(related to Problem: Changing Places)

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. $21\frac{57}{143}$ min. and $4$ hr. $16\frac{112}{143}$ min., and the last pair is $10$ hr. $59\frac{83}{143}$ min. and $11$ hr. $54\frac{138}{143}$ 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: — $$a \text{ hr }\frac{720b + 60a}{143} \text{ min. and }b\text{ hr }\frac{720a + 60b}{143} \text{ min.}$$

For the letter $a$ may be substituted any hour from $0, 1, 2, 3$ up to $10$ (where naught 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. $58\frac{106}{143}$ min. and $11$ hr. $44\frac{128}{143}$ min., the latter being the time when the minute hand is nearest of all to the point IX — in fact, it is only $\frac{15}{143}$ 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 h. $5\frac{5}{143}$ min. at the head of the first column, and $1$ hr. $0\frac{60}{143}$ min. at the head of the second column. Now, by successively adding $5\frac{5}{143}$ min. in the first, and $1$ hr. $0\frac{60}{143}$ min. in the second column, we get all the eleven pairs in which the first time is a certain number of minutes after naught 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 adding 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 $5\frac{5}{143}$ 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!

Thank you to the contributors under CC BY-SA 4.0!



Project Gutenberg

  1. Dudeney, H. E.: "Amusements in Mathematics", The Authors' Club, 1917

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