12 primes is found: Two solutions with 11 primes were also found:

8

Computer search solution. A quick and dirty Python script then identifies... Here is the script: from sympy import isprime solutions = [] def find_odd(odds, evens): if len(odds) >= 6: solutions.append((odds, evens)) for o in range(odds[-1]+2 if odds else 1, 101, 2): if not odds: print(f"\n{o: 2d} ", end="") if all(isprime(o + ...

8

Incorrect answer Oops! The "solution" below is wrong in the following way: Unfortunately it's 3:25am local time and I don't have time to attempt a proper fix. My apologies; hopefully someone else will do a less-hilariously-broken job. My incorrect solution follows, because I don't believe in hiding my screwups :-). First of all, Obviously the same holds ...

8

A solution for part B:

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She was The decorator put

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New high score :P List of possible primes is 43, 41, 37, 31, 29, 23, 19, 17, 13, 11, 7, 5, 3, 2. Distances between those primes are 2, 4, 6, 2, 6, 4, 2, 4, 2, 4, 2, 2, 1. Which? The key observation: So This means Let's see which options are there: That should be all. First one, number is in reverse

7

Partial answer, to be used as a springboard Parity The first thing to notice is that Then, of course, We notice also that Prime gaps The primes between 1 and 100 are: 2 3 5 7 11 13 17 19 23 29 31 37 41 43 47 53 59 61 67 71 73 79 83 89 97 Now look at Similarly, So far we have: I'm guessing further deductions can be made ...

7

I think the answer is Reasoning Example

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I think this works: And thus the pattern of sums and differences is: Basically it just The other N for which this occurs, of course, is

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For other values of N: Number of solutions (including full and partial reversals): First solution found in each case:

7

I don't know if this sufficiently answers your question but I've found an example which has alternating prime differences all the way around so there isn't an "unavoidable exception".

7

arbitrahj and Paul Panzer have already provided great analysis of this problem. I would like to demonstrate that Lemma Theorem Putting these together

7

Wrong answer Well, it might be the right answer for all I currently know, but my argument has multiple errors in it (pointed out by others in comments). I may fix it, but others should feel free to post correct solutions before I do :-).

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