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 $67^{24} \bmod 83=\left(67^{16} \bmod 83\right)\left(67^{8} \bmod 83\right) \bmod 83=(77)(36) \bmod 83=2272 \bmod 83 = 33$