Để 35 + x chia hết cho 5 thì:
\(\dfrac{35+x}{5}\) có giá trị nguyên mà:
\(\dfrac{35+x}{5}=\dfrac{35}{5}+\dfrac{x}{5}=7+\dfrac{x}{5}\) ⇒ \(\dfrac{x}{5}\) có giá trị nguyên
⇒ \(x\in B\left(5\right)=\left\{0;5;10;15;20;25;30;35;40;45;50;55;60;65;70;75;80;...\right\}\)
Mà: \(10< x< 60\)
\(\Rightarrow x\in\left\{15;20;25;30;35;40;45;50;55\right\}\)
(35 + x) ⋮ 5 khi x ⋮ 5
⇒ x ∈ B(5) = {0; 5; 10; 15; 20; ...; 55; 60; ...}
Mà 10 < x < 60
⇒ x ∈ {15; 20; 25; 30; 35; 40; 45; 50; 55}