Respostas
respondido por:
2
Usando o pequeno teorema de Fermat
##########################
p:primo
a:inteiro
a^p =a (mod p) (i)
a^(p-1)=1 (mod p) (ii)
##########################
5^60 /26 ...queremos o resto
*******63= 2*25+10
resto = (5^2)^25 * 5^10 (mod 26)
resto = (5^2)^25 * 5^10 (mod 26)
**usando (ii) resto = (5^2)^25 (mod 26) =5²
resto = 5² * 5^10 (mod 26)
******Para 5²/26 ==>resto = 5² (mod 26)=-1 ...25-26=-1
resto = (5²) * 5^10 (mod 26)
resto = (5²) * (5²)^5 (mod 26)
resto = (-1) * (-1) (mod 26) = 1
##########################
p:primo
a:inteiro
a^p =a (mod p) (i)
a^(p-1)=1 (mod p) (ii)
##########################
5^60 /26 ...queremos o resto
*******63= 2*25+10
resto = (5^2)^25 * 5^10 (mod 26)
resto = (5^2)^25 * 5^10 (mod 26)
**usando (ii) resto = (5^2)^25 (mod 26) =5²
resto = 5² * 5^10 (mod 26)
******Para 5²/26 ==>resto = 5² (mod 26)=-1 ...25-26=-1
resto = (5²) * 5^10 (mod 26)
resto = (5²) * (5²)^5 (mod 26)
resto = (-1) * (-1) (mod 26) = 1
respondido por:
1
Resposta:
1
Explicação passo-a-passo:
5^2 ≡ (- 1) mod(26)
(5^2)^30 ≡ (- 1)^30 mod(26)
5^(2 . 30) ≡ 1 mod(26)
5^60 ≡ 1 mod(26)
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