It is required to name the quotient of five or three lines of figures--each line consisting of five or more figures--only seeing the first line before the other lines are even put down. Any person may write down the first line of figures for you. How do you find the quotient? 86,214 42,680 57,319 62,854 37,145 ------ 286,212 When the first line of figures is set down, subtract 2 from the last right-hand figure, and place it before the first figure of the line, and that is the quotient for five lines. For example, suppose the figures are 86,214, the quotient will be 286,212. You may allow any person to put down the two first and the fourth lines, but you must always set down the third and fifth lines, and in doing so always make up 9 with the line above. Therefore in the annexed diagram you will see that you have made 9 in the third and fifth lines with the lines above them. If the person you request to put down the figures should set down a 1 or 0 for the last figure, you must say: “We will have another figure,” and another, and so on until he sets down something above 1 or 2. 67,856 47,218 52,781 ------ 167,855 In solving the puzzle with 3 lines, you subtract 1 from the last figure, and place it before the first figure, and make up the third line yourself to 9. For example: 67,856 is given, and the quotient will be 167,855, as shown in the above diagram. The Remainder. A very pleasing way to arrive at an arithmetical sum, without the use of either slate or pencil, is to ask a person to think of a figure, then to double it, then add a certain figure to it, now halve the whole sum, and finally to abstract from that the figure first thought of. You are then to tell the thinker what is the remainder. The key to this lock of figures is, that _half_ of whatever sum you request to be added during the working of the sum _is the remainder_. In the example given, 5 is the half of 10, the number requested to be added. Any amount may be added, but the operation is simplified by giving only even numbers, as they will divide without fractions. Think of 7 Double it 14 Add 10 to it 10 -- Halve it 2 ) 24 -- Which will leave 12 Subtract the number thought of 7 -- _The remainder_ will be 5 The Three Jealous Husbands. Three jealous husbands, A, B and C, with their wives being ready to pass by night over a river, find at the water-side a boat which can carry but two at a time, and for want of a waterman they are compelled to row themselves over the river at several times. The question is, how those six persons shall pass, two at a time, so that none of the three wives may be found in the company of one or two men, unless her husband be present?