Question 498322
<pre>
You couldn't use algebra unless you wanted to find it for x digit changes
instead of 27 didgit changes. To use algebra you need an unknown.  But 
here's how to do it with just basic math:

6:38 means {{{6&38/60}}} hours after 12:00.  Each digit change represents
1 munute or {{{1/60}}}th of an hour.  So after 27 digit changes that will be 
{{{27/60}}} of an hour added to {{{6&38/60}}},

{{{6&38/60}}}{{{""+""}}}{{{27/60}}}{{{""=""}}}{{{6&65/60}}}{{{""=""}}}{{{7&5/60}}} 
hours after 12:00 and since each minute is {{{1/60}}}th of an hour
that will be at 7:05, not 7:01.  It really is 7:05, not 7:01. Here is how my
computer program gave it: 


1 digit change later the clock will read 6:39
2 digit changes later the clock will read 6:40
3 digit changes later the clock will read 6:41
4 digit changes later the clock will read 6:42
5 digit changes later the clock will read 6:43
6 digit changes later the clock will read 6:44
7 digit changes later the clock will read 6:45
8 digit changes later the clock will read 6:46
9 digit changes later the clock will read 6:47
10 digit changes later the clock will read 6:48
11 digit changes later the clock will read 6:49
12 digit changes later the clock will read 6:50
13 digit changes later the clock will read 6:51
14 digit changes later the clock will read 6:52
15 digit changes later the clock will read 6:53
16 digit changes later the clock will read 6:54
17 digit changes later the clock will read 6:55
18 digit changes later the clock will read 6:56
19 digit changes later the clock will read 6:57
20 digit changes later the clock will read 6:58
21 digit changes later the clock will read 6:59
22 digit changes later the clock will read 7:00
23 digit changes later the clock will read 7:01
24 digit changes later the clock will read 7:02
25 digit changes later the clock will read 7:03
26 digit changes later the clock will read 7:04
27 digit changes later the clock will read 7:05

Edwin</pre>