Để `13n+50\vdotsn+2` thì `(13n+50)/(n+2)∈Z`
Ta có:
`(13n+50)/(n+2)=(13n+26+24)/(n+2)=(13(n+2)+24)/(n+2)=13+24/(n+2)`
Để `(13n+50)/(n+2)∈Z` thì `24\vdotsn+2`
`=>n+2∈Ư(24)`
`<=>n+2∈{-24; -12; -6; -4; -3; -2; -1; 1; 2; 3; 4; 6; 12; 24}`
`<=>n∈{-26; -14; -8; -6; -5; -4; -3; -1; 0; 1; 2; 4; 10; 22}`