Cho A= \(2^{15}+2^{17}+2^{19}\) . CMR : A chia hếch cho 21
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A = 2^15.(1+2^2+2^4) = 2^15.21 chia hết cho 21
=> ĐPCM
k mk nha
\(A=2^{15}+2^{17}+2^{19}\)
\(=2^{15}+2^{15}.2^2+2^{15}.2^4\)
\(=2^{15}\left(1+4+16\right)\)
\(=2^{15}.21⋮21\)
\(\Rightarrow A⋮21\)
Vậy \(A⋮21\)
a) Ta có: \(8^5+2^{11}\)
\(=\left(2^3\right)^5+2^{11}\)
\(=2^{15}+2^{11}\)
\(=2^{11}\left(2^4+1\right)\)
\(=2^{11}.17⋮17\left(đpcm\right)\)
Mình làm câu b)
\(A=2+2^2+2^3+..+2^{20}\)
\(A=\left(2+2^2\right)+\left(2^3+2^4\right)+...+\left(2^{19}+2^{20}\right)\)
\(=2\left(1+2\right)+2^3\left(1+2\right)+...+2^{19}\left(1+2\right)\)
\(=2.3+2^3.3+...+2^{19}.3=3\left(2+2^3+...+2^{19}\right)⋮3^{\left(đpcm\right)}\)
b,
A=(2+2^2)+(2^3+2^4)+...+2^20
=6+2^3.6+2^4.6+...2^19.6
=6.(1+2^3+2^4+...+2^19)÷3
Vì 6÷3 nên A÷3 (đmcm)
\(A=2^{15}+2^{17}+2^{19}=2^{14}\left(2+2^3+2^5\right)=2^{14}.42⋮21\)
\(A=2^{15}+2^{17}+2^{19}\)
\(=2^{15}\left(1+2^2+2^4\right)\)
\(=2^{15}.\left(1+4+16\right)\)
\(=2^{15}.21⋮21\)