18.16: Criptografía
- Page ID
- 110424
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1. \(\mathrm{ZLU ~ KZB ~ WWS ~ PLZ}\)3. \(\mathrm{SHRED ~ EVIDENCE}\)
5. \(\mathrm{O2H ~ DO5 ~ HDV}\)7. \(\mathrm{MERGER ~ ON}\)
9. \(\mathrm{MNI ~ YNE ~ TBA ~ AEH ~ RTA ~ TEA ~ TAI ~ LRE ~ A}\)
11. \(\mathrm{THE ~ STASH ~ IS ~ HIDDEN ~ AT ~ MARVINS ~ QNS}\)
13. \(\mathrm{UEM ~ IYN ~ IOB ~ WYL ~ TTL ~ N}\)
15. \(\mathrm{HIRE ~ THIRTY ~ NEW ~ EMPLOYEES ~ MONDAY}\)
17. \(\mathrm{ZMW ~ NDG ~ CDA ~ YVK}\)
19. a)\(3\) b)\(0\) c)\(4\)
21. Probamos todo\(n\) del 1 al 10
\ (\ begin {array} {|r|r|r|}
\ hline\ mathrm {n} & 4^ {\ mathrm {n}} & 4^ {\ mathrm {n}}\ bmod 11
\\ hline 1 & 4 & 4\
\ hline 2 & 16 & 5\\
\ hline 3 & 64 & 9\
\\ hline 4 & 256 y 3\
\ hline 5 y 1024 & 1\\
\ hline 6 & 4096 & 4\\
\ hline 7 & 16384 & 5\\
\ hline 8 & 65536 & 9\\
\ hline 9 & 262144 & 3\\
\ hline 10 & 1048576 & 1\\
\ hline
\ end {array}\)
Ya que tenemos repeticiones, y no todos los valores del 1 al 10 se producen (por ejemplo, no hay\(\left.n \text { is } 4^{n} \bmod 11=7\right)\), 4 no es un generador\(\bmod 11\).
23. \(157^{10} \bmod 5=(157 \bmod 5)^{10} \bmod 5=2^{10} \bmod 5=1024 \bmod 5=4\)
25. \(3^{7} \bmod 23=2\)
27. Bob enviaría\(5^{7}\) mod\(33=14\). Alice lo descifraría como\(14^{3} \bmod 33=5\)
31.
a.\(67^{8} \bmod 83=\left(67^{4} \bmod 83\right)^{2} \bmod 83=49^{2} \bmod 83=2401 \bmod 83=77\)
\(67^{16} \bmod 83=\left(67^{8} \bmod 83\right)^{2} \bmod 83=77^{2} \bmod 83=5929 \bmod 83=36\)
b.\(17000 \bmod 83=(100 \bmod 83)^{*}(170 \bmod 83) \bmod 83=(17)(4) \bmod 83=68\)
c.\(67^{5} \bmod 83=\left(67^{4} \bmod 83\right)(67 \bmod 83) \bmod 83=(49)(67) \bmod 83=3283 \bmod 83=46\)
d.\(67^{7} \bmod 83=\left(67^{4} \bmod 83\right)\left(67^{2} \bmod 83\right)(67 \bmod 83) \bmod 83=(49)(7)(67) \bmod 83=22981 \bmod 83=73\)
e.\(67^{24}=67^{16} 67^{8}\) entonces\(67^{24} \bmod 83=\left(67^{16} \bmod 83\right)\left(67^{8} \bmod 83\right) \bmod 83=(77)(36) \bmod 83=2272 \bmod 83 = 33\)