Special sums of numbers The natural numbers from 1 to 100 are written next to each other in any order. Now all sums of three adjacent numbers are calculated. How many of these sums can be odd at most?
I have 94 out. I did it by trial and error. Did you get something else out of it or did I do it right?
If you do odd-even-even $25$ times, followed by $25$ odds in a row, you get $97$ odd sums of three consecutive numbers. The only even sum comes from an even-odd-odd in positions $75, 76, 77$.
I started by looking at just the first 10 numbers.
$$
1, 2, 3, 4, 5, 6, 7, 8, 9, 10
$$
When adding numbers, if both of them are either even or odd, the answer will always be even. If one is even and one is odd, the answer will be odd. So, in these first 10, we see an alternating sequence when adding 3 at a time.
$$
(1 + 2 + 3)\ \%\ 2 = 0\\
(2 + 3 + 4)\ \%\ 2 = 1\\
(3 + 4 + 5)\ \%\ 2 = 0\\
(4 + 5 + 6)\ \%\ 2 = 1\\
$$
So now we can see this alternating sequence of even and odd answers. We also know there are 98 sums (since each one starts from the next natural number and we don't count 99 or 100). So just taking half of 98, 49 should be it.