Cho a, b ∈ N . Chứng tỏ rằng nếu 5a+3b và 13a+8b cùng chia hết cho 2018 thì a, b cũng chia hết cho 2018 . 07/12/2021 Bởi Isabelle Cho a, b ∈ N . Chứng tỏ rằng nếu 5a+3b và 13a+8b cùng chia hết cho 2018 thì a, b cũng chia hết cho 2018 .
@py Bài làm : `(5a+3b)⋮2018 ⇒8.(5a+3b)⋮2018` `(13a+8b)⋮2018 ⇒3.(13a+8b)⋮2018` `⇒8.(5a+3b)−3.(13a+8b)⋮2018` `⇒40a+24b−39a−24b⋮2018⇒a⋮2018` Nhưng `5a + 3b ⋮ 2018 ⇒ 3b ⋮ 2018` `⇒ b ⋮ 2018` Vậy `………………………..` Bình luận
$\text{(*)}$ $5a+3b$ $\vdots$ $2018$ $65a+39b$ $\vdots$ $2018$ $\text{(**)}$ $13a+8b$ $\vdots$ $2018$ $65a+40b$ $\vdots$ $2018$ $\text{Từ (*)(**)}$ $\text{=>} $b$ $\vdots$ $2018$ $\text{->}$ $3b$ $\vdots$ $2018$ $\text{->}$ $5b$ $\vdots$ $2018$ $ƯCLN(5,2018)=1$ $\text{=>}$ $a$ $\vdots$ $2018$ Bình luận
@py
Bài làm :
`(5a+3b)⋮2018 ⇒8.(5a+3b)⋮2018`
`(13a+8b)⋮2018 ⇒3.(13a+8b)⋮2018`
`⇒8.(5a+3b)−3.(13a+8b)⋮2018`
`⇒40a+24b−39a−24b⋮2018⇒a⋮2018`
Nhưng `5a + 3b ⋮ 2018 ⇒ 3b ⋮ 2018`
`⇒ b ⋮ 2018`
Vậy `………………………..`
$\text{(*)}$
$5a+3b$ $\vdots$ $2018$
$65a+39b$ $\vdots$ $2018$
$\text{(**)}$
$13a+8b$ $\vdots$ $2018$
$65a+40b$ $\vdots$ $2018$
$\text{Từ (*)(**)}$
$\text{=>} $b$ $\vdots$ $2018$
$\text{->}$ $3b$ $\vdots$ $2018$
$\text{->}$ $5b$ $\vdots$ $2018$
$ƯCLN(5,2018)=1$
$\text{=>}$ $a$ $\vdots$ $2018$