NUTS TO CRACK

Otago Witness, Issue 4048, 13 October 1931, Page 9

NUTS TO CRACK

By

T. L. Briton.

(Fob the Otago Witness.)

Readers with a little ingenuity will find In this column an abundant store of entertainment and amusement, and the solving 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 tn their solutions unless these are specially asked for, but to keep them for comparison with those published tn the Issue following the publication of the problems A POSER. It is quite possible that this cryptograph will be judged by the reader to be equally as difficult as the more formidable cipher, and perhaps to have earned its name, “ a poser.” It was shown to a lady who it said to be expert at this kind of puzzle problem, but she could make nothing of it after an hour’s study The item has been taken from the news columns of this paper, and, although in the construction of the code no extraneous matter has been introduced, the reader may have to don his best thinking cap before converting the passage into an intelligible sentence. ECPM VETAHT NEOYIE SEHE EEIUAG REFCS R NSNAS RTOTTN EBEIPE OITSE YANITI SXLRH TNIM HRWSHR DVATOOT IYESI OILEA EEHRPOL FGET OILAFL EAEFL TORSE RLIIC CVHTT RUTH UEDCNE NATHN CAEHT RTENAD SRAGTAU A GAME OF CARDS. After the strenuous efforts caused by the last problem, here is one that can well be solved while the reader reposes in his armchair. Three persons engage in a game of cards, each playing for himself and for a small stake. Before starting play each put his money upon the table, no further money being produced, and it was noted that X had ninepence more than Y and Z together, ail having different sums. X lost the first of the three games and, when paying over to the other two, he gave to each of them exactly ..s much as each of those players had, Y securing more points than Z. When Y lost the next game Z was the more fortunate of the two, the number of points made by him being four times as many as X secured. But the third and last game was lost by Z, and each of the others received from him as much as he had, both sums being the same amount. The question is, If when Y paid his losses to X and Z each of those two players had their money doubled, and at the end one of the players had lost three shillings and ninepence, how much did they start with respectively, all having equal amounts at the finish of the game?

A DISSECTING PUZZLE. Here is a useful dissecting geometrical problem that should interest the reader of a practical turn of mind. Make a diagram of two squares adjoining and letter them A, B, C, D and B, E, F, C, respectively, the line BC being common to both figures. After drawing two parallel diagonals, AC and BF, cut off the triangle BFE and a four-sided figure in the form of a trapezoid will be made, the four corners of which are ABFD. It is required to divide this quadrilateral into four parts of equal area and shape, and to find \vhat will be the size of each of those equal portions if the base of the original figure, namely, UF, is two feet in length. AN ALPHABETICAL SUM. Another test of the readers’s ingenuity should be provided by this alphabetical sum, which, after the experience of a similar kind some weeks ago, will not be altogether strange to him. Each letter used has the same numerical value throughout, and although all the digits do not come into use the nought may be considered to be one of these, should it be found necessary to employ it. The seven letters used form the words “ NOT EASY.” The sum is to divide TEAAO by NO, the answer to which is OSO. Proceeding to work it alphabetically the divisor multiplied by O gives TAO, and being subtracted leaves YY with the A brought down. The second multiplication gives YTH, the latter being the eighth letter used, and when subtracted in the proper manner leaves TA. It will be obvious to the reader that when the last letter 0 is brought down, making TAO, the last multiplied number word is the same, and therefore there is no remainder. Can the reader find the proper equivalents for these eight letters respectively? A CALENDAR PROBLEM. Here is an excellent calendar problem by a French mathematician, M. Fourrey, which, though of ancient origin, will perhaps be new to most readers. It is based on the present or Gregorian calendar, which, as the reader is aware, provides for the omission of the extra day in February (the 29th) three times in every 400 years. This is his problem. In the century and a-half between 1725 and 1875 the French won a certain battle on April 22 in a certain year, and exactly 4382 days afterwards, also

on the same day and the same month, they gained another victory. As the sum of the digits of those two victory years is 40, Fourrey asks, what are the respective dates of the two battles? There can be only one solution of this problem, although there the many twoyears within the period mentioned, the sum of whose digits is 40. SOLUTION OF LAST WEEK’S PROBLEMS. ON A DEFINITE PLAN. The passage is “ The finding of the Maori oven on the Stratford side of Mount Egmont has been followed by scientific investigations which reveal that Egmont erupted within the past 500 years.” AN EVERY-DAY OCCURRENCE. As the problem stated that there were more than two in the party, each received one at 3d per lb, and 21b at 2d per lb, six people altogether. THE “ LIMITED.” The distance between “ A ” and “ B ” is 180 miles. TAXING PETROL. As there could have been only one drum left, the 30 gallons vessel, 15s was payable as duty taxes. A BOX WITHOUT A LID. If two-inch-square pieces be cut from the four corners, one from each, the box would have the largest holding capacity possible with the material stated. ANSWERS TO CORRESPONDENTS. “ Capacity.”—(l) A litre measures less than a quart, being slightly less than one and three-quarter pints. (2) A ratio of one quantity to another is not altered when they are multiplied or divided by the same number, but it is when the same number is added to both. G. Mac D.— (1) There is one more example than the number you mention. (2) The “ cut ” problem tripped many others besides yourself.