Here's how I solved this (a long time ago and without all the dead ends):
1) Recognize that all odd factors k will generate a unique list. 2025/k will be the center of the list and the length will be k. Eg, 2025/3=675, so there will be a list of length 3 with 675 at the center. 674+675+676=2025 🎉
2) BUT, some of the lists will have negative #s. Eg, 2025/675=3, which would correspond with a 675 length list centered at 3. 😡 This list goes from 3-337 to 3+337 or -334, -333,...,333, 334, 335, 336, 337, 338, 339, 340. BUT, lots of opposites so this is just 335, 336,...,339, 340. New solution! ☺️
3) None of these lists can be the same. Why? The lists without negatives will all be different odd lengths (k), and the lists that start with negatives will all have an even number of terms because we will eliminate all of the negative pairs and 0. So all these lists are unique! 🤯
4) Now we just need to prove that we aren't missing anything. If we were missing an odd length list, the center would have to be a factor of 2025. ❌ Could we be missing some even length list? Lol, I can't remember how I proved this. 😭
5) So, 2025 has 15 factors. They are all odd. But careful! When k=1, we get a list of just 2025, which we don't want to include. So 14, or 1 less than the number of odd factors, is the final answer!
This is a cool one! It comes from the list centered at 1, with initially 2025 numbers in the list. So this gives you the list -1011, -1010, ..., 1010, 1011, 1012, 1013.
Btw, this problem lives in the domain of what I call "staircase numbers", one of my favorite problem spaces, and was part of a multi-day project back in the day when I taught 6th grade (the "big" problem we explored then was around what numbers can't be written as consecutive sums).
Comments
1) Recognize that all odd factors k will generate a unique list. 2025/k will be the center of the list and the length will be k. Eg, 2025/3=675, so there will be a list of length 3 with 675 at the center. 674+675+676=2025 🎉
Claude says 10
Gemini says 14
Copilot say 15
So I'm going to average all those and say the answer is 11.75.
This is how we AI, right?
n + n+1 = 2025
3 terms:
n + n+1 + n+2 = 2025
So for k terms, we will have:
kn + T(k)
where T(k) is the k^th triangular number
But many don't work! This is fun.
We also need to have k | [2025-T(k)]
kn + T(k) = 2025
so that:
kn = 2025 - T(k)
n = (2025 - T(k))/k
Choose number of terms k and check if:
(2025 - ((k-1)^2+k-1)/2)/k
is an integer.
That feels way less gross than it looks. And fits nicely in a spreadsheet