A computer simulation of a 12 hour analog clock keeps perfect time while it is running, and
has two hands—an hour hand and a minute hand—both of which move continuously around
the 12 hour dial. (For example, at 2:30, the hour hand will be exactly halfway between 2 and
3.) Because of careless programming, the minute hand looks exactly like the hour hand, so
that the two are indistinguishable.
On one day, the clock stops at some time after 12am and before 12pm. How many times could
it have stopped without it being possible after 12pm to tell what time it stopped?  Options
A) 11  B) 12  C) 66  D) 132  E) 708
A configuration of the hands is called “ambiguous” if you cannot use this broken clock to   distinguish the minute hand from the hour hand.   In the above diagram it must be the minute hand that is pointing to 6 and the hour hand that   is pointing midway between 2 and 3. (If the hour hand were to point to 6, the minute hand   would have to point to 12.) 2:30 is not an ambiguous configuration.   Let m 2 (0; 1) represent the fraction of the way around the clock face for the minute hand and   h 2 (0; 1) the corresponding fraction for the hour hand. For example, 2:30 hasm = 1=2 = 0:5   and h = 2:5=12 = 0:2083.   As the minute hand moves 12 times as fast as the hour hand, a configuration is ambiguous if   m and 12h have the same fractional part. Because the hands are indistinguishable, a time is   also ambiguous when h and 12m have the same fractional part. We express these conditions   as: 12h m( mod 1) and 12m h( mod 1). Thus 144h h( mod 1), and so 143h 0(   mod 1). There are 142 solutions to this equation in the open interval (0; 1): h 2 H1 := f k   143 :   k = 1; 2; : : : ; 142g.   Times when the hour and minute hand perfectly overlap (h = m) are in H1, but these are not   ambiguous times. These configurations occur when 12h h( mod 1). That is, 11h 0(   mod 1). There are 10 such times: h 2 H2 :=   k   11 : k = 1; 2; : : : ; 10   The number of ambiguous times is the cardinality of H1nH2. As H2 H1 the cardinality of   H1nH2 is 142 􀀀 10 = 132, which is answer (d).   correct answer 132
<result>D</result>
A computer simulation of a 12 hour analog clock keeps perfect time while it is running, and
has two hands—an hour hand and a minute hand—both of which move continuously around
the 12 hour dial. (For example, at 2:30, the hour hand will be exactly halfway between 2 and
3.) Because of careless programming, the minute hand looks exactly like the hour hand, so
that the two are indistinguishable.
On one day, the clock stops at some time after 12am and before 12pm. How many times could
it have stopped without it being possible after 12pm to tell what time it stopped?