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NT
0
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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NN
2
31 tháng 8 2015
Nếu \(n=0\to n^{1997}+n^{1975}+1=1\) không phải là số nguyên tố.
Xét \(n\) là số nguyên dương. Ta có \(n^{1997}-n^2=n^2\left(n^{3\times665}-1\right)\vdots\left(n^3\right)^{665}-1\vdots n^3-1\vdots n^2+n+1.\)
Suy ra \(n^{1997}-n^2\vdots n^2+n+1.\)
Tương tự, \(n^{1975}-n=n\left(n^{3\times658}-1\right)\vdots\left(n^3\right)^{658}-1\vdots n^3-1\vdots n^2+n+1.\)
Từ đó ta suy ra \(n^{1997}+n^{1975}+1=\left(n^{1997}-n^2\right)+\left(n^{1975}-n\right)+\left(n^2+n+1\right)\vdots n^2+n+1.\)
Vì \(n^{1997}+n^{1975}+1\) là số nguyên tố (chỉ có hai ước dương là 1 và chính nó) và \(n^2+n+1>1\), nên \(n^{1997}+n^{1975}+1=n^2+n+1.\) Suy ra \(\left(n^{1997}-n^2\right)+\left(n^{1975}-n\right)=0.\) Do \(n\)là số nguyên dương nên \(\left(n^{1997}-n^2\right)\ge0,\left(n^{1975}-n\right)\ge0.\) Vậy \(n=1.\)
Thử lại với \(n=1\to n^{1997}+n^{1975}+1=3\) là số nguyên tố.
Đáp số \(n=1.\)
NC
1
31 tháng 1 2021
Xét n=0n=0 không thỏa mãn.
Xét n≥1n≥1
Với n∈Nn∈N thì:A=n4+2n3+2n2+n+7=(n2+n)2+n2+n+7>(n2+n)2A=n4+2n3+2n2+n+7=(n2+n)2+n2+n+7>(n2+n)2
Mặt khác, xét :
A−(n2+n+2)2=−3n2−3n+3<0A−(n2+n+2)2=−3n2−3n+3<0 với mọi n≥1n≥1
⇔A<(n2+n+2)2⇔A<(n2+n+2)2
Như vậy (n2+n)2<A<(n2+n+2)2(n2+n)2<A<(n2+n+2)2, suy ra để $A$ là số chính phương thì
A=(n2+n+1)2⇔n4+2n3+2n2+n+7=(n2+n+1)2A=(n2+n+1)2⇔n4+2n3+2n2+n+7=(n2+n+1)2
⇔−n2−n+6=0⇔(n−2)(n+3)=0⇔−n2−n+6=0⇔(n−2)(n+3)=0
Suy ra n=2
TT
0
NT
0
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