[superprismatic]

Find a set of fifteen distinct

positive integers such that each

number in the set divides the sum

of all of the other fourteen numbers

in the set. Can you find more than

one such set?

[Guest]

2: 2, 16, 30, 32, 40, 64, 78, 128, 390, 780, 1560, 3120, 6240, 12480, 24960

3: 3, 24, 45, 48, 60, 96, 117, 192, 585, 1170, 2340, 4680, 9360, 18720, 37440

[Guest]

So it appears that the sum of these numbers must be divisible by each individual number... Perfect numbers! But which one(s)? After some thought I think any will do for a start and then just keep doubling each number to get a set of 15. For 6: 1,2,3,6,12,24,48,96,192,384,768,1536,3072,6144,12288.

I'm not going to bother working out the rest. But you can use any perfect numbers to generate a set.

[Guest]

there are many solutions

Any solution where I₁ & I₂ are chosen at random - then I_y=SUM_1->(y-1)(I_x) or I_y=2*I_y-1

1

2

3

6

12

24

48

96

192

384

768

1536

3072

6144

12288

[Guest]

{1} works trivially.

{1,2} works as a two element set.

{1,2,3} works as a set of three elements.

Then concatenating {6}.

{1,2,3,6}

then you concatenate {12}, the second smallest common multiple of the previous terms. We pick the second largest because the LCM is actually {6}. Continuing to concatenate the second smallest common multiple of the previous terms gives up to term fifteen the sequence,

{1,2,3,6,12,24,48,96,192,384,768,1536,3072,6144,1228}

It may be shown that, after the 4th term, by concatenating the second smallest common multiple, we are actually concatenating the double of the latest term (i.e. 2 times the LCM). For any term, the terms after can be divided by it, and the terms before sums to it. Thereby their sum can be divided by the term in question.

[Guest]

2: 2, 16, 30, 32, 40, 64, 78, 128, 390, 780, 1560, 3120, 6240, 12480, 24960

3: 3, 24, 45, 48, 60, 96, 117, 192, 585, 1170, 2340, 4680, 9360, 18720, 37440

any set of size n

= X * {Y₁,...Y_n}

such Y_n = 2Y_n-1 + 1 & Y₀ = 0

will work for any positive integer X

[Guest]

there are many solutions

Any solution where I₁ & I₂ are chosen at random - then I_y=SUM_1->(y-1)(I_x) or I_y=2*I_y-1

1

2

3

6

12

24

48

96

192

384

768

1536

3072

6144

12288

Nice, I think that's the solution, but just to point out you need to pick l₁ and l₂ so that one is divisible by the other, you picked 1 and two but you can't pick for example 2 and 3...