MATH135: various fixes

Closes #25
Closes #26
Closes #27
Closes #28
Rebuild PDFs for #29
Closes #30

Co-authored-by: Linda Wang <61102300+ylw311@users.noreply.github.com>
Co-authored-by: Max Luo <90939473+nyannnnn@users.noreply.github.com>
Co-authored-by: evelina <91142355+135ze@users.noreply.github.com>
Co-authored-by: Christina Li <87205170+christinaly19@users.noreply.github.com>

MATH135/EP09.tex

@@ -54,7 +54,7 @@
           \end{center}
           and conclude that $d=265$ is a solution to our LDE\@.
           Since $1 < 265 < 1656$, it is in fact the decryption key.
-          Therefore, Alice's privkey is $(265,1379)$.
+          Therefore, Alice's privkey is $(265,1739)$.
         \end{proof}
   \item Suppose Alice wishes to send Bob the message $M = 20$.
         Bob's public key is $(23, 377)$ and Bob’s private key is $(263, 377)$.

MATH135/EP10.tex

@@ -41,7 +41,7 @@
   Note that $1 = 1\cis0$.
   Applying the CRNT, we have that the five roots are given by
   $\sqrt[5]{1}\cis\left(\frac{2k\pi}{n}\right)$ for $k=0,1,2,3,4$.
-  These values are $\{ 1, \cis\frac\pi5, \cis\frac{4\pi}{5}, \cis\frac{6\pi}{5}, \cis\frac{8\pi}{5} \}$.
+  These values are $\{ 1, \cis\frac{2\pi}{5}, \cis\frac{4\pi}{5}, \cis\frac{6\pi}{5}, \cis\frac{8\pi}{5} \}$.
   I am too lazy to learn \verb|tikz| to draw the diagram.
 \end{proof}
 
@@ -397,11 +397,11 @@
                   By the Division Algorithm, $j = qn + r$ for integers $q$ and $0 \leq r < n$.
                   Then, $1 = z^j = z^{qn+r} = z^{qn}z^r = (z^n)^q z^r = 1^q z^r = z^r$.
 
-                  If $r=0$, then $j = qn$ and $j \mid n$.
+                  If $r=0$, then $j = qn$ and $n \mid j$.
                   Otherwise, we have $1 \leq r \leq n-1$ and $z^r = 1$, which is a contradiction
                   to the fact that $z$ is a primitive $n$-th root of unity.
 
-                  Therefore, $r = 0$ and $j \mid n$.
+                  Therefore, $r = 0$ and $n \mid j$.
 
                   ($\Larr$) If $n \mid j$ and $j = nk$ for an integer $k$,
                   then $z^j = z^{nk} = (z^n)^k = 1^k = 1$.

MATH135/EP11.tex

@@ -80,8 +80,9 @@
           To find roots of this cubic, the RRT gives candidates $x=1,-1,\frac13,-\frac13$.
           In fact, $q(-\frac13)=0$.
           Dividing $q(x)$ by $(3x+1)$, we obtain the factor $(x^2+3x+1)$.
-          The discriminant of this quadratic is negative, so it is irreducible in $\R[x]$.
-          Therefore, $f(x) = (x+1)(3x+1)(x^2+3x+1)$.
+          The discriminant of this quadratic is positive
+          and it has roots $-\frac32 \pm \frac{\sqrt{5}}{2}$.
+          Therefore, $f(x) = (x+1)(3x+1)(x-\frac32+\frac{\sqrt{5}}{2})(x-\frac32-\frac{\sqrt{5}}{2})$.
         \end{proof}
   \item $x^4 + 27x \in \C[x]$
         \begin{proof}[Solution]

MATH135/FE2020W.tex

@@ -28,7 +28,7 @@
 \prob{Determine the units digit (i.e., the ones digit) of $7^{202}$.}
 \begin{sol}
   We must evaluate $7^{202} \pmod{10}$.
-  Since $7^2 \equiv 49 \ equiv -1 \pmod 10$, it follows that $7^{202} \equiv (7^2)^{101} \equiv (-1)^{101} \equiv -1 \equiv 9 \pmod 10$.
+  Since $7^2 \equiv 49 \equiv -1 \pmod{10}$, it follows that $7^{202} \equiv (7^2)^{101} \equiv (-1)^{101} \equiv -1 \equiv 9 \pmod{10}$.
 
   Therefore, the last digit is 9.
 \end{sol}

index.md

@@ -285,6 +285,11 @@ Thanks to everyone on the list who've helped me make this resource better for ev
 30. **MATH135/EP01/WE1(b)** 28 + 3 = 25 (Imran, [#20](https://github.com/RetroCraft/problems/pull/20))
 31. **MATH135/EP11/RP03(e)** ...another factoring typo (Linda, [#24](https://github.com/RetroCraft/problems/issues/24))
 32. **MATH135/EP10/RP12** Wrong expansion of $$(\cos\theta + i\sin\theta)^4$$ (Jessica and Evelina)
+33. **MATH135/EP10/RP15** Backwards divisibility notation (Linda, [#25](https://github.com/RetroCraft/problems/issues/25))
+34. **MATH135/EP11/RP03** Did not fully factor quadratic (Jessica & Evelina, [#26](https://github.com/RetroCraft/problems/issues/26))
+35. **MATH135/EP10/WE04** 2 × 1 = 1 (Christina, [#27](https://github.com/RetroCraft/problems/issues/27))
+36. **MATH135/EP09/RP01** Flipped digits in final answer (various, [#28](https://github.com/RetroCraft/problems/issues/28))
+37. **MATH135/EP10/RP01** Lost a negative (various, [#29](https://github.com/RetroCraft/problems/pull/29))
 
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