Number TheoryChallengeHard Number Theory ShareHard Number Theory · Challenge problemEvaluate ∑k=112⌊k213⌋\displaystyle\sum_{k=1}^{12}\left\lfloor \frac{k^2}{13} \right\rfloork=1∑12⌊13k2⌋.Check Show answer Walk through it step by step Practice more like this Review this concept