\(\overline{2x7}\) ⋮ \(\overline{x1}\) ( x # 0)
⇔ 200 + 10x + 7 ⋮ 10x + 1
⇔ (10x +1) + 206 ⋮ 10x + 1
⇔ 206 ⋮ 10x + 1
206 = 2.103
Ư(206) = { 1; 2; 103; 206}
10x + 1 \(\in\) {1; 2; 103; 206}
x \(\in\) { 0; \(\dfrac{1}{10}\); \(\dfrac{51}{5}\); \(\dfrac{41}{2}\)}
Vì x \(\in\) N nên x = 0 mà x #0 vậy S = \(\varnothing\)