Page:Amusements in mathematics.djvu/166

154 hours by 11 we get 1 hr. 5 min. $27 3⁄11$ sec, and this is the time after twelve o'clock when they are first together, and also the time that elapses between one occasion of the hands being together and the next. They are together for the second time at 2 hr. 10 min. $54 6⁄11$ sec. (twice the above time); next at 3 hr. 16 min. $21 9⁄11$ sec.; next at 4 hr. 21 min. $49 1⁄11$ sec. This last is the only occasion on which the two hands are together with the second hand" just past the forty-ninth second." This, then, is the time at which the watch must have stopped. Guy Boothby, in the opening sentence of his Across the World for a Wife, says, "It was a cold, dreary winter's afternoon, and by the time the hands of the clock on my mantelpiece joined forces and stood at twenty minutes past four, my chambers were well-nigh as dark as midnight." It is evident that the author here made a slip, for, as we have seen above, he is 1 min. $49 1⁄11$ sec. out in his reckoning.

61.—CHANGING PLACES.

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. $$\frac{720b+60a}{143}$$ min. and $$b$$hr. $$\frac{720a+60b}{143}$$ min.

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. $58 106⁄148$ 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 0 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!

62.—THE CLUB CLOCK.

positions of the hands shown in the illustration could only indicate that the clock stopped at 44 min. $5 1143⁄1427$ sec. after eleven o'clock. The second hand would next be "exactly midway between the other two hands" at 45 min. $52 496⁄1427$ sec. after eleven o'clock. If we had been dealing with the points on the circle to which the three hands are directed, the answer would be 45 min. $22 106⁄1427$ sec. after eleven; but the question applied to the hands, and the second hand would not be between the others at that time, but outside them.

63.—THE STOP-WATCH.

time indicated on the watch was $5 5⁄11$ min. past 9, when the second hand would be at $27 3⁄11$ sec. The next time the hands would be similar distances apart would be $54 6⁄11$ min, past 2, when the second hand would be at $32 8⁄11$ sec. But you need only hold the watch (or our previous illustration of it) in front of a mirror, when you will see the second time reflected in it! Of course, when reflected, you will read XI as I, X as II, and so on.

64.—THE THREE CLOCKS.

a mere arithmetical problem this question presents no difficulty. In order that the hands shall all point to twelve o'clock at the same time, it is necessary that B shall gain at least twelve hours and that C shall lose twelve hours. As B gains a minute in a day of twenty-four hours, and C loses a minute in precisely the same time, it is evident that one will have gained 720 minutes (just twelve hours) in 720 days, and the other will have lost 720 minutes in 720 days. Clock A keeping perfect time, all three clocks must indicate twelve o'clock simultaneously at noon on the 720th day from April 1, 1898. What day of the month will that be?

I published this little puzzle in 1898 to see