[tex]898 = 2\cdot(449)[/tex]
[tex]2^{101} +898^6 = 2^6\cdot\left(2^{95}+449^6\right)[/tex]
[tex]2^3 \equiv 1 \mod(7) \\~\\2^{93} \equiv 1 \mod(7) \\~\\2^{95} \equiv 4 \mod(7)[/tex]
Pelo pequeno teorema de Fermat:
[tex]449^{7-1} \equiv 1 \mod(7) \\~\\449^{6} \equiv 1 \mod(7)[/tex]
Temos que:
[tex]2^6\cdot\left(2^{95}+449^6\right) \equiv \left( 2^3\right)^2 \cdot\left(2^{95}+449^6\right) \mod(7) \\~\\2^6\cdot\left(2^{95}+449^6\right) \equiv 1^2 \cdot\left(2^{95}+449^6\right) \mod(7) \\~\\2^6\cdot\left(2^{95}+449^6\right) \equiv 2^{95}+449^6\right \mod(7) \\~\\2^6\cdot\left(2^{95}+449^6\right) \equiv 4+1\right \mod(7) \\~\\2^6\cdot\left(2^{95}+449^6\right) \equiv 5\right \mod(7)[/tex]
A divisão dá resto 5.
