Editing Location arithmetic (section)

===Addition===
First, lay out a binary number on a row using counters to represent the 1s in the number. For example, 29 (= 11101 in binary) would be placed on the board like this:
{| border="0" cellpadding="0" cellspacing="1" style="text-align:center; background-color: white; color: black;"
|+ '''11101 (= 29) on one row'''
|- style="height:30px; background-color: silver;"
| style="width:30px;"|&nbsp;
| style="width:30px;"|[[Image:Location arithmetic one counter.svg]]
| style="width:30px;"|[[Image:Location arithmetic one counter.svg]]
| style="width:30px;"|[[Image:Location arithmetic one counter.svg]]
| style="width:30px;"|&nbsp;
| style="width:30px;"|[[Image:Location arithmetic one counter.svg]]
|- style="height:30px; background-color:white; color: black;"
| style="width:30px;"|'''0'''
| style="width:30px;"|'''1'''
| style="width:30px;"|'''1'''
| style="width:30px;"|'''1'''
| style="width:30px;"|'''0'''
| style="width:30px;"|'''1'''
|}
The number 29 is clearly the sum of the values of the squares on which there are counters. Now overlay the second number on this row.  Say we place 9 (= 1001 in binary) on it like this.
{| border="0" cellpadding="0" cellspacing="1" style="text-align:center; background-color: white; color: black;"
|+ '''Overlay 1001 (= 9)'''
|- style="height:30px; background-color: silver;"
| style="width:30px;"|&nbsp;
| style="width:30px;"|[[Image:Location arithmetic one counter.svg]]
| style="width:30px;"|[[Image:Location arithmetic two counters.svg]]
| style="width:30px;"|[[Image:Location arithmetic one counter.svg]]
| style="width:30px;"|&nbsp;
| style="width:30px;"|[[Image:Location arithmetic two counters.svg]]
|- style="height:30px; background-color:white; color: black;"
| style="width:30px;"|'''0'''
| style="width:30px;"|'''0'''
| style="width:30px;"|'''1'''
| style="width:30px;"|'''0'''
| style="width:30px;"|'''0'''
| style="width:30px;"|'''1'''
|}
The sum of these two numbers is just the total value represented by the counters on the board, but some of the squares have more than one counter.  Recall however, that moving to the left of a square doubles its value. So we replace two counters on a square with one counter to its left without changing the total value on the board. Note that this is the same idea used to abbreviate
location numerals. Let's start by replacing the rightmost pair of counters with a counter to its left, giving:
{| border="0" cellpadding="0" cellspacing="1" style="text-align:center; background-color: white; color: black;"
|- style="height:30px; background-color: silver;"
| style="width:30px;"|&nbsp;
| style="width:30px;"|[[Image:Location arithmetic one counter.svg]]
| style="width:30px;"|[[Image:Location arithmetic two counters.svg]]
| style="width:30px;"|[[Image:Location arithmetic one counter.svg]]
| style="width:30px;"|[[Image:Location arithmetic one counter.svg]]
| style="width:30px;"|←
|}
We still have another square with two counters on it, so we do it again:
{| border="0" cellpadding="0" cellspacing="1" style="text-align:center; background-color: white; color: black;"
|- style="height:30px; background-color: silver;"
| style="width:30px;"|&nbsp;
| style="width:30px;"|[[Image:Location arithmetic two counters.svg]]
| style="width:30px;"|←
| style="width:30px;"|[[Image:Location arithmetic one counter.svg]]
| style="width:30px;"|[[Image:Location arithmetic one counter.svg]]
| style="width:30px;"|&nbsp;
|}
But replacing this pair created another square with two counters on it, so we replace a third time:
{| border="0" cellpadding="0" cellspacing="1" style="text-align:center; background-color: white; color: black;"
|+ '''Result 100110 = 38'''
|- style="height:30px; background-color: silver;"
| style="width:30px;"|[[Image:Location arithmetic one counter.svg]]
| style="width:30px;"|←
| style="width:30px;"|&nbsp;
| style="width:30px;"|[[Image:Location arithmetic one counter.svg]]
| style="width:30px;"|[[Image:Location arithmetic one counter.svg]]
| style="width:30px;"|&nbsp;
|- style="height:30px; background-color:white; color: black;"
| style="width:30px;"|'''1'''
| style="width:30px;"|'''0'''
| style="width:30px;"|'''0'''
| style="width:30px;"|'''1'''
| style="width:30px;"|'''1'''
| style="width:30px;"|'''0'''
|}
Now each square has just one counter, and reading off the result in binary 100110 (= 38) gives the correct result.