Fixes emphasis in 2017 READMEs

Author: romellem

2017/13/README.md

@@ -168,7 +168,7 @@ Your plan is to hitch a ride on a packet about to move through the firewall. The
 
 In this situation, you are _caught_ in layers `0` and `6`, because your packet entered the layer when its scanner was at the top when you entered it. You are _not_ caught in layer `1`, since the scanner moved into the top of the layer once you were already there.
 
-The _severity_ of getting caught on a layer is equal to its _depth_ multiplied by its _range_. (Ignore layers in which you do not get caught.) The severity of the whole trip is the sum of these values. In the example above, the trip severity is `0*3 + 6*4 = _24_`.
+The _severity_ of getting caught on a layer is equal to its _depth_ multiplied by its _range_. (Ignore layers in which you do not get caught.) The severity of the whole trip is the sum of these values. In the example above, the trip severity is <code>0\*3 + 6\*4 = <em>24</em></code>.
 
 Given the details of the firewall you've recorded, if you leave immediately, _what is the severity of your whole trip_?
 

2017/15/README.md

@@ -56,8 +56,8 @@ In the interest of trying to align a little better, the generators get more pick
 
 They still generate values in the same way, but now they only hand a value to the judge when it meets their _criteria_:
 
-*   Generator A looks for values that are multiples of `_4_`.
-*   Generator B looks for values that are multiples of `_8_`.
+*   Generator A looks for values that are multiples of _`4`_.
+*   Generator B looks for values that are multiples of _`8`_.
 
 Each generator functions completely _independently_: they both go through values entirely on their own, only occasionally handing an acceptable value to the judge, and otherwise working through the same sequence of values as before until they find one.
 

2017/17/README.md

@@ -12,7 +12,7 @@ If you don't move quickly, fixing that printer will be the least of your problem
 
 This spinlock's algorithm is simple but efficient, quickly consuming everything in its path. It starts with a circular buffer containing only the value `0`, which it marks as the _current position_. It then steps forward through the circular buffer some number of steps (your puzzle input) before inserting the first new value, `1`, after the value it stopped on. The inserted value becomes the _current position_. Then, it steps forward from there the same number of steps, and wherever it stops, inserts after it the second new value, `2`, and uses that as the new _current position_ again.
 
-It repeats this process of _stepping forward_, _inserting a new value_, and _using the location of the inserted value as the new current position_ a total of `_2017_` times, inserting `2017` as its final operation, and ending with a total of `2018` values (including `0`) in the circular buffer.
+It repeats this process of _stepping forward_, _inserting a new value_, and _using the location of the inserted value as the new current position_ a total of _`2017`_ times, inserting `2017` as its final operation, and ending with a total of `2018` values (including `0`) in the circular buffer.
 
 For example, if the spinlock were to step `3` times per insert, the circular buffer would begin to evolve like this (using parentheses to mark the current position after each iteration of the algorithm):
 

2017/24/README.md

@@ -44,7 +44,7 @@ With them, you could make the following valid bridges:
 
 (Note how, as shown by `10/1`, order of ports within a component doesn't matter. However, you may only use each port on a component once.)
 
-Of these bridges, the _strongest_ one is `0/1`\--`10/1`\--`9/10`; it has a strength of `0+1 + 1+10 + 10+9 = _31_`.
+Of these bridges, the _strongest_ one is `0/1`\--`10/1`\--`9/10`; it has a strength of <code>0+1 + 1+10 + 10+9 = <em>31</em></code>.
 
 _What is the strength of the strongest bridge you can make_ with the components you have available?
 
@@ -59,6 +59,6 @@ In the example above, there are two longest bridges:
 *   `0/2`\--`2/2`\--`2/3`\--`3/4`
 *   `0/2`\--`2/2`\--`2/3`\--`3/5`
 
-Of them, the one which uses the `3/5` component is stronger; its strength is `0+2 + 2+2 + 2+3 + 3+5 = _19_`.
+Of them, the one which uses the `3/5` component is stronger; its strength is <code>0+2 + 2+2 + 2+3 + 3+5 = <em>19</em></code>.
 
 _What is the strength of the longest bridge you can make?_ If you can make multiple bridges of the longest length, pick the _strongest_ one.

2017/4/README.md

@@ -6,7 +6,7 @@
 
 ## --- Day 4: High-Entropy Passphrases ---
 
-A new system policy has been put in place that requires all accounts to use a _passphrase_ instead of simply a pass_word_. A passphrase consists of a series of words (lowercase letters) separated by spaces.
+A new system policy has been put in place that requires all accounts to use a _passphrase_ instead of simply a pass<em>word</em>. A passphrase consists of a series of words (lowercase letters) separated by spaces.
 
 To ensure security, a valid passphrase must contain no duplicate words.