NUTS TO CRACK

Otago Witness, Issue 4032, 23 June 1931, Page 65

NUTS TO CRACK

By

T. L. Briton.

(For the Otago Witness.)

Readers with a little Ingenuity will And 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 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 tile problems A SMART NEWSBOY. A quick reply given by a diminutive 10 year-old newsboy to his mate who had asked a derisive question as to the price of his papers revealed that the youngste' was very smart with figures, at any rate when they concerned his own affairs. It is a curious fact concerning the typ>known as “ calculating prodigies ” that, when they are taught the elementary rules of arithmetic, they seem to lose their mysterious power. “ How much are you charging for the papers to-day," his mate. “ Same as the office charge." the newsboy replied, " and if I paid a» much as I sell them for the office would receive from me for (50 dozen exactly 40 more threepenny pieces than the num ber of papers I would receive from the office and sell for two of the ’ best ’ ” — by “ best" meaning sovereigns. The reader will no doubt find no use for pen or pencil in solving this little armchair problem, the question being how much per dozen were the papers? BOAT ROWING. Here is a little problem in combinations which, though not presenting any difficulty, should give the reader five minutes of good mental exercise, particu larly if he adopt the “ trial ” method of discovering the correct answer. Fifteen members of a rowing club practised regularly every day last week, including Sunday. Every one went out at precisely the same time. There were always three in a boat, including the coxswain, but never more, and during the period referred to no two rowers went out in a boat together more than once, and not one of the 15 went twice in the same boat. Although there were plenty of suitable boats to choose from, it so happened that during this practice week the number of different boats used by the rowers was the smallest number possible under the conditions described. Can the reader say how many different boats were used in that week?

PRACTICAL DISSECTION. Draw a square on a sheet of paper and letter it A, B, C, D, reading round the perimeter clockwise, from the top left-hand corner. Delete the line BC, but not the letters, and produce the line DC to E, making CE equal to a side of the square. Then, join the points BE. The quadrilateral thus described will be the original square and a-half, and the interesting little geometrical problem is to divide this figure into four equal parts all of exactly the same shape, using not more than four lines for the purpose. It requires no mathematical knowledge beyond the elementary but some ingenuity will be needed in order to find how four symmetrical foursided figures of similar shape and size can be formed out of the trapezoid. The problem is both interesting and useful. IN ANOTHER WAY. Here is another way of proceeding to construct a magic square. Arrange 15 counters in a 16 square in regular sequence, beginning with one at the top left hand corner and reading horizontally to the cell in the bottom row nearest the right hand corner, which contains counter 15. By moving (perpendicularly or along the horizontal rows) the counters one at a time to the adjoining vacant cell, it is required so to arrange the whole 15 to make a magic square in which all rows, columns and long diagonals will add up the same With plenty of patience this can be achieved without any very serious mental strain, but if it is also required that the number of moves taken to accomplish this result must be the fewest number possible under the conditions, some thought, in addition to the reader’s patience, will be essential to success. What is the fewest number required? It is immaterial which cell is vacant at the finish, but none must be occupied by two counters at the same time. A MOVING STAIRCASE. The moving staircase has come to New Zealand, and most readers are conversant with their mechanism. Here is a little problem concerning one of these structures on the London Tube Railway propounded by the well-known mathematician, H. Dudeney. A person walking down 26 steps of the staircase from the top requires 30 seconds altogether to get to the bottom, but if he makes 34 steps instead of 26 he will need only 18 seconds to get to the bottom. Can the reader calculate the height of the staircase measured by steps if the time be taken from the moment the top step begins to descend to the time that the person steps off the last step at the bottom on to the level platform?

SOLUTIONS OF LAST WEEK’S PROBLEMS. ANOTHER CODE. The art of their construction, intelligible to those who know the key and otherwise to those who do not, has been studied for centuries, and during war or rebellion their correct interpretation is of great service to the opposite side if discovered. WITH 10 COUNTERS. 715 multiplied by 46 will give the remaining figures constituting the nine digits and cipher, namely, 32890. A SIMPLE GEOMETRICAL PUZZLE. Six straight lines or cuts will divide any circle into 22 parts, every cut to intersect another, but more than two lines must not intersect at the same point. Rule: Multiply the number of lines or cuts by that number, plus 1. Divide the result by 2 and add 1. IN A NINE SQUARE. 1.9.2 2.1.9 2.7.3 3.8.4 4.3.8 5.4.6 5.7.6 6.5.7 8.1.9 UP AND DOWN HILL. Six and three-quarter miles. ANSWERS TO CORRESPONDENTS. B. C. F.—The essential feature of cryptographs is that the communication may be given to the world, yet be unintelligible to everyone except those who possess the key “Beta.”—Thanks; will be used in due course provided that, upon examination, its solution does not involve a process unsuitable to the average reader. “Equilibrium.”—Yes; resting in equilibrium under its own weight, the centre of gravity must be vertically above its base. “ Election Figures.”—The statement of the problem assumed the cipher to be a digit, and there were only nine figures involved. Thanks for “ road ” item.