Fix typos

Advanced/Pidgeonhole Problems/main.tex

@@ -380,7 +380,10 @@
 		In a year, there are $364$ clean breaks. This leaves $499 - 364 = 135$ ``dirty'' breaks. \par
 		We therefore have $135$ places to start a block on a dirty break, and $135$ places to end a block on a dirty break. This gives us a maximum of $270$ dirty blocks. \par
 
-		However, there are $401$ possible blocks, since we can start one at the $1^{\text{st}}, 2^{\text{nd}}, ..., 401^{\text{st}}$ espresso. \par
+		% spell:off
+		However, there are $401$ possible blocks,
+		since we can start one at the $1^{\text{st}},2^{\text{nd}}, ..., 401^{\text{st}}$ espresso. \par
+		% spell:on
 
 		Out of $401$ blocks, a maximum of $270$ can be dirty. We are therefore guaranteed at least $131$ clean blocks. This completes the problem---each clean block represents a set of consecutive, whole days during which exactly 100 espressos were consumed.