hint 1

hint 2

hint 3

orl wrote: Determine for which positive integers $ k$ the set \[ X = \{1990, 1990 + 1, 1990 + 2, \ldots, 1990 + k\} \] can be partitioned into two disjoint subsets $ A$ and $ B$ such that the sum of the elements of $ A$ is equal to the sum of the elements of $ B.$ Let $ S = 1990(k+1) + \frac{k(k+1)}{2}$ be the sum of the elements in $ X$. So obvoiusly $ S$ is even which implies $ k \equiv 0,3 \bmod 4$ For $ k \equiv 3 \bmod 4$ we let $ a = \{1990, 1990+1, .., 1990 + \frac{k-3}{4}, 1990 + 3 \cdot \frac{k+1}{4}, ..., 1990 + k\}$ And $ B = X - A$ For $ k \equiv 0 \bmod 4$ we notice that $ X$ has an odd number of elements. (we set $ k = 2n$) Thus is the sum of the smallest subset at most: $ 1990 + (n+1) + \cdots + 1990 + 2n$ = $ 1990 \cdot n + \frac{2n(2n+1)-n(n+1)}{2} = 1990n + \frac{n(3n-1)}{2}$ But then is $ 1990n + \frac{n(3n-1)}{2} \ge 1990(2n+1) + \frac{n(n+1)}{2}$ Which is equivalent to $ n^2 - 1991n - 1 \ge 0$ Which yields $ n \ge 1992$ So if $ 4 \mid k$ we have the condition $ k \ge 3984$ And I'm too lazy to write down the proof that we can partion it we $ k \ge 3940$ and $ 4 \mid k$.... LOL... Just realized, that this was the exact same solutions a in the hints..

Mathias_DK wrote: But then is $ 1990n + \frac{n(3n-1)}{2} \ge 1990(2n+1) + \frac{n(n+1)}{2}$ Which is equivalent to $ n^2 - 1991n - 1 \ge 0$ Which yields $ n \ge 1992$ There is a mistake in calculations. Should be $$ 1990n + \frac{n(3n+1)}{2} \ge 1990(n+1) + \frac{n(n+1)}{2}$$and therefore $n^2\geq 1990$, so $n\geq 45$. For $k$ one gets $k=92+4s$. Using the argument of discrete continuity one can construct valid example for this case.

Sum of elements in a and b is : k*(k+1)/2 + (k+1)*1990. Now this has to be divisible by so 4|k(k-1) so k is 0 or 3 mod 4.Contruction for 3 mod 4 is trivial.For k = 4a we prove that k>=90.If k<90 then difference of numbers of elemnets in A and B can't be greater than 1(if it were greater than 1 if all elements 1 2 3 .. k were on one side sum would be less than 4278(excluding all 1990)contradiction (differece is odd)) and then if we leave 1990 on one side(wlog right) and all numbers on the left their difference is less than 1990 (it is at most k^2/2 which is less than 1990). Contruction for k=92 is : 1990+(1990+2)+(1990+3)+...+(1990+46)+(1990+64)=(1990+1)+(1990+47)+(1990+48)+...+(1990+64)+(1990+65)+..+(1990+92) ,next by induction if we add 4 number 4l+1,4l+2,4L+3,4l+4 we add 4l+1 and 4l+4 ,and other two on the other side we saves equality and we are done.