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\(A=10^{1998}-4=100...000\) \(-4=999...996\)

Ta có: 999...996 : 3 = 333...332

999...996 : 9 =111...110\(\dfrac{2}{3}\)

Vậy A chia hết cho 3 mà không chia hết cho 9

\(A\in N\Leftrightarrow\dfrac{8n+193}{4n+3}\in N\Leftrightarrow8n+193⋮4n+3\)

Ta có : 8n+193 = 2(4n+3)+187

Để 8n+193 \(⋮\) 4n+3 thì 187 \(⋮\) 4n+3

\(\Rightarrow4n+3\inƯ\left(187\right)\)

\(\Rightarrow4n+3\in\left\{\pm1;\pm11;\pm17;\pm187\right\}\)

Ta có bảng sau :

Vì \(n\in N\) nên \(n\in\left\{2;46\right\}\)

Bài 1:

a.Ta có :

\(2^{225}=\left(2^3\right)^{75}=8^{75}\)

\(3^{150}=\left(3^2\right)^{75}=9^{75}\)

Vì \(8^{75}< 9^{75}\) nên \(2^{225}< 3^{150}\)

b. Ta có :

\(2^{91}=\left(2^{13}\right)^7=8192^7\)

\(5^{35}=\left(5^5\right)^7=3125^7\)

Vì \(8192^7>3125^7\) nên \(2^{91}>5^{35}\)

c. Ta có :

\(3^{4000}=\left(3^2\right)^{2000}=9^{2000}\)

Vì \(9^{2000}=9^{2000}\) nên \(3^{4000}=9^{2000}\)

Bài 2:

a. \(\dfrac{120^3}{30^3}=\dfrac{\left(30.4\right)^3}{30^3}=\dfrac{30^3.4^3}{30^3}=4^3=64\)

b. \(\dfrac{45^{10}.5^{20}}{75^{15}}=\dfrac{\left(5.3^2\right)^{10}.5^{20}}{\left(3.5^2\right)^{15}}=\dfrac{5^{10}.3^{20}.5^{20}}{3^{15}.5^{30}}=\dfrac{5^{30}.3^{20}}{3^{15}.5^{30}}=3^5=243\)

c. \(\dfrac{390^4}{130^4}=\dfrac{\left(130.3\right)^4}{130^4}=\dfrac{130^4.3^4}{130^4}=3^4=81\)