Sat May 22 10:42:30 2004 : Will Zero Knowledges proofs be covered? What about SHA-1 and El Gamal?

Sat May 22 21:25:33 2004 : Is this class graded on a curve?

Sun May 23 09:40:47 2004 : When will the syllabus be up?

Sun May 23 10:56:14 2004 : John: Thank you for your questions! (1) I will probably cover El Gamal (and the discrete logarithm problem), but SHA-1 and Zero Knowledge proofs will probably not make the cut. That doesn't mean that you can't read about them yourself! :) (2) The course will definitely *not* be graded on a curve. (3) And I should have the syllabus up within a week or two--I'm still hashing out some details. Cheers!

Sun May 23 12:07:17 2004 : Will Tex files be provided or will you help us for those of us wanting to learn LaTeX?

Mon May 24 09:48:56 2004 : John: :) This is not a class in LaTeX, and certainly no one will be expected to type up their homework... but those files that I distribute I will also give out in TeX for your edification. Cheers!

Tue May 25 11:05:34 2004 : Is the textbook for this class finalized? That is the probablity that this text won't be used is about 0.000000000000000000000000000000001?

Tue May 25 11:56:18 2004 : John: Yes, the textbook is finalized, and the probability is a very small epsilon > 0 that the textbook won't be used. :) I plan to follow the text generally while supplementing it as much as I can with relevant and fun material.

Tue May 25 12:48:19 2004 : hello; when I try to look at the syllabus, I get error 'file not found'

Tue May 25 13:30:46 2004 : The syllabus isn't up becasue John hasn't put it up! We must really be so bored and can't wait for math 115 to begin!

Fri May 28 11:57:55 2004 : John: I just put up the syllabus. I'm still working out the homework assignments. :)

Fri May 28 12:27:59 2004 : On the syllabus for week 3, it says Fermat's Last Theorem. It should be Fermat's Little Theorem, as I don't think we could cover the entire proof of Fermat's Last Theorem in one lecture. What is the grading scale for this class? Will it be determined later, or is an A 85% an above, etc.., as some Math profs have done here? Finally, is a 50 minute midterm enough time to test us on the material? It seems to be more of a speed test than a test of our understanding.

Fri May 28 16:29:28 2004 : John: Thanks for your comments! (1) I fixed the typo in the syllabus. (2) I reserve the right to adjust the grading scale for the class to my liking. It is usually only possible in retrospect to know whether or not an exam was a good one or not. This is an upper-division math class and a grading scale based on rigid percentages seems out of place. (3) The material covered in Weeks 1 through 3 is relatively basic, therefore I expect that the exam will be very doable in 50 minutes. This can always be changed, of course, since the class is 2 hours long. :)

Thu Jun 3 08:15:10 2004 : How did John get that Cal icon on the browser's web address on his home page?

Thu Jun 3 08:58:17 2004 : John: I don't know. Which icon? It doesn't appear on my browser...

Sat Jun 5 11:38:39 2004 : Ah, I figured out it was favicon.ico For example and I see this also on John's main page

Tue Jun 8 18:49:50 2004 : Will we possibly have time to cover section 4.5, 10.1, 10.2, and anything from chapter 12?

Wed Jun 9 11:11:57 2004 : John: We will definitely say a word about ElGamal (when we talk about orders of elements, we will talk about discrete logs); right now, I have chosen to talk about Diophantine equations rather than continued fractions, because that's closer to my heart; and as for systems of linear congruences, I'll check to see how comfortable the class is with linear algebra and if so, hopefully we will have a chance to talk about that, too. But better to cover some things well rather than too many things not-so-well! :)

Sat Jun 19 09:22:24 2004 : On hw #1, section 1.1, problem 4c, do we consider the case where we are multiplying 0 with an irrational number?

Sat Jun 19 12:05:21 2004 : On hw #1, problem 1.1A (b), it says that "S cannot contain a decreasing sequence of distinct real numbers" But we don't actually have sequences, but just elements of that set that appear to just be increasing?

Sat Jun 19 17:02:57 2004 : John: Geez luweez, someone's getting an early start! :) For 1.1.4(c), yes, 0 is a valid thing to consider. But can you find a less trivial counterexample? For 1.1A(b), we say that a set S _contains_ a sequence of elements if every element of the sequence comes from S. (This exercise is meant for you to practice with the well-ordering property, as well as to practice a little with logic and proofs--it's not supposed to be hard.)

Sun Jun 20 09:54:59 2004 : Shouldn't homework #2 be Due July 6 because there is no school on July 5?

Sun Jun 20 10:19:02 2004 : John: *Thanks*! Yes, there is no class Monday, July 5, so the homework will be due July 6 instead.

Thu Jun 24 14:11:27 2004 : John: In Problem 18 in section 1.4, the problem is incorrect as stated. (You should restrict to 0 <= r <= b/2 and keep the e = +/- 1, but why?) The problem should read: Show that if a and b are positive integers, then there are unique integers q, r such that a = bq+r where -b/2 < r <= b/2. Also, on Problem 10 in section 3.1, you may use the fact: if p is a prime and p divides a product of primes q_1 q_2 ... q_n, then p is equal to one of the primes p = q_i.

Sat Jun 26 11:12:38 2004 : In section 4.1, problem 16 does "final digit" mean the least significant digit?

Sun Jun 27 19:46:14 2004 : John: Yep, "final digit" means least significant digit, the "ones" digit.

Mon Jun 28 16:57:46 2004 : What, if anything, can we bring to the midterm?

Tue Jun 29 18:26:50 2004 : John: Just your brains and a pencil is all you will need. There will be some problems that will require you to do a small amount of basic arithmetic, but I will keep the numbers small enough to be done by hand. No calculators will be allowed, but I think this is in your best interests, since then (1) you need not go out and invest in high technology and (2) the problems will be simpler.

Tue Jun 29 20:53:26 2004 : Suppose that I have the equation x^2 + 1 = 0. Then x = plus or minus i But if I apply Theorem 3.17, it says that any root of x^2 + 1 should either be an integer or an irrational. Instead, I have an imaginary number! What condition is x^2 + 1 violating such that Theorem 3.17 doesn't apply?

Wed Jun 30 09:18:02 2004 : John: Good question! I was implicitly assuming that the underlying root was a real number. We said to start the proof that the root was either rational or irrational; of course this does not apply to complex roots, which by definition are not irrational.

Wed Jun 30 21:47:57 2004 : Will hws 3 and 4 be up over this weekend?

Wed Jun 30 22:26:50 2004 : John: Sure! I have them 90% planned out, I may add or subtract a few problems, but I'll have them up this weekend. :)

Thu Jul 1 21:17:35 2004 : What other kind of nonlinear diophantine equation will we study?

Thu Jul 1 22:40:58 2004 : John: We have many great things to cover! In Week 8, I hope to cover some more Diophantine topics... but I will take your input on how best to spend that time (where we can cover fun, advanced things not appearing on the final). :)

Sat Jul 3 15:04:24 2004 : For hw #3, can we get a hint on problem #14 in section 4.3?

Sat Jul 3 15:54:02 2004 : Also, what does problem 4.3A have to do with the Chinese Remainder Theorem? It seems this problem belongs in section 4.4?

Sun Jul 4 12:26:56 2004 : John: Hello, I hope everyone is enjoying the long weekend! For #4.3.14, this is a juicy problem, we'll talk about it in class. And Problem #4.3A really is a relatively straightforward application of the Chinese Remainder theorem... (And I'm still waiting for all of the typos you can find!)

Sun Jul 4 15:24:38 2004 : Well I suppose another typo is the typo in the Typo handout. For Typo (6), you have two \lfloor. You are missing a \rfloor

Mon Jul 5 17:46:55 2004 : Another typo is probably on the math 115 web page which says HW 3 is due *June* 14

Mon Jul 5 17:58:39 2004 : John: Thanks!

Thu Jul 8 15:01:43 2004 : John: I just finished writing up the solutions for HW #3, and it is a bit long, isn't it? Don't worry, this will be as long as the assignments get, and next week the assignment is very light (since the midterm is coming up). Two comments: For #14 in Section 6.1, what they want you to do is to find the least positive residue of 3^100 mod 7. I haven't talked about "base 7 expansions" yet, I will next week. For #16 in Section 6.1, I suggest breaking it up into two cases: if n = ab, where a,b > 1, look at the cases where a > b and if a = b. In the latter, n = b^2, and show that b and 2b both divide (n-1)!.

Sat Jul 10 15:33:59 2004 : Problem #3 in section 6.2 for hw #4 doesn't seem interesting at all because there really isn't any way of making use of Fermat's Little Theorem. The only way is just actually using repeated squaring to compute this which simply isn't interesting at all (unlike Carmichal numbers, which are way more interesting!)

Sat Jul 10 15:42:18 2004 : John: Carmichael numbers are interesting, and we'll talk about them on Wednesday. I didn't want to give too much homework this time, and I wanted to give you one (easy) computational example. Also, I'm not sure I ever asked you to practice with repeated squaring. Here's one hint to simplify the calculation: use Euler's theorem, since it applies. Then you needn't do too many squarings! :)

Mon Jul 12 18:22:55 2004 : The lecture on p-adic numbers was very fun today! Are we responsible for p-adic numbers on the final exam? I didn't see any practice problems on the hw for this week.

Mon Jul 12 18:52:07 2004 : John: I'm glad you enjoyed it! :) You will not be responsible for anything that I cover this week either on the midterm or on the final. (E.g. you should know Hensel's lemma, but you needn't know what's "really going on" vis-a-vis the p-adics.)

Tue Jul 13 13:04:03 2004 : John: Thanks everyone for your valuable feedback! There were 24 respondents, and here are the results. For (1) pace of the class, the score averaged 8/10; most people thought the pace was fine, and an equal number thought that it should move more slowly and more quickly. It's hard to make everyone happy, here, so I'll do my best to get to interesting topics by the end and will also spend more time when covering difficult proofs. For (2) group work, the score was 7/10, with most people finding it helpful only occasionally (when y'all are not too tired). I will spend some time thinking of other ways to use this time! For (3) and (4) well-prepared and clear/concise, the scores were 9/10 and 8.5/10; you are right that I often times talk and write too fast! It's the debater in me, I'll work on this. (I do like to write as much as possible on the blackboard, but that means more writing for you, too!) As for other comments, there were a few people that thought the homeworks were too long, and I don't mean to work you too hard... I will readjust and give you more hints on the harder problems. There were a few people who requested topics, and I would love it to hear how you would like to spend the last week. Please post here or send me an e-mail and I will do my best to allow time for it. Thanks everyone!

Wed Jul 14 14:33:07 2004 : John, the review problems today....covered Hensel's Lemma, Newton's method, Chinese Remainder theorem, as well as other topics. Are we to assume these are the only topics on the midterm? Make our lives easier and say they are :)

Wed Jul 14 14:34:08 2004 : Could you explain problem 1 of the review again? I didn't quite get it...

Wed Jul 14 15:22:40 2004 : John: Good afternoon, all! Short of telling you the problems on the midterm, I'm not sure I can say much more. :) I know this isn't the first midterm you have taken in your lives. As for the problems, I'll post in just a second the review sheets and the solutions written out. Cheers!

Thu Jul 15 13:08:41 2004 : Will you be posting the midterm solutions over the weekend, or do we have to wait until Monday?

Thu Jul 15 14:28:32 2004 : #4 was hard! what is the trick here?

Thu Jul 15 15:04:56 2004 : John: I just put up solutions to the midterm and the score breakdowns. No tricks in #4: just check for divisibility by 2, 3, 5, 7... by computing n modulo each of these primes (and use FLT). The first prime that works is 11.

Thu Jul 15 16:07:14 2004 : So a C is 20?

Thu Jul 15 17:12:00 2004 : John: Yep, I assigned 20 a C letter grade and 19 a C-, but as I mentioned, these are only meant to be a rough guideline, and they don't include your homework score. Have a great weekend!

Fri Jul 16 14:04:11 2004 : hi john, you're doing a great job as our instructor. I think you'll make a great professor. One request though, can you slow down a bit on the harder sections of this course? Thanks.

Fri Jul 16 17:26:12 2004 : On problem 41 of section 7.1, why does the book's proof make use of a theorem covered in section 7.2? Is there a way of proving the result without making use of the theorem covered in section 7.2?

Sat Jul 17 12:03:38 2004 : On problem 6 in section 7.3 are we allowed to write up a program that figures this out?

Sun Jul 18 14:40:32 2004 : Hi that a brilliant piece of abstract art in the upper left hand corner, or is it an upside-down dog? = )

Sun Jul 18 19:13:58 2004 : John: Thank you for the feedback! Like I said, I will do my best to make sure to cover difficult proofs at an approachable pace. Always interrupt and ask questions if something is unclear. I just got back from Tahoe, I'll answer your questions about the homework when I get into my office tomorrow. Oh, and yes, that's an upside-down Scottish terrier. :)

Mon Jul 19 12:23:41 2004 : John: In Section 7.4, Problem 17, you might have the multiplicative function f(n) = 0 for all n, in which case f(1) = 0. You may either assume that f(1) = 1, or adjust the formula to read sum(d|n) mu(d)f(d) = (f(1)-f(p_1)) ... (f(1)-f(p_k)). As for your questions (which are good ones!): For Section 7.1, Problem 41, all you need to know is that lambda is multiplicative--you can either rely upon the result in Section 7.2, or you can prove it directly! :) (Actually, as Problem 40 indicates, lambda is completely multiplicative.) For Section 7.3, Problem 6, one possible approach is indeed the brutal one, and hey, a computer program would be fine. I'll be talking about this a lot in class and we'll talk about odd perfect and abundant numbers. Here's an alternative approach which requires less computation: we will prove in class that if N is odd and perfect, then N must have at least 3 prime factors (and these proofs will use inequalities, so it will also imply that if N is odd and abundant, then N has at least 3 prime factors). Now you just need to test odd numbers with at least three prime divisors, and there are even more tricks! Just so you don't think the problem is too hard, the answer is less than 1000. :)

Tue Jul 20 17:28:11 2004 : For factoring 2^30 - 1, I understand that since 5 divides 30, then 2^5-1 = 31 divides 2^30 - 1, so after that step, then how do you go on to factoring the remaining term?

Tue Jul 20 17:51:06 2004 : John: Thanks for your question! The idea in factoring n=2^30-1 is to find as many factors as possible using this trick and then hope that the remaining product is "easy". For example, the prime divisors of 30 are 2, 3, and 5, and this gives the divisors: 2^2-1=3, 2^3-1=7, 2^5-1=31. Now using this trick doubly we can use the composite divisors. Start with 2^6-1=63; then already we know that 2^2-1=3 and 2^3-1=7 divides this, and indeed the cofactor is 63/21=3, so in fact 3^2.7 divides n. Similarly 2^10-1=3.31.11, and 2^15-1=7.31.151, where 151 is prime. Putting these together, we have the factors 3^ dividing n, and the remainder, 331, turns out to be prime. Neat, huh? For more practice, try factoring n=3^12-1. [N.B. The reason we are treating this is that the numbers 2^p-1 will come up already tomorrow!]

Tue Jul 20 18:13:39 2004 : I was looking at hw #1 of the Math 195 cryptography web page, and I'm not sure what the hidden message is for problem 2.1 . Any hints?

Wed Jul 21 08:39:35 2004 : John: It took me fo-evah to figure that one out. Hint: Look at the last words of each line.

Thu Jul 22 12:04:42 2004 : John: One small typo in Problem 17 in section 7.4: You need to know that f(1)=1 in order for the formula to be correct. You can either assume this, or replace 1 by f(1) in the formula. :) Have a great weekend!

Fri Jul 23 11:28:15 2004 : Can we make use of the h function that we talked about in class and its properties on the hw? Also, can we make use of problem 32a in section 7.3 since we proved it in class?

Fri Jul 23 23:00:29 2004 : John: Yes, and yes! Please feel free to use any results that we prove in class, just quote them properly. :)

Sun Jul 25 16:16:54 2004 : John: Attention all students! Reverse-linking this website, I have found at least two other number theory courses which have linked to this page and are using (or will be using) our errata sheet. :) We are about to enter typo-laden chapters, and this is your best chance to add your name to the list of typo warriors! E-mail me your typo today!

Sun Jul 25 21:36:23 2004 : Where does John pick up the phrase "good times", or "cool beans" ? It sounds so funny.

Tue Jul 27 18:47:23 2004 : John: In solving 9.1.15, there is no indication where r comes from. (It is possible that r = 0!) To be precise, you should start with an element r with gcd(r,p)=1 so you can even get off the ground.

Tue Jul 27 19:07:48 2004 : Can you post solutions to worksheet 7?

Tue Jul 27 20:08:46 2004 : John: There is no worksheet #7... do you mean the latest one, worksheet #6? If so, we will go over these problems on Thursday, don't worry. :)

Wed Jul 28 13:18:02 2004 : John: Hello, everyone! I have been trying to keep a slower pace, so I will only be able to treat a small number of topics from primality testing tomorrow. As a result, I will ask the reader not to grade the problems from Section 9.5.

Fri Jul 30 14:55:02 2004 : Can we write a program to solve Problem 16 in Section 9.3, or is there a different approach to this problem?

Sat Jul 31 00:47:03 2004 : John: This is the problem which asks you to find the smallest prime p with a primitive root r which is not a primitive root modulo p^2, right? I thought since the answer is small (p < 30), it would require not much hand calculation. Turns out it does require verifying a number of congruences, and writing a program is a good way to do this! At least that's what I did in the solutions. I guess it's good to have at least one harder computationally-oriented problem in the course. :)

Sat Jul 31 17:53:13 2004 : If we used a program to help aid us in the computation, then how would we justify that we found the smallest such odd prime? It seems that we would need to justify this to the reader, and would require the reader to have faith that our program found the smallest such odd prime.

Sun Aug 1 00:36:59 2004 : John: I guess then that it would be up to the reader to judge the correctness of your program! :)

Sun Aug 1 12:00:57 2004 : John, are you going to be a GSI for any upper div math classes in the Fall?

Sun Aug 1 21:12:40 2004 : John: I will not be teaching in the fall, I have to write my dissertation! I will however be teaching in the spring, hopefully an interesting upper-division class. See you tomorrow. :)

Sun Aug 1 21:27:59 2004 : What is your dissertation about?

Sun Aug 1 22:49:03 2004 : John: One chapter of my thesis will be on my work on quadratic forms. You can check it out on my website:

Mon Aug 2 13:04:23 2004 : John: Hello everyone--the votes tallied thus far give the most favorite topics as: elliptic curves, quadratic forms, and the Jacobi symbol. I'll take ballots until tomorrow, but this sounds good to me. :) Also, we will do a fun ciphertext challenge on Tuesday second hour, so you may want to bring a calculator! See you then.

Thu Aug 5 11:57:01 2004 : Thanks so much for the Harvard link for the problem in 8.4. It worked great!!!

Thu Aug 5 19:37:49 2004 : what topic were we going to cover on Monday?

Thu Aug 5 23:29:43 2004 : John: We will cover pythagorean triples and sums of squares on Monday, and a friendly introduction to elliptic curves on Tuesday. Review on Wednesday. :) Have a great weekend!

Sat Aug 7 22:51:32 2004 : Hi JOhn, I have a few questions about this week's homework #7: Problem 8.1A: I've concluded that ;48 = THE. Where do I go from here? Please help. 11.1#12: Which quadratic congruences (mod2) have solutions to ax^2+bx+c=0(mod p) p doesn't divide a: What does the quadratic congruence look like? is it a polynomial? Im having trouble transitioning from integers.Please help. 11.2#4: I don't know where to begin, could you explain what it means to find the primes where 5 is a quadratic residue? I looked at theorem 11.6 where they showed which primes is 2 a quadratic residue but I don't understand that proof either.

Sat Aug 7 22:57:30 2004 : John: Happy to help. For 8.1A: the other high frequency characters are 5,6,d,*,) so they are likely to be r,n,i,o,a,s in some order. Now guess! For For 11.1.12(a), a quadratic congruence modulo 2 only depends on what the coefficients are modulo 2, so there are only four possibilities. For 11.2.4, check out the solution to Exercise 3. Use quadratic reciprocity, and the fact that the Legendre symbol only depends on what p is modulo 5. What are the quadratic residues modulo 5?

Sun Aug 8 15:07:11 2004 : 8.1A: John I still can't figure this out. I've tried all the letters and I'm getting frustrated. Is there an algorithm or something? trial and error is killing me.

Sun Aug 8 18:58:08 2004 : Are you going to answer the question above? I need help too.

Sun Aug 8 20:49:13 2004 : John: OK, I guess frustration is part and parcel with cryptanalysis! For those of you still puzzling, the message starts out, "A good glass in the bishop's hostel..." The message appears to be the where, when, and how for a sharpshooter. Is that enough of a hint? See you tomorrow!

Sun Aug 8 21:52:03 2004 : I can't believe after I finally solved 8.1A you gave it up to the class. I feel that I just went through all the decrypting for the way what does this message mean????

Sun Aug 8 21:55:02 2004 : I finally solved the 8.1 message (without the previous message)...but what does it mean?

Sun Aug 8 22:19:46 2004 : haha.

Sun Aug 8 22:46:33 2004 : John: Hey--no reason for anyone to suffer from a cipher! If you solved it yourself, no one can take that satisfaction away. :) As I said before, the message appears to be the where, when, and how for a sharpshooter.

Sun Aug 8 23:17:09 2004 : Hint on the meaning of the answer to 8.1A: type it into Google.

Mon Aug 9 09:50:19 2004 : Two questions: is the second half of Wednesday's lecture going to be a review session? Are we responsible for anything covered this week on the final? Thanks.

Tue Aug 10 15:27:13 2004 : Does the final include the section on Carmichael numbers and pseudoprimes?

Tue Aug 10 15:41:33 2004 : John: Nope, nothing on pseudoprimes or Carmichael numbers.

Tue Aug 10 17:54:25 2004 : Does the final include being able to do Pollard rho or proving why it works? What about Hensel's Lemma?

Tue Aug 10 18:40:28 2004 : John: No Pollard rho. Hensel's lemma might prove useful on one or two of the problems.

Tue Aug 10 20:05:20 2004 : what about quadratic residues/non residues? On Cryptography, how much should we know? Caesar, affline?

Tue Aug 10 20:11:44 2004 : John: You definitely should know about quadratic residues and nonresidues--the reciprocity law is one of the main results in the class! As for cryptography, I can't say more. As with all of the subjects, study what you think is most relevant.

Tue Aug 10 20:51:51 2004 : I'm not trying to easy my way out, but could it be possible if you could tell us what to study more in-depth? Some of this stuff is hard and it'd be alot easier to know some topics more thoroughly than to know bits and pieces of all topics am I right?

Tue Aug 10 21:02:15 2004 : On problem 8 of section 13.1, how would you show that there are infinitely many such solutions?

Tue Aug 10 21:15:18 2004 : John: If you have come to class and have done the homework, you'll know which things are important since they reappear. You can also see what is more important in the review session tomorrow.

Tue Aug 10 21:37:09 2004 : What topics are you covering in the review session?

Tue Aug 10 21:47:28 2004 : Wilson's theorem: suppose we have a congruence mod n, two numbers must have the same gcd with n right? What is the converse of wilson's theorem?

Tue Aug 10 21:49:32 2004 : Can you help me with a problem? Prove that -1 is not a square modulo the prime p if p is congruent to 3 mod 4.

Wed Aug 11 09:00:07 2004 : John: I'm not exactly sure what your question is about Wilson's theorem. (Are you asking two questions? It is true that if a = b (mod n), then gcd(a,n) = gcd(b,n).) Wilson's theorem states that (p-1)! = -1 (mod p) if p is prime. The converse states that if p is not prime, then (p-1)! <> -1 (mod p). As for -1 not being a square, this says that (-1/p) = -1 if p = 3 (mod 4). Either use Euler's criterion, or the argument that if x^2 = -1 (mod p), then x has order 4 modulo p, so 4 | (p-1).

Wed Aug 11 12:38:25 2004 : John: To solve #13.1.8, you should try to repeat the proof of the parametrization of Pythagorean triples for the new equation. Start by writing it 2y^2 = z^2-x^2 = (z-x)(z+x)... and see if you can find formulas. (It's not easy, but it's not too hard!)

Wed Aug 11 21:26:59 2004 : Is the "point at infinity" used for elliptic curves related to the "infinity" of the Riemann sphere?

Wed Aug 11 22:31:34 2004 : John: Excellent question! In one sense, yes, both the Riemann sphere and the "projective" plane are ways of "compactifying" the ordinary complex plane. But it turns out that they actually have very different geometries, so in another sense, no. (The Riemann sphere is actually a real surface with no holes, and an elliptic curve together with the "point at infinity" makes a complex torus, or a donut.)

Thu Aug 12 12:45:19 2004 : Will the solutions to the final be posted?

Thu Aug 12 14:22:45 2004 : What will the grading scale be?

Mon Aug 16 09:23:21 2004 : When will grades appear on bearfacts? Also when and how can we view our final exams?

Mon Aug 16 10:35:55 2004 : John: Hello all! I thought I had submitted the grades on Thursday afternoon, but there has been a technical glitch in the receipt of these grades by the registrar. I hope it will be resolved in due course. Have a great summer!