Difference between revisions of "2024 SSMO Relay Round 2 Problems/Problem 2"
(Created page with "==Problem== Let <math>T = TNYWR.</math> If <cmath>a = \sum_{n=1}^{N}n(n+1)(n+2),</cmath> find the last three digits of <math>a.</math> ==Solution==") |
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==Solution== | ==Solution== | ||
| + | We seek to find the last three digits of <math>\sum_{n=1}^{19}n(n+1)(n+2).</math> 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 | ||
Latest revision as of 14:43, 10 September 2025
Problem
Let 
If ![\[a = \sum_{n=1}^{N}n(n+1)(n+2),\]](http://latex.artofproblemsolving.com/b/0/a/b0a3529373cf9111c8d51770317f44364d29fd90.png)
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
