don't tell me it's really 2005+...+4009 If yes, this topic should be placed in intermediate forum.

Well don't blame me, blame Centro!! So i think that, as an Olympiad problem, this problem deserves a place here, but if it's more convenient to place it in the intermediate forum, it doesn't matter

Jutaro wrote: Well don't blame me, blame Centro!! So i think that, as an Olympiad problem, this problem deserves a place here, but if it's more convenient to place it in the intermediate forum, it doesn't matter Oh no! I was tricked by the question! We really need some checking to see how many "a"s are there to make -a-(a-1)-....+0+... is positive.

$(k-1002)+(k-1001)...+k+...(k+1001)+(k+1002)=2005k$ so every number of the form $2005k$ can be written in that form, and only such numbers do. hence the answer to the first problem is $2005^2$ As for the secound question we have: then sum of 2004 consecutive integers is the sum of $2005$ minus the last term. so these sums are equal to $2005k-k-1002=1002(2k-1)$ So numbers that can be written in both ways are those numbers of the form $(2005)(1002)(2t-1)$ so the rest is easy.

Pascual2005 wrote: $(k-1002)+(k-1001)...+k+...(k+1001)+(k+1002)=2005k$ so every number of the form $2005k$ can be written in that form, and only such numbers do. hence the answer to the first problem is $2005^2$ As for the secound question we have: then sum of 2004 consecutive integers is the sum of $2005$ minus the last term. so these sums are equal to $2005k-k-1002=1002(2k-1)$ So numbers that can be written in both ways are those numbers of the form $(2005)(1002)(2t-1)$ so the rest is easy. I think the answer to the problem is $2005 \cdot 2004$ not $2005^2$, I assume, you missed the case $k=0$

panamath wrote: Pascual2005 wrote: $(k-1002)+(k-1001)...+k+...(k+1001)+(k+1002)=2005k$ so every number of the form $2005k$ can be written in that form, and only such numbers do. hence the answer to the first problem is $2005^2$ As for the secound question we have: then sum of 2004 consecutive integers is the sum of $2005$ minus the last term. so these sums are equal to $2005k-k-1002=1002(2k-1)$ So numbers that can be written in both ways are those numbers of the form $(2005)(1002)(2t-1)$ so the rest is easy. I think the answer to the problem is $2005 \cdot 2004$ not $2005^2$, I assume, you missed the case $k=0$ $k=0$ gives 0, which is not a positive integer.