Answer :

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:

a
hr |
720b + 60a |
min.
and b hr. |
720a + 60b
min. |

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!*