For Wednesday, September 21

For Wednesday:

  1. BRING A LAPTOP TO CLASS, WE WILL DO A WORKSHOP DAY.
  2. In class you received your Module 2 assessment back, along with solutions (on paper).  Please compare your solutions to the published solutions and take time to understand all your (and my!) errors.
  3. There is an opportunity to replace one grade from among the Part B questions (four questions each worth 10 points):
    1. If you wish to replace a grade of 7, 8, or 9, I will provide a similar problem and ask you to hand in a solution on canvas.  I will announce this after I have finished the in-person replacements for other students.
    2. If you wish to replace a grade less than or equal to 6, then we will meet in person, and I will ask you to solve a similar problem.  If you wish to do this, then please schedule a time with me (zoom or in person).  This time can be anytime in the next week or so, but you must contact me to schedule it before Wednesday.  Just email me to set this up.
  4. Just some extra info.  I have a very special place in my heart for the euclidean algorithm.  I have written up my explanation from class in the overleaf course notes (available in “Notes/Videos” in the top bar).  I also have two videos about it under “Notes/Videos” in the top bar.  These videos explain the same algorithm, but not exactly the same (more “coats of paint” to your understanding) and you may wish to watch them for review at some point.
  5. Compute gcd(183,105).  Show the process by hand.
  6. Solve the Diophantine equation (meaning, find all integer solutions, show the process by hand):
    1. 183x + 105y = 0
    2. 183x + 105y = 3
    3. 183x + 105y = 1
  7. Looking at your work above, what is the gcd of 61 and 35 (it should not require new computations, it’s there for you to see)
  8. Looking at your work above, what is the full set of solutions to $35x + 61y = 1$?  (it should not require new computations, it’s there for you to see in your work above)
  9. Find the inverse of 35 modulo 61.  This should not require new computations either; it should be discernible from your work above.   Hint:  To find the inverse of $a$ modulo $n$, you need to solve $ax \equiv 1 \pmod n$, which is to say, $ax + ny = 1$.