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NT
1
15 tháng 9 2018
Xet \(n=3k\)
\(\Rightarrow3^{6k}+3^{3k}+1\equiv3\left(mod13\right)\)
Xet \(n=3k+1\)
\(\Rightarrow3^{6k+2}+3^{3k+1}+1\equiv9+3+1\equiv0\left(mod13\right)\)
Xet \(n=3k+2\)
\(\Rightarrow3^{6k+3+1}+3^{3k+2}+1\equiv3+9+1\equiv0\left(mod13\right)\)
Vậy vơi mọi n tự nhiên và n không chia hêt cho 3 thì
\(3^{2n}+3^n+1⋮13\)
VT
1
21 tháng 8 2018
3^2n+3^n=9^n+3^n⋮12 đồng dư 12 mod 13
⇒3^2n+3^n+1⋮13⇒3^2n+3^n+1⋮13
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
8 tháng 12 2023
Bài 1:
cho a2 + b2 ⋮ 3 cm: a ⋮ 3; b ⋮ 3
Giả sử a và b đồng thời đều không chia hết cho 3
Vì a không chia hết cho 3 nên ⇒ a2 : 3 dư 1
vì b không chia hết cho b nên ⇒ b2 : 3 dư 1
⇒ a2 + b2 chia 3 dư 2 (trái với đề bài)
Vậy a; b không thể đồng thời không chia hết cho ba
Giả sử a ⋮ 3; b không chia hết cho 3
a ⋮ 3 ⇒ a 2 ⋮ 3
Mà a2 + b2 ⋮ 3 ⇒ b2 ⋮ 3 ⇒ b ⋮ 3 (trái giả thiết)
Tương tự b chia hết cho 3 mà a không chia hết cho 3 cũng không thể xảy ra
Từ những lập luận trên ta có:
a2 + b2 ⋮ 3 thì a; b đồng thời chia hết cho 3 (đpcm)
NT
4
TT
Thầy Tùng Dương
VIP
26 tháng 9 2018
Nhận thấy A = 3n + 4n +1 chia hết cho 2 với mọi n tự nhiên, để A chia hết cho 10 ta cần A chia hết cho 5 là đủ.
Nhận xét: 34 \(\equiv\)1 (mod 5), ta sẽ xét các trường hợp: n = 4k, n = 4k+1, n = 4k+2, n = 4k+3 với k là số tự nhiên.
TH1: n = 4k.
A = 34k + 4.(4k) + 1 = 81k + 16k +1 \(\equiv\)1 + k + 1 \(\equiv\)2+k (mod 5)
Để A chia hết cho 5 thì k phải có dạng 5h + 3, với h là số tự nhiên. Vậy n = 4.(5h+3) = 20h +12 thì A chia hết cho 10.
Tương tự với các trường hợp sau bạn giải tiếp nhé!
4 tháng 1 2020
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