"NUTS!"

Evening Post, Volume CXIII, Issue 61, 14 March 1927, Page 17

"NUTS!"

[intellect sharpeners I = . No. XVIIL f

| (By T. L. Briton.) |

I All rights reserved. | Eeaders 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 the "nuts" may appear harder . ' than others, it will be found that none will require a sledge-hammer to crack them. COMPLICATED. A correspondent has sent rue particulars of a queer relationship. It is not a hypothetical case, as the correspondent knows intimately the parties concerned. The head of the family was one of the leading men of the district where he resided many years, being the storekeeper, postmaster, Justice of the Peace, Coroner, and Registrar—a veritable Pooh-Bah. P.B. was married and had a son, S, his wife dying shortly afterwards. A year or so later P.B. married E, the issue being two daughters, D and C. Now B had a younger sister, X, who afterwards married S, and they had a son, Y. My informant asks me -to let him know the respective relationships of the family, but as that would deprive the reader of an interesting little problem, I shall limit myself to stating what seems to be obvious, viz., that P.B. and S, besides being father and son, are also brothers-in-law, and that D and C are really stepsisters to their aunt X. Can the reader say how V and D stand in point of relationship? HALVES AND THIRDS. The following three rows of figures include the nine digits, obviously none being repeated. It will be noted that the top row is a number equal to one-half of the one in the second row, and exactly one-third of the bottom number. There is a low limit to the number of different arrangements that can produce the same result, being fewer than half a dozen, where the nine digits are used once and once only. Can the reader discover them? This problem makes an excellent little game of patience, without the inconvenience of haying to carry a pack of cards or other equipments 2 7 3 5 4 6 8 1 0 THE DENTIST'S GOLD. The accountant of a firm of dentists, after checking the invoice of five parcels of gold received from the wholesale' suppliers, and verifying the weights separately, decided to weigh them in combinations of two, making ten operations of the process. The latter weights were 30, 32, 33, 34, 35, 36, 37, 38, 40, and 41 pennyweights inclusive of the wrappers. These figures were then handed to the assistant, the accountant explaining what they were, and asking him to calculate with that data, the individual weights of the packets. After a little time the assistant produced his results which agreed with the actual weighings made by the accountant. What were the individual weights of the five packets? COINS. It is recognised that the intrinsic value of a sovereign is its nominal value of one pound, and that if one be cut into halves each would be worth ten shillings, or, at any rate, ten shillings less the cost of remintage. A silver or copper coin, however, would be rendered practically valueless if* dissected. But in the following problem let it be assumed that the intrinsic value of a shilling is one-twentieth of a pound or sovereign, and that a threepenny piece and a penny is one-quarter and one-twelfth of a shilling respectively; so that if either of the silver or copper coins be broken in parts would retain their proportionate value. The problem then is that, supposing the same fractional part of a shilling, a threepenny piece, and a penny be broken off each, what would be the value of the remaining parts of each coin, if together they were worth exactly one shilling? ■:-.--■,-. ■:• CHANCES. The chances that the average person is prepared to take when it comes to a matter of sport, generally bear an inverse ratio to the probabilities of the speculation being a profitable one, because in such cases the mathematical aspect of the venture is seldom if ever looked into. Here is a little problem, however, of which the mathematical solution is sought, instead of one based on any theory of probabilities. A box contains three half-sovereigns and one sixpence. What ia the correct amount that should be paid for permission to draw for "keeps" any one of them, assuming that the proportion of gold coins drawn is in the .same ratio to the silver coin as they are in the box? SOLUTIONS OF LAST WEEK'S PROBLEMS. Two Walkers.—The two gentlemen G and H met exactly one quarter of a mile from X. Budding Mathematicians.—The pear cost twopence, being 2Mi per cent, of the price of three and one-third dozens, viz., 6s Bd. Ten dozen would therefore cost half aa many shillings aa the number of pears that could be purchased for the amount of the bill, which was 6s Bd. The Draughtboard.— The pieces should be placed on the following squares:—lo 2E, 38, 4H, SA, 6G, 7D, and BP, when it will be found that no two of them are m the same line, vertical}-, horizontally, or diagonally. By starting off with the eight pieces in the top row, the correct positions can be attained in 23 moves, one square at a time. Everyone Shared.—There were ten men who received £1 10s, fifty women getting ±5 between them, and one hundred and torty children who received £3 10s, being £10 amongst 200 people. Chronometers.—The first time after midday on Ist January, 1899, when the three chronometers showed exactly similar time was at noon on 12th December, 1902. The pitfall that tiio reader was liable to encounter was that 1900, though divisible by 4, was not a leap year. Answers to Correspondents.—"J.C.L.": Acknowledged through the post to avoid delay. "Miss M.S.": It was clever of you, as it was correct where quite a number «mr™ „triPß?,d "P- "I'-0'8.," "Melb.," J..J.. : Vyill appear next week.