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This might be a stupid question but:

Assume you start with 1 stack of *n* items and want to end with *n* stacks with 1 item in each. To do this you successively divide it into smaller stacks with *a* and *b* items in each and create a product *ab*. (*n* is a natural number) Prove that the sum of all *ab* is equal to *n*(*n*\-1)/2.

Can I assume that the sum of all *ab* is the same no mater how you divide the stacks.

example *n* = 4

1. (4)
   1. both a and b= 2 , ab(1) = 4
   2. a -> 1+1, ab(2) =1
   3. b -> 1+1, ab(2) =1
   4. 4+1+1=6
2. (4)
   1. a(1) = 3,  b(1) = 1,   ab(1) = 3
   2. a(1) -> a(2) = 2,  b(2) =1,  ab(2) = 2
   3. a(2) -> a(3) = 1,  b(3) =1,  ab(3) = 1
   4. 3+2+1 = 6
3. 4(4-1)/2= 6

Can I do this or do I have to write some proof for it always being equal?
ago
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no you have to prove it. that's the whole point of this problem.
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You have to prove it. This sounds like a good problem.
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