INTELLECT SHARPENERS

Otago Daily Times, Issue 21668, 11 June 1932, Page 18

INTELLECT SHARPENERS

Written for the Otago Daily Times. By T. L. Buitox. HALVING THE CIRCUMFERENCE, It was the practice of a market gardener who specialised in the cultivation of asparagus to have the vegetable made up in fairly large bundles of size to sell at one shilling per bundle. These were more or less cylindrical in form, but for the purposes of a little problem it may be assumed that their circumference was spherical and exactly 18 inches round. The gardener, in view of the slump, decided to have the asparagus made in smaller bundles to sell at half the price, and with that object part of his crop was made up in the same shaped bundles but was only half the circumference of those’ he sold at one shilling each. There was no difference in the make-up as all the asparagus sold by the gardener was tightly and uniformly packed both in the large and smaller bundles. In view of this, can the reader say whether the purchaser of the original-sized bunches 18 inches in circumference at one shilling per bundle received better, lower, or as good value than hig neighbour who bought two bunches each nine inches in circumference for the same money? AGISTMENT OF HORSES. Three horse dealers, X, Y, and Z, had between them 50 horses for which they could get no bids in the sale yards. A well-grassed paddock wag therefore leased for the animals' agistment, the agreement being that they could have the grass rights for £24 for the 50 horses. The firstnamed dealer was the owner of only five of the 50 horses and these were kept in the paddock for six months,, Y’s 15 animals 'remained there for eight months, while the others, namely, 30 horses belonging • to Z, were taken out of the field at the end of three months, and it was agreed between them that the amount of £24 should ,be paid by the three dealers upon that basis. Before the owner of the paddock gent in his account for the rental, the three men had gone bankrupt, and the question is how much the lessor received from the three men, if X paid 10s in the £_ and the other t>vo only 10s in the £ on their respective accounts? A PARADE OF THOROUGHBREDS. A number of thoroughbreds on competitive exhibition was being paraded on a circular track within the local show grounds, each class being shown seperately. In the 16 hands’ section there were fewer than 30 horses. There were several of each of the usual colours — hays, blacks, browns and chestnuts—hut only one grey amongst the lot. They were paraded iu single file with a uniform distance between, and from the few details following, the reader will he able to determine the exact number of horses paraded. Five-ninths of the number of horses in front of the grey horse added to one-half the number of those behind that particular thoroughbred represents the number that were being shown of that class. The question is how many were there? The correct answer should be quickly found without using pen or pencil, and if the would-be solver will adopt this method, he will find that the question affords an excellent opportunity for enjoyable mental exercise, for it is quite liltelv to demand a few moments of hard thinking. s A COUNTER PUZZLE. An interesting little counter puzzle which involves, a simple arithmetical caleolation is played .by two persons with 15 counters. These are placed upon the table in one heap and each player takes at a time olie. two or three, the winner being the one who has the odd number when all the counters have been drawn. This is, of course, a simple enough procedure if the result is left to chance by haphazard drawing, any number of counters being taken at random, hut it is quite possible that an, investigation by a player will reveal a reliable rule enabling him to make a certainty of winning every time, particularly if he has the first draw. This puzzle-game will he found well worth a little study with the object of finding a method or system by the use of which the first player can never lose, simple ns it may appear to block him. According to Competitions, a sennniatliematical journal published in London, Disraeli was fond of this kind of mental relaxation and greatly enjoyed introducing this particular puzzle-game to an uninitiated friend. » 4 SEA SCOUTS. A correspondent, “ One of Them,” has sent the following little question;—During an encampment of the Sea Scouts, most of the sporting events were performed on the water and their few flatbottomed boats, none of which was ever out of daylight commission, were the chief equipment in use. None of these craft would hold more than three boys, and the® correspondent asks what is the fewest number of them that are necessary for 15 Scouts to row out at the same time on every day in one week, exclusive of Sunday, three Scouts being in each boat. A stipulation is that no two Scouts should be together in any boat more than once, and that no one should go out more than once in the same boat. The correspondent says that this formed- one of the many puzzles given by the Scoutmaster to solve around the camp fire at night, before “ lights out ” was sounded. The problem is one of the Well-known “Kirkmari’s schoolgirl problems,” which have been interesting much older people than Sea Scouts for a very long time, and are always new.i .SOLUTIONS OF LAST WEEK’S PROBLEMS. A SWIMMING RACE. It depends upon the width of the stream. Less than one hundred and sixteen feet would enable the hoy to win, hut if over that distance his sister would heat him. A FISHING EXPEDITION. “B ” group won the distinction with 42, “ A ” securing half that number, whilst the other two groups tied with 35.. INCOMES OF TWO PARTNERS. £3OO and £250. THE WIDTH OF A ROAD. The road was one chain wide and therefore two and a-half acres in area, its value being £25. BY PROCESS OF REASONING. As indicated last week here is the method of deduction. In the first sum it is clear that 396 must he the difference of 99a and 99c, so that “ A ” minus “ C ” equals 4. In the second sum it will he noted that 297 is the difference of 99a and 99h, so that “A” minus “ B ” equals 3. Therefore the minimum value of ” A ” is 4, and if we take this or any other value and increase “ A ” by unity, the effect is merely to increase the value of each group by 111 in its turn, so that the results of subtraction remain the same. The possible values of ABC are therefore 410, 521. 632, 743, 854, or 905. ANSWERS TO CORRESPONDENTS. B. T.—Yes, it is an interesting little tangle which has cropped up before in other guises. “ Curious.”—lt is hardly a mathematical question, hut your view is the one generally accepted as correct. jf course, there is almost an unlimited number of different ways of varying it. “Homer.”—Yes, the context infers that the stakes between A and B in the Phnr Lap-Gloaming contests are level. D. IT. W,— (1) Not a fallacy; (2) 10, 1, and 80. J. S.—Thanks. Other item sent on. “ Mark.” —Appreciated.