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There's two answers here: a real answer, and a (slightly) cheating answer. Real Answer First, we can put an upper bound on our answer. At most, we have a two digit number. This is because the smallest three digit number (100) is an order of magnitude more than the maximum number of displayed segments ($21 = 7\times3$). Through a process of elimination, ...

23

I suppose this is also cheating?

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I don't know if you can get smaller than

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There are three valid patterns: Or, as graphics:

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Here is one way of making the rectangular grid arrangement that Bass suggested: Only the digit 2 in the middle is flipped over. By the way, this kind of packing puzzle shape is called a Polystick.

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My digital clock is in Assuming we can't pad the answer to infinity with unnecessary zeroes, this seems to be the maximal answer, because Edit: @WChargin asks in the comments: Very devious approach! However, it doesn't turn out to yield a higher answer, or any valid answer at all. I can't think of a tidy number-theory way to address this, but I wrote ...

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This assumes you cannot null-pad the number, since if you can null-pad it, the highest number is infinite. For a n-digit number, the maximum number of lines visible is 7n. Therefore, the number cannot be greater than 2 digits, since 21<100 for n = 3, and the number of lines increases linearly with n while the number increases exponentially with n. ...

11

Here's my solution, with the following features: Distinct 2 and 5 Distinct 6 and 9 Rotations only (Mirror images of numbers don't look all that nice) No upside-down digits (max. rotation 90 degrees) This was way more difficult than expected. There may be other non-mirror solutions apart from this one and its symmetries, but if there are, they are few and ...

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The answer is Why? Much easier than usual from you, Tryth! :-)

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You said - "The clock is digital, and some of the seven bars on its units digit are broken." and then at the end you say - "In each of these four scenarios, what is the maximum number of broken bars, and which might it/they be?" Firstly this should mean that the answers to all the four questions should be an integer in the interval [1,7] ...

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Here's one to start off: 1 E - 9 8 7 6 5 $=10^{-98765}$

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Building up on f'' s answer: ### ### ### ### ### ### # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # ### ### ### ### ### ### # # # # # # # # # # # ...

7

I will use the A-G bar notation given in the puzzle. I am also assuming that I only need to find one time that works, based on comments. Case 1: 4 Broken, A, B, D, G. James walks in and E is lit. There are only 4 digits that use E (0, 2, 6, 8). The digit must be a 2 because the other 3 numbers use at least 6 bars, which would mean that at least 2 bars would ...

7

I think I got it. I started by counting the possible middle segments of the digits, there are 11. This means only one possible location for a digit isn't used. Taking this into account, the one is as good as placed: that digit's shape is extremely unfriendly to neighbours on the left and the bottom. From there, the 7, 8 and 6 are just about forced, but one ...

6

Interesting problem! I was tempted to use brute-force with a "x - y + z" general format, just to see what happens. I'm sharing bellow my results. Problem Formulation find(1202, 2021) [Executed in 12.7s] find(2022, 2202) [Finished in 12.6s] Finally, for the sake of curiosity, i also tried a x * y + z model:

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I think I have found the answer for the first question

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Partial solution:

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Using actual digits (instead of letters or concepts), I present or $9^{11111111111111}$

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Regarding only the maximum of 28 elements and 4 numbers, the highest element is 6 (without leading zeroes) and 22 (with leading zeroes : 00:22). Only regarding decimal system.

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If we're restricted to finite numbers, I submit... ggggggg9. That's the (((((((9th number in Graham's series)th number in Graham's series)th number in Graham's series)...)))) where Graham's series (notated as gn for some n) is defined under "definition" here.

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I submit ALEPH 999. It's infinite, 998 'steps' above countable infinity.

3

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TREE(99) from Kruskal's tree theorem.

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As I hinted in the comments, and since there's no no-computers tag, there are a lot of solutions.

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Explanation:

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Answer to part (a) Just like my answer here, I converted this problem into a SAT instance. Answer to part (b)

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If we can rearrange all the 49 crayons independently, then we can make this shape: Since we know that any square shaped bunch of unit squares is automatically the optimal way to maximally overlap the unit square sides, and continuing from a square to a rectangle by adding a single row also gives maximal crayon overlap, we know that this pattern is an ...

2

After finding and writing up the solution to this slightly related problem I decided to impose similar constraints for this problem, formulate it as a SAT instance and solve: No flipping The digits can only lie in one of two orthogonal orientations; this implies that 6 and 9 cannot be represented by the same segments The end result of this was Edit:

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If I understand correctly what you're asking, then Proof: Similarly, for the second problem, Proof:

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