- fix 4.4.20 #17

src/Epp.tex

@@ -12136,7 +12136,7 @@ \subsubsection{Exercise 20}
 The sum of any three consecutive integers is divisible by 3.
 
 \begin{proof}
-True. Assume $a,b,c$ are any three consecutive integers. By definition of consecutive, $a = n, b = n+1, c = n+2$ for some integer $n$. Then $a+b+c = n+n+1+n+2 = 3n+6 = 3(n+2)$. Let $t = n+2$. Then $t$ is an integer (being a sum of integers). So $a+b+c = 3t$ where $t$ is an integer. So by definition of divisibility, $a+b+c$ is divisible by 3.
+True. Assume $a,b,c$ are any three consecutive integers. By definition of consecutive, $a = n, b = n+1, c = n+2$ for some integer $n$. Then $a+b+c = n+n+1+n+2 = 3n+3 = 3(n+1)$. Let $t = n+1$. Then $t$ is an integer (being a sum of integers). So $a+b+c = 3t$ where $t$ is an integer. So by definition of divisibility, $a+b+c$ is divisible by 3.
 \end{proof}
 
 \subsubsection{Exercise 21}
@@ -26823,8 +26823,8 @@ \subsubsection{Exercise 6}
 \(B \subseteq A\)
 
 \begin{proof}
-Suppose \(y \in B\). Then \(y = 10b - 3\) for some integer $b$. Then \(y = 10b - 5 + 5 - 3 = 5(b-2) + 2\) where $b-2$
-is an integer. Let \(a = b-2\). Therefore \(y = 5a + 2\) for some integer $a$, therefore \(y \in A\).
+Suppose \(y \in B\). Then \(y = 10b - 3\) for some integer $b$. Then \(y = 10b - 5 + 5 - 3 = 5(2b-1) + 2\) where $2b-1$
+is an integer. Let \(a = 2b-1\). Therefore \(y = 5a + 2\) for some integer $a$, therefore \(y \in A\).
 This proves \(B \subseteq A\).
 \end{proof}
 
@@ -42584,7 +42584,7 @@ \subsubsection{Exercise 33}
 How many students checked \#2 but neither of the other two?
 
 \begin{proof}
-The number of students who checked \#2 but not \#1 or \#3 is \(N(C) - N(C \cap (H \cup D)) = 17 - (6+2+1) = 8\).
+The number of students who checked \#2 but not \#1 or \#3 is \(N(C) - N(C \cap (H \cup D)) = 26 - (6+2+1) = 17\).
 \end{proof}
 
 \subsubsection{Exercise 34}