Divide the Pyramid

The $n^{th}$ square pyramidal number can be found by using $\sum_{i = 1} ^{n}i^2 = \frac{(n)(n+1)(2n+1)}{6}$ . Prime factorzing $49307698992$ leads to $49307698992 = 6*2018^{3}$ $n = 6*2018^{3}$ Thus let $P = 6*2018^{3}$ pyramidal number which means $P =\frac{(6*2018^{3})(6*2018^{3}+1)(2(6*2018^{3})+1)}{6}$ Let $m$ be the maximum number whereby $2018^{m} \:| \;P$ . Simplifying P $P =(2018^{3})(6*2018^{3}+1)(2(6*2018^{3})+1)$ We then want to find $2018^{m} \; | \; (2018^{3})(6*2018^{3}+1)(2(6*2018^{3})+1)$ Clearly, the highest power of $2018$ that divides $P$ is $\boxed{3}$ because of the first term. The second and third term both are not divisible by $2018$ .