Using an exponential cipher system, encipher the word HALT. where $p = 29, k = 11$, and $m = 1$. Using an exponential cipher system, encipher the word HALT. where $p = 29, k = 11$, and $m = 1$.
The progress I have made so far:
H A L T
$07, 00,11,19$
Since, $m =1$, we break this up into $2*m$ pieces, giving us $2$.
$0700$ and $1119$
From here we have to calculate the C value for each which is congruent to $p^k \pmod p$
The two congruences I get are:
a.) $0700^{11} \pmod 29$
b.) $1119^{11} \pmod 29$


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*After this I began to get a bit lost, my professor used repeated squares but I can't quite get it. Any help is greatly appreciated. 


From what I make of the question, this might be helpful for you...
Note that $11 = 2^3 + 2^1 + 2^0$ re-writing the exponent as the sum of powers of $2$.
Then the modular equation becomes:
$700^{11} \equiv 700^{2^3 + 2^1 + 2^0} \equiv 700^{2^{3}} \cdot 700^{2^1} \cdot 700^{2^0} $mod $29$
Now proceed with Square and Reduce method (I'll show you the first couple of steps):
$700^{2^0} \equiv 700 \equiv 4$ mod $29$
$700^{2^1} \equiv 4^{2} \equiv 16$ mod $29$ 
$700^{2^2} \equiv 16^{2} \equiv 24$ mod $29$
Say $700^{2^3} \equiv x$ mod $29$
Then $700^{13} \equiv x \cdot 16 \cdot 700$ mod $29$
Hope this helps
