18.16: Cryptography
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. We test out all \(n\) from 1 to 10
\(\begin{array}{|r|r|r|}
\hline \mathrm{n} & 4^{\mathrm{n}} & 4^{\mathrm{n}} \bmod 11 \\
\hline 1 & 4 & 4 \\
\hline 2 & 16 & 5 \\
\hline 3 & 64 & 9 \\
\hline 4 & 256 & 3 \\
\hline 5 & 1024 & 1 \\
\hline 6 & 4096 & 4 \\
\hline 7 & 16384 & 5 \\
\hline 8 & 65536 & 9 \\
\hline 9 & 262144 & 3 \\
\hline 10 & 1048576 & 1 \\
\hline
\end{array}\)
Since we have repeats, and not all values from 1 to 10 are produced (for example, there is no \(\left.n \text { is } 4^{n} \bmod 11=7\right)\), 4 is not a generator \(\bmod 11\).
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 would send \(5^{7}\) mod \(33=14\). Alice would decrypt it as \(14^{3} \bmod 33=5\)
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. \(67^{24}=67^{16} 67^{8}\) so \(67^{24} \bmod 83=\left(67^{16} \bmod 83\right)\left(67^{8} \bmod 83\right) \bmod 83=(77)(36) \bmod 83=2272 \bmod 83 = 33\)