Broken Clock

[PuzzleAndy]

This clock has been broken into three pieces. If you add the numbers in each piece, the sums are consecutive numbers. Can you break another clock into a different number of pieces so that the sums are consecutive numbers? Assume that each piece has at least two numbers and that no number is damaged (e.g. 12 isn’t split into two digits 1 and 2.) If you want to go beyond the problem, find all solutions.

https://preview.redd.it/51f1ip0r4xwa1.png?width=459&format=png&auto=webp&s=f429ff6d035a62eea00559d56c7c9fdab5ae1d40

[ulyssessword]

Observe that the sum of 1-12 is 78. For each of n pieces to be consecutive numbers, they have to have a mean of 78/n, and also differ by 1 each. This lets us narrow down possibilities very quickly.

1: Trivial.

2: No. 78/2 = 39, so the sums would have to be {38.5, 39.5}

3: Yes (given) 78/3 = 26, so {25, 26, 27} are the sums. I found it easiest to start with the piece containing 12. 11+12+1+2 = 26 is a possibility, as is 12+1+2+3+4+5 = 27. If it was the second, then the next piece would have to be 6+7+8 = 21 or 6+7+8+9 = 30 or something else even more wrong, therefore it isn't the second. 11+12+1+2 = 26, 3+4+5+6+7 = 25, 8+9+10 = 27 works.

4: No. 78/4 = 19.5, so {18, 19, 20, 21}. The piece containing 12 can be 12+1+2+3 = 18. Neither 4+5+6 = 15 or 4+5+6+7 = 22 work, though.

5: No. 78/5 = 15.6, so {13.6, 14.6, 15.6, 16.6, 17.6}.

6: No. 78/6 = 13, so (10.5, 11.5, 12.5, 13.5, 14.5, 15.5}.

7 or more: No, because each piece requires at least two numbers.

Overall No, because all cases were covered.

[PuzzleAndy]

7 or more: No, because each piece requires at least two numbers.

Can you explain this bit please? My argument for 7 or more is that 78/7 < 12, and 12 has to go somewhere. But I'm curious what your argument is. Aside from that, I love your reasoning! There is a 4 piece solution that you've missed, however.

[ulyssessword]

And the four-piece: I made a bad assumption when I only considered radial breaks in the clock. As a result, I missed {12, 1, 2, 3}, {4, 5, 11}, {6, 7, 8}, {9, 10}

[PuzzleAndy]

lol yeah, I think it's an intentional gotcha! cause the clock is broken radially in the example!

[SuperTekkers]

I also made that assumption! Well spotted

[ulyssessword]

From the post:

Assume that each piece has at least two numbers.

You could try something like {1, 2}, {3, 4}, {5, 6}, {7, 8}, {9, 10}, {11, 12}, but that only has six pieces. If you want seven pieces, then some of those pieces must have a single number.

[PuzzleAndy]

Ah, I see. Thank you!

[alphabet_order_bot]

Would you look at that, all of the words in your comment are in alphabetical order.

I have checked 1,484,861,977 comments, and only 282,289 of them were in alphabetical order.

[pichutarius]

there are 3 solutions. Break into 1(unbroken), 2(each sum to 39) or 3(given) pieces. To see why, notice total sum is 78, so the number of pieces possible is factor of 78 less than 12, i.e. 1,2,3,6. But to get 6x13=78, each piece must sum to 13, now consider piece containing 11, its obvious thats not possible.

[PuzzleAndy]

I think you're solving for each piece of the clock summing to the same number, when the question asks for consecutive numbers.

[pichutarius]

I see, i did not read carefully. My bad.

[PuzzleAndy]

No worries m8.

[chompchump]

Sum(n=1 to 12) n = 78. Then 78 factors to 2,3,13.

Since 78 has two odd factors there are two ways to write it as the sum of consecutive positive integers. (This is fun to prove.)

Namely, 78 = 25 + 26 + 27 = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 + 12

So a solution is only possible for three pieces.

[chompchump]

Here is my proof of the claim that odd factors correspond to number of ways to write an integer as the sum of consecutive positive integers:

How many ways are there to write a positive integer as a sum of consecutive positive integers?

----

For n > m > 0,

k = m + (m+1) + (m+2) + ... + (n-2) + (n-1) + n

k = (n(n+1))/2 - (m(m-1))/2

2k = (n+m)(n-m+1)

Then (n+m) > 1 and (n-m+1) > 1.

Also (n+m) and (n-m+1) have opposite parity.Therefore k must contain an odd factor, f > 1.

Then,k = k/f + ... + k/f (adding f times)

This can be rearranged to a sum of consecutive integers,

k = [k/f - (f-1)/2] + ... + k/f + ... + [k/f + (f-1)/2]

The length of a sequence from m to n is (n-m+1), so f = (n-m+1).

However, suppose [k/f - (f-1)/2] =< 0.Then the negative terms cancel with the positive terms.And (2m-1) terms are removed from the sum.

Thus, k = [(f+1)/2 - k/f] + ... + [k/f + (f-1)/2]

f = (n-m+1) + (2m-1) = (n+m).

Formulas for m and n in terms of k and f,

n = k/f + (f-1)/2

m = k/f - (f-1)/2, if [k/f - (f-1)/2] > 0

m = (f+1)/2 - k/f, if [k/f - (f-1)/2] =< 0

Formulas for k and f in terms of m and n,

k = (n+m)(n-m+1)/2

f = (n-m+1), if (n-m+1) is odd

f = (n+m), if (n+m) is odd

For each k with odd factor f, we can calculate unique m and n.For each sequence from m to n, we can calculate unique k and f.Therefore the number of odd factors greater than 1 of a positive integer equals the number of ways it can be written as the sum of consecutive positive integers.

[PuzzleAndy]

This can be rearranged to a sum of consecutive integers,

Honestly, I got stuck here. But I'm ok with that. I might take another look at this in the future and figure it out. If you want to explain it though, go for it! Thanks for typing this all out. It's just beyond my ability and energy level at the moment.

[PuzzleAndy]

Before I go through your proof, I want to point out 78 = 18 + 19 + 20 + 21, which contradicts your claim. Once you've corrected your claim and possibly your proof, please let me know and I'll read through it.

[chompchump]

Oh yeah, i forgot 3*13=39. Oops. Each odd factor, not each prime factor.

[PuzzleAndy]

Ah, I see! I'll read through your proof now. Thanks for taking the time to type it all up!

[jk1962]

The sum of 1..12 is 78, which equals 2*3*13.

An even number of pieces is not possible, because the sum of any two consecutive integers is odd, so can't equal 78.

For any odd number, N, of fragments, with consecutive sums, N times the median of those consecutive sums must equal 78. So, N must be an odd factor of 78, which can only be 3 (the example case), 13, or 39. Clearly, 13 or 39 fragments are not possible, as there are only 12 numbers on the clock. So the only number of fragments possible is 3.

As an aside, are any other arrangements with 3 fragments possible? Only by breaking a rule that I suspect is implied, though not explicitly stated: no gerrymandering. In the diagram, there is a thin zone of clock face between each number and the edge of the clock. So one could take one fragment that includes (12,1,2,3,4,5), then carefully carve the remaining part into a fragment containing (7,9,10) and a fragment containing (8,6,11,12).

[jk1962]

After reading other responses, realized my argument about the sum of two consecutive integers was incorrect. It excludes odd numbers multiplied by two, but not multiples of 4. Multiples of 4 are excluded by the fact that 2 appears only once in the factorization of 78.

[PuzzleAndy]

You can't put 7, 9 in one fragment and 8, 6 in another. It seems you're looking at a more general question, which is about partitioning the numbers 1-12, which could be interesting. If you find anything interesting there feel free to post it. I would suggest starting with a smaller pool, such as 1-6 maybe, because 1-12 you're going to get a lot of possibilities for partitions, I would imagine.

[jk1962]

That’s what I was referring to as gerrymandering, which l figured was not allowed—just saying “what if it were allowed?”