* Chứng minh \(4^a+a+b\equiv0\left(mod2\right)\)
Ta có:
\(a+1+b+2007=a+b+2008\equiv a+b\equiv0\left(mod2\right)\)
\(\Rightarrow4^a+a+b\equiv0\left(mod2\right)\)
* Chứng minh \(4^a+a+b\equiv0\left(mod3\right)\)
Ta có:
\(a+1+b+2007=a+b+2008\equiv1+a+b\equiv0\left(mod3\right)\)
\(\Rightarrow a+b\equiv2\left(mod3\right)\)
\(\Rightarrow4^a+a+b\equiv1+a+b\equiv1+2\equiv0\left(mod3\right)\)
Vì 2, 3 nguyên tố cùng nhau nên \(4^a+a+b\equiv0\left(mod6\right)\)
bài này không đúng với \(a=5\) bn à
\(a+1⋮6\Rightarrow a+1-6⋮6\Rightarrow a-5⋮6\Rightarrow a=6m+5\left(m\in N\right)\)
\(b+2007⋮6\Rightarrow b+2007-2010⋮6\Rightarrow b-3⋮6\Rightarrow b=6n+3\left(n\in N\right)\)
Do đó 4a + a + b chia hết cho 2 (1)
Ta có:
\(4^a+a+b\)
\(=4^{6m+5}+6m+3+6n+5\)
\(=4^{6m+5}+6\left(m+n\right)+8\)
Do 4 có dạng 3k + 1 nên 46m + 5 = 3h + 1. Do đó:
\(4^a+a+b=3h+1+6\left(m+n\right)+8=3\left(h+2n+2m+3\right)⋮3\left(đpcm\right)\)