a, Ta có: 4n+3$\vdots$n-2
⇒4(n-2)+11$\vdots$n-2
⇒n-2∈Ư(11)={±1;±11}
n-2=1⇒n=3
n-2=-1⇒n=1
n-2=-11⇒n=-9
n-2=11⇒n=13
Vậy n∈{3;1;-9;13}
b, Ta có: n+4$\vdots$n+1
⇒n+1+3$\vdots$n+1
⇒n+1∈Ư(3)={±1;±3}
n+1=1⇒n=0
n+1=-1⇒n=-2
n+1=3⇒n=2
n+1=-3⇒n=-4
Vậy n∈{0;-2;2;-4}
c, Ta có: 6x+11$\vdots$31
⇒6(x+7y)-31y$\vdots$31
Vì: 6(x+7y)-31y$\vdots$31
⇒6(x+7y)$\vdots$31
⇒x+7y$\vdots$31