a) Ta có: $m = 3 + 3^2 + 3^3 +... + 3^{99}+3^{100}$
Nhận xét: $3^5 \equiv 3 \pmod{5}$, do đó $3^{100} \equiv 3^4 \equiv 1 \pmod{5}$.
Suy ra: $m \equiv 3 + 3^2 + 3^3 +... + 3^{99}+1 \equiv \dfrac{3(3^{100}-1)}{2}+1 \equiv \dfrac{3-1}{2}+1 \equiv 2 \pmod{5}$.
Vậy m không chia hết cho 5.
Ta có: $m = 3 + 3^2 + 3^3 +... + 3^{99}+3^{100}$
Nhận xét: $3^2 \equiv 9 \equiv -3 \pmod{12}$, do đó $3^{100} \equiv 9^{50} \equiv (-3)^{50} \equiv 3^{50} \equiv 9^{25} \equiv (-3)^{25} \equiv -3^{25} \equiv -3^{24} \cdot 3 \equiv 9^{12} \cdot 3 \equiv (-3)^{12} \cdot 3 \equiv 3^{12} \cdot 3 \equiv 9^6 \cdot 3 \equiv (-3)^6 \cdot 3 \equiv 3^6 \cdot 3 \equiv 9^3 \cdot 3 \equiv (-3)^3 \cdot 3 \equiv -3^3 \cdot 3 \equiv -27 \cdot 3 \equiv -9 \pmod{12}$.
Suy ra: $m \equiv 3 + 3^2 + 3^3 +... + 3^{99}-9 \equiv \dfrac{3(3^{100}-1)}{2}-9 \equiv \dfrac{3-1}{2}-9 \equiv -8 \equiv 4 \pmod{12}$.
Vậy m chia hết cho 12.
b) Ta có: $2m + 3 = 2(3 + 3^2 + 3^3 +... + 3^{99}+3^{100}) + 3 = 3(3 + 3^2 + 3^3 +... + 3^{99}+3^{100}) + 3 = 3m + 3$.
Suy ra: $m = \dfrac{3^n-3}{3} = 3^{n-1}-1$.
Do đó: $2m + 3 = 2(3^{n-1}-1) + 3 = 2 \cdot 3^{n-1} + 1 = 3^n$.
Suy ra: $n = \log_3(2m+3)$.