M= 35+36+37+38+39+310.
CMR M chia hết cho 91.
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\(S=\left(1+3\right)+...+3^8\left(1+3\right)=4\left(1+...+3^8\right)⋮4\)
\(S=\left(1+3+3^2\right)+...+3^7\left(1+3+3^2\right)\)
\(=13\left(1+...+3^7\right)⋮13\)
\(S=1+3+3^2+3^3+3^4+3^5+3^6+3^7+3^8+3^9\)
\(S=\left(1+3\right)+\left(3^2+3^3\right)+\left(3^4+3^5\right)+\left(3^6+3^7\right)+\left(3^8+3^9\right)\)
\(S=4+3^2\left(1+3\right)+3^4\left(1+3\right)+3^6\left(1+3\right)+3^8\left(1+3\right)\)
\(S=4+3^2.4+3^4.4+3^6.4+3^8.4\)
\(S=4\left(3^2+3^4+3^6+3^8\right)\)
\(4⋮4\\ \Rightarrow4\left(3^2+3^4+3^6+3^8\right)⋮4\\ \Rightarrow S⋮4\)
\(S=3+3^2+3^3+3^4+3^5+3^6+3^7+3^8+3^9\\ =\left(3+3^2+3^3\right)+3^3.\left(3+3^2+3^3\right)+3^6.\left(3+3^2+3^3\right)\\ =39+3^3.39+3^6.39\\ =-39.\left(-1-3^3-3^6\right)⋮\left(-39\right)\)
S = 3 + 32 + 33 + 34 + 35 + 36 + 37 + 38 + 39
S = ( 3 + 32 + 33 ) +34 + 35 + 36 + 37 + 38 + 39
S = 39 + 34 + 35 + 36 + 37 + 38 + 39
Vì 39 ⋮ -39
<=> S ⋮ -39
\(S=1.\left(1+3\right)+3^2\left(1+3\right)+3^4\left(1+3\right)+...+3^8\left(1+3\right)\)
\(S=4x\left(1+3^2+...+3^8\right)\)
Vì 4 chia hết cho 4 nên S chia hết cho 4
Gọi A= 3638+4143
Để A chia hết cho 77 thì A phải chia hết cho 11 và 7
*Cm A chia hết cho 7
\(36\equiv1\left(mod7\right)\Rightarrow36^{38}\equiv1^{38}\left(mod7\right)\Leftrightarrow36^{38}\equiv1\left(mod7\right).\)
\(41\equiv-1\left(mód7\right)\Rightarrow41^{43}\equiv-1^{43}\left(mod7\right)\Leftrightarrow41^{43}\equiv-1\left(mod7\right)\)
=> 3638+4143 \(\equiv1+\left(-1\right)\left(mod7\right)\) <=> 3638+4143 \(\equiv\)0 ( mod 7 ) => 3638+4143 chia hết cho 7 (1)
*Cm A chia hết cho 11
\(36\equiv3\left(mod11\right)\Rightarrow36^{38}\equiv3^{38}\left(mod11\right)\)
\(41\equiv-3\left(mod7\right)\Rightarrow41^{43}\equiv-3^{43}\left(mod7\right)\) => -343 = -338.-35
=> 3638+4143 \(\equiv\)(-338+338 ).-35 ( mod 7 )
3638+4143 \(\equiv\) 0 (mod 7) 3638+4143 chia hết cho 11 (2)
Từ (1) và (2) suy ra 3638+4143 chia hết cho 77 => btđcm
3^5+3^6+...+3^10=3^5(1+3+3^2+...+3^5)=3^5.364 có 364 chia hết cho 91 nên M chia hết cho 91.