Toán Lớp 6: tìm số tự nhiên n sao cho 4n+1 chia hết n-3
Leave a reply
Register Now
Login
Lost Password
Lost your password? Please enter your email address. You will receive a link and will create a new password via email.
Toán Lớp 6: tìm số tự nhiên n sao cho 4n+1 chia hết n-3
Comments ( 2 )
4n + 1 \vdots n – 3
n – 3 \vdots n – 3
=> [(4n + 1) – 4(n – 3)] \vdots n – 3
=> [(4n + 1) – (4n – 12)] \vdots n – 3
=> 13 \vdots n – 3
=> n – 3 in Ư(13) = {1; -1; 13; -13}
=> n – 3 in {1; -1; 13; -13}
=> n in {4; 2; 16; -10}
Vậy n in {4; 2; 16; -10}
4n + 1 ⋮ n – 3
Ta có : $\left \{ {{4n + 1 ⋮ n – 3} \atop {n – 3 ⋮ n – 3}} \right.$ ⇒ $\left \{ {{4n + 1 ⋮ n – 3} \atop {4n – 12 ⋮ n – 3}} \right.$
⇒ ( 4n + 1 ) – ( 4n – 12 ) ⋮ n – 3 ⇔ 13 ⋮ n – 3 ⇔ n – 3 ∈ Ư ( 13 ) = { – 1 ; 1 ; -13 ; 13 } ⇔ n ∈ { 2 ; 4 ; 16 ; -10 }
Vậy với n ∈ { 2 ; 4 ; 16 } thì 4n + 1 ⋮ n – 3