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!_ 62.--THE CLUB CLOCK. The positions of the hands shown in the illustration could only indicate that the clock stopped at 44 min. 51+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. The 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.