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`36^{4}.16^{2}`
`=6^{8}.2^{8}`
`=(6.2)^8`
`=12^8`
`4^{6}.27^{4}`
`=2^{12}.3^{12}`
`=(2.3)^12`
`=6^12`
2 câu này không rõ đề
\(36^4\cdot16^2=1296^2\cdot16^2=429981696\)
\(4^6\cdot27^4=2^{12}\cdot3^{12}=2176782336\)
\(24^{54}.54^{24}.2^{10}\\ =8^{54}.3^{54}.27^{54}.2^{54}.2^{10}\)
\(=2^{162}.3^{54}.3^{72}.2^{54}.2^{10}\\
=2^{226}.3^{126}\\
=2^{3.63+37}.3^{2.63}\\
=8^{63}.9^{63}.2^{37}\\
=72^{63}.2^{37}\)
Dễ thấy \(72^{63}.2^{37}⋮̸72^{63}\)
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)
\(24^{54}.54^{24}.2^{10}=3^{54}.2^{162}.2^{24}.3^{72}.2^{10}=3^{126}.2^{196}\)
ta có: \(72^{63}=9^{63}.8^{63}=\left(3^2\right)^{63}.\left(2^3\right)^{63}=3^{72}.2^{108}\)
ta có: \(\frac{3^{126}.2^{196}}{3^{72}.2^{108}}=3^{54}.2^{88}\)
suy ra \(3^{126}.2^{196}\) chia hết cho \(3^{72}.2^{108}\)
suy ra \(24^{54}.54^{24}.2^{10}\) chia hết cho \(72^{63}\)