Ta có
`A= ( 4x +3y )` $\vdots$ `13`
⇔ `5( 4x +3y )` $\vdots$ `13`
Do `13` $\vdots$ `13` , `x` , `y` $\in$ `Z` nên `13( x + y )` $\vdots$ `13`
Xét hiệu
`5( 4x +3y ) - 13( x + y )`
= `20x+15y−13x−13y`
= `7x + 2y`
Mà `5( 4x +3y ) - 13( x + y )` $\vdots$ `13` ( do `5( 4x +3y )` $\vdots$ `13` , `13( x + y )` $\vdots$ `13`)
⇔`7x + 2y` $\vdots$ `13`
Vậy `A` = `(4x +3y )` $\vdots$ `13` thì `B` =`7x + 2y` $\vdots$ `13`