There are thirty-six pairs of times when the hands exactly change places
between three p.m. and midnight. The number of pairs of times from any
hour (n) to midnight is the sum of 12 - (n + 1) natural numbers. In
the case of the puzzle n = 3; therefore 12 - (3 + 1) = 8 and 1 + 2 + 3
+ 4 + 5 + 6 + 7 + 8 = 36, the required answer.
The first pair of times is 3 hr. 21+57/143 min. and 4 hr. 16+112/143
min., and the last pair is 10 hr. 59+83/143 min. and 11 hr. 54+138/143
min. I will not give all the remainder of the thirty-six pairs of times,
but supply a formula by which any of the sixty-six pairs that occur from
midday to midnight may be at once found:--
720b + 60a 720a + 60b min.
a hr ---------- min. and b hr. ---------------
143 143
For the letter a may be substituted any hour from 0, 1, 2, 3 up to 10
(where nought stands for 12 o'clock midday); and b may represent any
hour, later than a, up to 11.
By the aid of this formula there is no difficulty in discovering the
answer to the second question: a = 8 and b = 11 will give the pair 8 hr.
58+106/143 min. and 11 hr. 44+128/143 min., the latter being the time
when the minute hand is nearest of all to the point IX--in fact, it is
only 15/143 of a minute distant.
Readers may find it instructive to make a table of all the sixty-six
pairs of times when the hands of a clock change places. An easy way is
as follows: Make a column for the first times and a second column for
the second times of the pairs. By making a = 0 and b = 1 in the above
expressions we find the first case, and enter hr. 5+5/143 min. at the
head of the first column, and 1 hr. 0+60/143 min. at the head of the
second column. Now, by successively adding 5+5/143 min. in the first,
and 1 hr. 0+60/143 min. in the second column, we get all the _eleven_
pairs in which the first time is a certain number of minutes after
nought, or mid-day. Then there is a "jump" in the times, but you can
find the next pair by making a = 1 and b = 2, and then by successively
adding these two times as before you will get all the _ten_ pairs after
1 o'clock. Then there is another "jump," and you will be able to get by
addition all the _nine_ pairs after 2 o'clock. And so on to the end. I
will leave readers to investigate for themselves the nature and cause of
the "jumps." In this way we get under the successive hours, 11 + 10 + 9
+ 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1 = 66 pairs of times, which result agrees
with the formula in the first paragraph of this article.
Some time ago the principal of a Civil Service Training College, who
conducts a "Civil Service Column" in one of the periodicals, had the
query addressed to him, "How soon after XII o'clock will a clock with
both hands of the same length be ambiguous?" His first answer was, "Some
time past one o'clock," but he varied the answer from issue to issue. At
length some of his readers convinced him that the answer is, "At 5+5/143
min. past XII;" and this he finally gave as correct, together with the
reason for it that at that time _the time indicated is the same
whichever hand you may assume as hour hand!_