We know that:
e∗d=1(modϕ)⇒e∗dp=1(modp−1).

Guessing that e = 65537 (although we could iterate through to check all e if needed) we can recover p in the following way:
e∗dp=1+kp(p−1)

for some integer kp
. As e is fairly small, so will kp

. We can rearrange the above for:
p=e∗dp−1kp+1

with the only unknown being kp
.  The letter c should appear at least 21 times.
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