Ta có :

Áp dụng tính chất luỹ thừa : $\left( {{a}^{n}}+{{b}^{n}} \right)\vdots (a+b)$ với $n$ lẻ ta có :

$A=\underbrace{\left( {{1}^{2017}}+{{99}^{2017}} \right)}_{chia\,h\hat{e}t\,cho\,100}+\underbrace{\left( {{2}^{2017}}+{{98}^{2017}} \right)}_{chia\,h\hat{e}t\,cho\,100}+...+\underbrace{{{50}^{2017}}}_{chia\,\,h\hat{e}t\,\,cho\,100}+\underbrace{{{100}^{2017}}}_{chia\,\,h\hat{e}t\,\,cho\,100}$ $\Rightarrow A\vdots 50$


$A=\underbrace{\left( {{1}^{2017}}+{{100}^{2017}} \right)}_{chia\,h\hat{e}t\,cho\,101}+\underbrace{\left( {{2}^{2017}}+{{99}^{2017}} \right)}_{chia\,h\hat{e}t\,cho\,101}+...+\underbrace{\left( {{50}^{2017}}+{{51}^{2017}} \right)}_{chia\,h\hat{e}t\,cho\,101}$ $\Rightarrow A\vdots 101$

Vì $(50;101)=1$ nên $A\vdots (50.101)$ hay $A\vdots B$.

B = 1 + 2 + 3 + ... + 100 = 101.50