2024 SSMO Relay Round 2 Problems/Problem 2

Problem

Let If find the last three digits of

Solution

We seek to find the last three digits of Note that \begin{align*} \sum_{n=1}^{19}n(n+1)(n+2)&=\sum_{n=1}^{19}(n^3+3n^2+2n)\\ &=\sum_{n=1}^{19}n^3+3\sum_{n=1}^{19}n^2+2\sum_{n=1}^{19}n\\ &=\left(\frac{19\cdot20}{2}\right)^2+3\left(\frac{19\cdot20\cdot39}{6}\right)+2\left(\frac{19\cdot20}{2}\right)\\ &=190^2+(19\cdot39)\cdot10+380\\ &=36100+7410+380\\ &\equiv100+410+380\\ &\equiv\boxed{890}\pmod{1000}. \end{align*}

~SMO_Team