Verified answer

a)

[tex]a + 8b \equiv r \pmod{47}\\100(a + 8b)\equiv 100r \pmod{47}\\100a + 100 \cdot 8b \equiv 100r \pmod{47}\\100a + (47 \cdot 2 + 6) 8b \equiv (47 \cdot 2 + 6) r \pmod{47}\\100a + 47 \cdot 2 \cdot 8b + 6 \cdot 8b \equiv 47 \cdot 2r + 6r\pmod{47}\\100a + 48b \equiv 6r\pmod{47}\\100a + (47 + 1)b\equiv 6r\pmod{47}\\100a + 47b + b \equiv 6r\pmod{47}\\100a + b \equiv 6r\pmod{47}[/tex]

b)

[tex]2750422 = 100a + b\\2750422 = 100 \cdot 27504 + 22\\a = 27504\\b = 22[/tex]

Então:

[tex]100a + b \equiv 6(a + 8b) \pmod {47}\\2750422 \equiv 6(27504 + 8 \cdot 22) \pmod {47}\\2750422 \equiv 6(27680) \pmod {47}[/tex]

27680 também pode ser reescrito:

[tex]27680 = 100a + b\\27680 = 100 \cdot 276 + 80\\a = 276\\b = 80[/tex]

Logo:

[tex]100a + b \equiv 6(a + 8b) \pmod {47}\\27680\equiv 6(276+ 8 \cdot 80) \pmod {47}\\27680\equiv 6(916) \pmod {47}[/tex]

916 também pode ser reescrito...

[tex]916 = 100 \cdot 9 + 16\\a = 9\\b = 16\\\\916 \equiv 6(9 + 8 \cdot 16) \pmod {47}\\916\equiv 6(137) \pmod {47}\\916\equiv 6(43) \pmod {47}\\916 \equiv 3 \cdot 86 \pmod {47}\\916 \equiv 3 \cdot 39 \pmod {47}\\916 \equiv 117 \pmod {47}\\916 \equiv 23 \pmod {47}[/tex]

Retornando:

[tex]27680\equiv 6(916) \pmod {47}\\27680\equiv 6 \cdot 23 \pmod {47}\\27680\equiv 2 \cdot 69 \pmod {47}\\27680\equiv 2 \cdot 22 \pmod {47}\\27680\equiv 44 \pmod {47}[/tex]

[tex]2750422 \equiv 6(27680) \pmod {47}\\2750422 \equiv 6 \cdot 44 \pmod {47}\\2750422 \equiv 3 \cdot 88\pmod {47}\\2750422 \equiv 3 \cdot 41 \pmod {47}\\2750422 \equiv 123 \pmod {47}\\2750422 \equiv 29 \pmod {47}[/tex]

Portanto, 2750422 é congrente a 29, módulo 47.