Lời giải:

\((0,125)^{100}.8^{102}=(\frac{1}{8})^{100}.8^{102}=\frac{8^{102}}{8^{100}}=8^{102-100}=8^2=64\)

64 nha

64 nhé

a) S = 2 + 2² + 2³ + ... + 2¹⁰⁰

ta có: (2+2²) + (2³+2⁴)+...+(2⁹⁹+2¹⁰⁰)

          chc 3  + chc 3 +....+  chc 3

=> S chia hết cho 3

b) S = 2 + 2² + 2³ + ... + 2¹⁰⁰

ta có: (2 + 2² + 2³ + 2⁴) + .... + (2^{97 +} 2⁹⁸ + 2⁹⁹ + 2¹⁰⁰)

                chc 15          +.......+    chc 15

=> S chia hết cho 15

chc nghĩa là chia hết cho nhak

Lời giải:
a)

Ta có:

\(1991\equiv 1\pmod {10}\Rightarrow 1991^{1997}\equiv 1^{1997}\equiv 1\pmod {10}(1)\)

\(1997\equiv 7\pmod {10}\Rightarrow 1997^{1996}\equiv 7^{1996}\pmod {10}(2)\)

Mà \(7^2\equiv -1\pmod {10}\Rightarrow 7^{1996}\equiv (-1)^{998}\equiv 1\pmod {10}(3)\)

Từ \((1);(2);(3)\Rightarrow 1991^{1997}-1997^{1996}\equiv 1-1\equiv 0\pmod {10}\) (đpcm)

b)

\(2^9+2^{99}=2^9(1+2^{90})\)

Ta thấy $2^{10}=1024\equiv -1\pmod {25}$
$\Rightarrow 2^{90}\equiv (-1)^9\equiv -1\pmod {25}$

$\Rightarrow 1+2^{90}\equiv 0\pmod {25}$ hay $1+2^{90}\vdots 25$

Mà $2^9\vdots 4$

Do đó:

$2^9+2^{99}=2^9(1+2^{90})\vdots 100$ (đpcm)

\(\text{a) }a+b\text{ chia hết cho 3}\)

\(\Rightarrow a^3+b^3=\left(a+b\right)\left(a^2-ab+b^2\right)\) chia hết cho 3

CMR 11¹⁰⁰ -1 chia hết cho 1000