Ta có : M=2+22+23+...+2107+2108
=(2+23+25)+(22+24+26)+...+(2104+2106+2108)
=2(1+22+24)+22(1+22+24)+...+2104(1+22+24)
=2.21+22.21+...+2104.21 chia hết cho 21
Vậy M chia hết cho 21.
Ta có : M = 2 + 22 + 23 + 24 .... + 2107 + 2108
Ta có:\(M=\)\(2+2^2+2^3+...+2^{107}+2^{108}\)
\(\Rightarrow M\)\(=\left(2+2^2+2^3+2^4\right)+\left(2^5+2^6+2^7+2^8\right)+...+\left(2^{105}+2^{106}+2^{107}+2^{108}\right)\)
\(\Rightarrow M\)\(=2\left(1+2+2^2\right)+2^5\left(1+2+2^2\right)+...+2^{105}\left(1+2+2^2\right)\)
\(\Rightarrow M\)\(=2.7+2^5.7+...+2^{105}.7\)
\(\Rightarrow M\)\(=7\left(2+2^5+...+2^{105}\right)\)
Vì \(21⋮21\Rightarrow21\left(1+2^2+2^3+...+2^{105}\right)\)
\(\Rightarrow M⋮21\)
M=2+22+23+.....+2107+2108
M=(2+22+23+24+25+26)+......+(2103+2104+2105+2106+2107+2108)
M=(2+22+23+24+25+26)+......+2102.(2+22+23+24+25+26)
M=126+......+2102.126
M=126.(1+......+2102)
M=21.6.(1+....+2102)\(⋮\)21
Vậy M\(⋮\)21
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