NUTS TO CRACK

Otago Witness, Issue 4029, 2 June 1931, Page 31

NUTS TO CRACK

by

T. L. Briton.

(For the Otago Witness.) Readers with a little Ingenuity will find In this column an abundant store of entertainment and amusement, and the solring of the problems should provide excellent mental exhilaration. While some of th. " nuts " may appear harder than others, it will be found that none will require a sledge-hammer to crack them. Solutions will appear tn our next issue, together with some fresh “ nuts." Readers are requested not to send In their solutions unless these are specially asked for, but to keep them for comparison with those published in the issue following the publication of the problems FOXES AND GEESE. A correspondent has asked for the publication of Sam Loyd's “ Foxes and Geese ” puzzle which amused and otherwise entertained problem lovers a generation ago. A square section of land has a small enclosure at each of its four corners, A, B, C, and D, and a lane running right round the boundaries inside gives access from one to the other. Situated in approximately the centre of the section are four more enclosures, E, F, G, and 11, but these have no direct road connection with each other. From A there are two roads of access, one to E, the other direct to F, while from B a road leads direct to H, which is closer to B than to any other corner, just as F’s nearest corner is A. A road runs direct also from B to F. An other road leads from enclosure H to the south-eastern corner C, and the latter connects also by road with G, which is situated immediately below enclosure H The fourth corner, D, has also two road communications to the centre, one to G and the other to E, thus making in addition to the boundary lanes eight roads to the centre enclosures from the corners. A fox is in each of the yards F and E, while a goose is located in each of the enclosures G and H, and the problem is to transfer the foxes to the goose enclosures and the geese to the foxes’ quarters in the fewest number of moves using any of the eight internal roads or those along the boundaries. But no enclosure should contain two animals at the same time. WITH FOUR FIGURES. There are only four examples of a number containing four figures, all different, where two of them multiplied by the other two give a result consisting of the same four figures employed. For instance, 27 (twenty-seven) multiplied by 81 (eighty-one) make a product consisting of the same digits, not necessarily in the same order. There are only two examples of a four-digit number in which three of them multiplied by the fourth give identically the same figures in the product. Four hundred and seventy-three multiplied by eight equals 3784. Can the reader find the other three examples in the former case, and the only other one where the multiplier is only one figure and the multiplicand three? But in all cases the number must comprise four different digits.

THREE PIECES OF GOLD PLATE. Draw three circles of the respective sizes of a shilling, a sixpence, and a threepenny piece. These coins are selected because their diameters would probably make a right-angle triangle, that of the shilling forming its hypotenuse, and the diameter of the threepenny piece the perpendicular. If the sizes of the coins do not exactly conform to the making of the figure geometrically, let it be taken for granted that they do, the suggestion to use them being prompted by the fact that the equipment for this problem is ready at hand. Three solid plates of pure gold are in the form described, and the ■ interesting and practical problem is to cut them into the fewest number of pieces possible, so that they may be equally shared between four people, both in weight and superficial area. This is merely a question of practical geometry, and as the thickness of the plate is uniform the “ weight ’’ phase of the problem does not require any calculation. For the information of the would-be solver it may be stated that the number of pieces into which the three plates may be cut in order to achieve the desired result is fewer than eight. PARTNERS. Here is a little question in every-day finance which should interest the reader in so far that it may give him a few moments of hard thinking. Jones and Brown are partners in a Stock Exchange business, and when the profits of last year were divided between them Jones received 5 per cent, on the capital he had invested in the concern, and Brown 4 per cent, on his. Both partners received equal amounts in this way, and the perplexing little question arises, How much did each of them receive as profitf for the year mentioned if the joint capital invested was £22,500? This little question came to hand from “ Student,” who possibly met the problem in an examination paper, though he does not say so, the only comment made being that he and his friends think that it has insufficient data. But perhaps the reader will join issue with the young gentlemen on that point. ELECTION FIGURES. During the progress of the recent municipal the figures obtained from

time to time during the count showed, in several instances, some remarkable coincidences, obvious to any observer who looks for curiosities in figures. One board showed the respective positions of three candidates at the stage when each had polled under 1000 votes, but more than 100. The names were in alphabetical order, and the figures against the name at the top, added to the number polled by candidate No. 2, made a total which equalled the exact figures shown against the name of the third contestant. Here is a little armchair problem arising from this coincidence. Supposing that in these nine figures no digit was repeated, and that the total in the bottom row is the smallest one possible under these conditions, can the reader say what votes the respective candidates had polled at the stage when the board indicated such position? The cipher may be considered a digit, provided it is given a place of arithmetical value. SOLUTIONS OF PREVIOUS WEEK'S PROBLEMS. WITH A SEVEN SQUARE. 1A 2B 30 4D 5E GF 7G 4F 5G 6A 7B 1C 2D 3E 7D IE 2F 3G 4A 5B 6C 3B 4G 5D 6E 7F 1G 2A GG 7A IB 2C 3D 4E 5F 2E 3F 4G 5A GB 70 ID 50 GD 7E IF 2G 3A 4B A STRAIGHT STAIRCASE. Thirty-two. To first step then return to floor, then to first, second, third return to second; then third, fourth, fifth, and return to fourth; fifth, sixth, seventh, return to sixth, and so on. ARITHMETICAL SIGNS. 12 3 minus 4 5 minus 6 7 plus SO equals 100. DIVIDING THE RESIDUE. Six miles of subdivision fences and eight miles on the boundaries. QUERIED. The loss was £7B as already published. The dealer used the hotelkeeper's money, not his own, to give the change. That is where some solvers tripped who queried the result. ANSWERS TO CORRESPONDENTS. “ Taumarunui.”—Yes, there is no solution with the six as has been already stated, but an opportunity to test your ingenuity with a seven square appeared last Tuesday. (2) The answer is 408; see issue of May 9. “ Curious.” —(1) In Belgium and Switzerland the principal coin is called a franc. (2) The “ angel,” an obsolete British coin, had a value equal to the present half sovereign. “ Colenso.”—Much obliged.