## anonymous 5 years ago how many natural solution dose the equation x1+x2+...+x10=1000 have, which noone of xi is multiple of 3?

Right away, I don't see any clever solution. Brute force would be $N = \sum_{\substack{x_1 = 0\\3\nmid x_1}}^{1000} \sum_{\substack{x_2 = 0\\3\nmid x_2}}^{1000-x_1}\cdots\sum_{\substack{x_{10} = 0\\3\nmid x_{10}}}^{1000-\sum_{i=1}^{9} x_i} 1$ $= \sum_{\substack{x_1 = 0\\3\nmid x_1}}^{1000} \sum_{\substack{x_2 = 0\\3\nmid x_2}}^{1000-x_1}\cdots\sum_{\substack{x_{10} = 0\\3\nmid x_{9}}}^{1000-\sum_{i=1}^{8} x_i} \left(1 + \left\lfloor \frac{1000 - \sum_{i=1}^9 x_i}{3} \right\rfloor \right)$