REVELS WITH ROPES

New Zealand Listener, Volume 3, Issue 77, 13 December 1940, Page 55

REVELS WITH ROPES

climbing the rope we can't hold back L. Wilson’s rope problem for a moment longer than necessary. All readers require to solve it satisfactorily is a Boy Scout book about knots, two or three hundred feet of one-inch cable, a pirate galleon, and the captain’s gig. PROBLEMS Rope Trick ~- It is suggested that readers give themselves a practical model to play with for this one, It is from L. Wilson, Christchurch, A pirate captured more prisoners than his craft would carry, so he put some in a boat to be towed behind. He made sure they had nothing that would cut the rope, and then bent on a double painter, which was looped round the after thwart of the boat, with a clove hitch round the forward one, and thence went through a ring in the bow to a bitt on the ship. The prisoners escaped. How? The rope was not cut and we presume the thwarts were not broken. A the monkey is still Harvest A farmer and his son agree to share equally in the labour of cutting the corn crop in a square field. Father comes in and cuts a strip a rod wide right round the outside of thé field. He tells his son that he’s done his share and walks off. If he was right, what was the area of the field?-(From R.C.J.M.) Savings A parent puts in a child’s money box every birthday 2/6 for every year of its age. How old would the child be when the money in the box totalled £17?(From X.G.T.) Sharpeners These three come from A.E.V. (Hamilton). Digits Again: A number of two digits is such that if we divide it by the sum of its digits we get a quotient three and remainder three. Also, it is less than the number secured by reversing the digits by four and a half times the sum of the digits. What is the number? Number Please: Given that two numbers have an L.C.M. of 1859, an H.C.F, of 13, and that one is 169, find the other. Wheat: A farmer sells a certain number of bushels of wheat at 7/6 a bushel and 200 bushels of barley at 4/6 a bushel. He receives altogether as much as if he had sold both wheat and barley at the rate of 5/6 a bushel. How much wheat did he sell? ANSWERS (See issue of November 29) Bar: The publican neither lost nor gained, He was paid for just as many drinks as he served.-(Problem from P, Mora.) Tote: R.G. comments on P. Mora’s problem: Mathematically, neither. The tote really (and usually) did in practice. The fact that the totals of the sums

left equalled one pound more than the amount invested is immaterial. By selecting different amounts hundreds of different results may be obtained; in fact, every whole number from 1 to 1540. Age: Everyone agrees on 78 years.(Problem from Sylvia, who, womanlike, omitted to supply the answer.) Professionals: ‘By 'Thursday of last week the mail had brought only two answers to Laurence Hayston’s problem. R.G. said: Green, humorist; Brown, poet; White, essayist; Black, novelist; Grey, playwright; Pink, historian. P.J.Q. said: Grey, essayist; Brown, novelist; Green, humorist; Pink, historian; White, playwright; Black, poet. Keep it Down: Just keep up keeping it down. Smokes: 2/5, says R.C.J.M. Exchange: The problem set by R.C.J.M.-Move the Money, on September 13, established the principle for working out the moves in problems such as Q.E.D.’s. Keep trying. Matrimonial: A.G. says: " Yes-by a previous marriage. He marries one woman, and after her death he marries her sister and then he dies. His second wife is his widow, his first wife is his widow’s sister," Metalwork: X.G.T. says: Base-3 ft. 734 inches square; height-1 ft. 10 in. More Geometry: There was insufficient space last issue to give A,.G.’s detailed proof of his answer. Here it is now: Let EC be x Then AC.CE = 15,000 x X = BC’. Therefore 15,000 x X is a perfect square. Therefore 6X is a perfect square. Therefore X is of form 6y’ where y is an integer. ;

The highest value of y is given by: AE.EC = EB’. Therefore (15,000 — 6y*) 6y*® = 5,000’ -I, Therefore y* = 327 as a maximum. From I (15,000 — 6y*) 6 must be a perfect square, Therefore (50°- y’) must be a perfect square, and its lowest value is 50* — 327 = 2173 = 46.7? Therefore 50° — y* must be either 49’ or 48° or 47°. Therefore y* = 50° — 40? = 99 or 50? — 48* = 14" or 50 -47 =3 x 97. The only solution for y’ to be a perfect Square is y = 14, Therefore BC* = 15,0007 x 6y*. Therefore BC-= 4200, Similarly BE = 4032 EF = 343 AB = 14,400 This proves there is only one solution.