All integers from 1 to 73 are recorded in a sequence such that each number

Question

All integers from 1 to 73 are recorded in a sequence such that each number (from the second onwards) divides the sum of all previous numbers.

c) What numbers can be in the third place and why?

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This question is from a Russian contest. See page 11, question C6. Тhis is the final task of one of the training test to prepare for the USE in Russia. The complexity of this task traditionally corresponds to the level of a good math Olympiad.

Previous questions in this problem (more obvious):

a) Could the last number be 5?

b) What numbers could be last?

In the b) it's hard enough to build an example for 1 in last place.

The problem is that we cannot use a computer. ) 240 minutes was given for solution and careful recording of all 20 tasks. Tools - pen and paper.

Answer 1

This is not an answer, but too long for a comment. For people who are looking for some examples, here are a few I generated.

72 36 54 27 63 42 49 7 70 60 48 66 33 57 38 19 39 65 13 26 68 56 28 37 29 58 40 50 25 51 34 20 69 23 64 32 16 44 22 15 9 6 30 10 2 41 43 21 3 61 31 62 8 24 14 35 59 12 4 5 55 11 67 17 45 52 46 53 47 18 71 1 73

72 36 27 45 60 48 32 64 24 68 34 51 33 66 55 65 52 26 39 69 46 44 22 49 23 50 40 62 42 56 70 35 43 18 58 29 57 38 4 73 25 10 31 61 16 8 19 15 67 1 2 13 7 20 53 41 54 63 21 28 14 3 17 71 5 30 12 6 47 11 59 9 37

70 35 21 63 27 72 48 56 49 9 50 25 15 60 40 64 44 68 51 17 52 39 65 26 41 3 37 31 62 20 45 29 58 24 59 5 10 1 71 22 11 55 66 33 53 34 4 23 69 42 47 43 6 30 28 7 61 12 46 14 54 18 36 8 16 32 19 57 38 67 13 2 73

60 30 45 27 54 72 48 56 49 63 42 39 65 50 70 55 33 66 44 22 18 36 58 38 57 21 29 43 15 9 73 19 37 13 52 26 59 3 14 35 47 12 71 25 20 10 61 31 62 32 24 68 34 2 16 40 11 67 17 51 6 28 5 53 46 69 23 7 41 64 8 4 1

42 1 43 2 44 66 33 21 63 45 72 54 27 57 38 32 64 22 11 67 12 68 52 39 65 40 60 30 26 46 69 23 58 48 36 41 37 14 56 29 19 8 24 4 28 62 31 59 16 7 49 20 55 5 34 61 35 70 10 50 25 15 3 71 17 13 47 53 6 51 9 18 73

Computation suggests that every number from 1 to 73 can be the third element of a sequence of the desired form, but this is not the case for other instances. For example, if we change 73 to 6, the numbers 1 and 3 are not possible as third elements of the sequence.

Answer 2

"All integers from 1 to 73 are recorded in a sequence such that each number (from the second onwards) divides the sum of all previous numbers."

Here is such a sequence, obtained by hand:

72 2 37 3 38 4 39 5 40 6 41 7 42 8 43 9 44 10 45 11 46 12 47 13 48 14 49 15 50 16 51 17 52 18 53 19 54 20 55 21 56 22 57 23 58 24 59 25 60 26 61 27 62 28 63 29 64 30 65 31 66 32 67 33 68 34 69 35 70 36 71 1 73

I trust the pattern is clear, and clearly works for all $n$ (though there are tiny differences depending on the parity of $n$).