The sum of the odd numbers from $$1$$ to $$2019$$ both inclusive, is divisible by

Solution

There are $$1010$$ odd numbers from $$1$$ through $$2019$$, and their sum is $$1010^2=1020100$$. This number is divisible by $$100$$ and also by $$101$$ because $$1020100=100\times101\times101$$. Hence it is divisible by both.

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