Question

[HAU TRAN]

Tìm các số tự nhiên a,b sao cho (2008.a+3.b+1).(2008^a+2008.a+b) = 225

Answer 1

[Phùng Hữu Kiên]

Đặt \(A=\left(2008a+3b+1\right)\left(2008^a+2008a+b\right)\)

Nếu \(a\ge1\Rightarrow A\ge\left(2008.1+3b+1\right)\left(2008^1+2008.1+b\right)\)

\(\Rightarrow A\ge\left(2009+3b\right)\left(4016+b\right)>225\)

Vậy \(a=0\Rightarrow A=\left(3b+1\right)\left(1+b\right)=225\)

\(3b+1\)chia 3 dư 1 \(\Rightarrow3b+1=25\)

\(\Rightarrow3b=24\Rightarrow b=8\)

Vậy \(\left(a;b\right)\in\left\{\left(0;8\right)\right\}\)

Answer 2

[co gai cau kinh]

doi ti