We should have some homework. I can see how I am doing and you can show me how clever you are! 
The number is the difference between two numbers that are equivalent
mod 12345.
I chose this path because of your lesson in class before. As for finding it, I had MAPLE search through all numbers consisting entirely of 1's (up to 12346 digits because of the pigeonhole principle) for one that had a remainder of 1 when divided by 12345. It returned a number that consited of 6577 1's. Subtracting 1 from this number yields my number. Since MAPLEAnswer=1 mod 12345 and 1=1 mod 12345, MAPLEAnswer1=11 mod 12345 = 0 mod 12345. Thus, the resultant number is a multiple of 12345. MAPLE confirms this both by returning an integer for the division and 0 for the mod. 
Answers to the NvJ
question
People answering the first question included
Emily Carey,
Andy Hurwich,
Mackenzie Esch,
the team of
Olivia Nnadi and Nancy Twu and Rachel Laskin,
Ryan Measel, and
Sudipta Bandyopadhyay. Slightly less satisfactory answers were
received from the team of Joseph Gregg and Daniel Wojcik, the team of
Dana Andre and Mike Cobby, and
Alfred Chi.
A sum
Try to explain why the sum of n^{2}/2^{n} as n
runs from 1 on up is 6. I think this can be done with reasoning
similar to (but not identical with!) what I did in class for
n/2^{n}.
Answers to the sum
question
This is a more subtle question. Such tools as Maple
and the TI89 both will return the answer 6 when the above sum is
requested. However, I was careful both in class and here to phrase my
request as "explain why ..." and not just as "What is the sum?"
There are various ways to explain why, including some using techniques
from calculus. The team of Eugene Shvarts and Moulin Chokshi gave an
excellent explanation using a method similar to what I used when I
analyzed the sum of n/2^{n}. Also Jonathan Lin gave a very
nice answer. Please note that (amazingly!) the sum of
n^{k}/2^{n} is an integer for every positive
integer k. I don't at all believe this is immediately obvious! You can
find out much
more about these numbers.
Generally, the OnLine
Encyclopedia of Integer Sequences is an amazing resource for
looking up and investigating integer sequences.
The word "animal" is used with some
generality. Last year, a bicycle was one of the animals. When asked if a bicycle was really an animal, I replied, "Well, it has two wheels, doesn't it?" 
Emily uses RSA. Her public key includes the following:

Alice and Bob decide to use the DiffieHellman protocol to
exchange information. This is what happens:
The message is the sum mod their prime number of the secret and the name of an "animal" using the method described below. What is the animal? 
Fred uses RSA. His public key includes the following:

Remember, in cryptanalysis, you can make guesses, etc. Any "tricks" are legitimate.
Entries in the wonderful ANIMAL
contest
Sudipta Bandyopadhyay entered and correctly identified Emily's
animal as the radish. The team of Mustafa Engin and Matt
Donahoe also correctly identified Emily's animal.
The fourperson team of Daniel DeMarco, Eugene Shvartz, Jonathan Lin, and Parshant Mittal identified both Emily's animal and the fierce animal of Alice and Bob, the celery.
Finally, the team of Joseph Gregg and Daniel Wojcik identified all three of the animals (even Fred's chicken!). I congratulate them. All of their answers arrived first. Please note that a direct factoring attack on Fred's N via Maple is not practical. One must be a bit devious and look at the number. Consider that it might be a product of two numbers of the form 10^{n}+m, and then factor 11773 (it is 61·193) and try to find the "correct" powers.
Very appropriate awards were given to all participants.
The secret, although a number, also represents an English word. I have made a word using a simple process which the following table may help explain:
A  B  C  D  E  F  G  H  I  J  K  L  M  N  O  P  Q  R  S  T  U  V  W  X  Y  Z 
01  02  03  04  05  06  07  08  09  10  11  12  13  14  15  16  17  18  19  20  21  22  23  24  25  26 
The English word SMILE becomes 19 13 09 12 05 which is more usually written 1913091205. This isn't a very convenient way of writing words, but it will be o.k. for what we do. Notice that you should always pair the digits from right to left, so that the number 116161205 will become 01 16 16 12 05 which is APPLE. The "interior" 0's will be clear but we need to look for a possible initial 0.
How to win this contest and get a valuable
prize
Tell me the secret word. Also identify all members of the group who have contributed to your
solution of this problem. I would prefer
email because the earliest solution wins. Please
give me a short description of the process you used.
Comment I would use Maple of course, and have
Maple create and remember certain polynomials for me and do
all of the arithmetic.
Written 7/1/2003
Entries in the TOAD contest
The Grand Prize winners are the team consisting of
Alfred Chi,
Michael Cobby,
Robert Hohenstein,
Jason Tchuo, and
Jeff Wen who submitted the answer
2345x^{4}+3456x^{3}+4567x^{2}+5678x+20150104
at 19:07:18.
The constant term, 20150104, "translates" to TOAD.
There are several Less Grand Prize winners. The six
member team of Ryan Measel,
Eric Hu,
Dipen Patel,
Landon Glover,
Mike Cole, and
Josh Verdeck submitted their answer at 21:04:13 with the best
exposition of any entry.
At 21:04:58 a team of valiant losers composed of
Emily Carey,
Ashley Freeman,
Elizabeth Strauss,
Katherine Wu, and
Debra Perrone sent a message.
Other
Less Grand Prize winners were the team (22:28:51) of
Lucy Xu,
Andreia Da Costa,
Myrtha Glaude,
Caitlin Costaney, and
Amy Q. Lin and the team
(23:19:04) of
Jonathan Lin,
Eugene Shvarts,
Parshant Mittal,
Daniel De Marco, and
Frederick Hardenbrook.
The
Weirdness Prize was won with the entry
submitted at 00:40:01 by the team of
Sarah Davies,
Daniel[le] Jones,
Angelica Harris, and
Nancy Girgis (yes, a team of only four!) whose answer was
7.552162243·10^{14}. Sigh.
Written 7/3/2003
Added 7/7/2003 Email was not perfect. Another team submitted an entry on Wed 7/2/2003 6:00 PM and hence should be considered the winners. Therefore I will today award a GrandER Prize to the team of Joseph Gregg, Chris Richards, Dan Wojcik, Ilia Izmolov, and Margeret Scholtz. I congratulate them!