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NUTS TO CRACK

By

T. L. Brison.

(For 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 the “ 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 the problems

SCOPE FORMNG ENUITY.

Here is a puzzle in figures, which besides its novelty, should prove interesting to the reader, with full scope for his ingenuity, for, without the exercise of that qualification with a modicum of patience, the puzzle will be difficult to solve. It has been taken from an American monthly publication of 1921, but a copy of the following issue with the solution is not to hand. The puzzle is in the form of a long division sum from which most of the figures are mising, and the reader is asked to restore it to its original state which had been completed and was mathematically correct. There are three figures in the divisor, six in the dividend and three in the quotient, the whole twelve being missing. The result of the first multiplication of the divisor by the first figure of the quotient reads x 0 x x, the x representing a deleted digit. After subtraction the figures read x x x x, the last being the figure brought down from the dividend in the prescribed manner. The result of the next multiplication is x 5 0 x, the remainder after subtraction being x x x also including the figure brought down. The product of the last multiplication is x 4 x, and there is no remainder. Can the reader restore the figures of the sum? HORSES IN THEIR STALLS. A horse-trainer had seven stalls in a circular shaped building, each occupied by a horse in training. The order of the horses in the boxes on Monday night was T. U. V. W. X. Y. Z. (identifying each by a letter, (in stalls 1,2, 3,4, 5, 6, and 7 respectively, the last numbered stall being between No. 6 and No. 1. The stable boys did not reserve the same box for the same horse every night, the practice being to put any animal that was brought in in any box that was empty at the time. But suppose that on Monday they were in the order given, and that neither on Tuesday night nor Wednesday night did any horse have as his neighbour a horse that had previously occupied a stall next can the reader say how the stalls were occupied on the two latter nights, with the following additional stipulation—that the horse “ U ” occupied a box as near as possible to “T ” and “Z ” one as far away from him as possible under the condition as previously laid down, viz., that no horse should have the same equine as his neighbour more than once in the three nights. AN ELEMENTARY QUESTION. Here is an elementary question conrerning the respective ages of two brothers, but it is one that may puzzle the reader to answer correctly right off. The difference between the ages in years of Jack and his brother Fred is four more than the age of the latter and twothirds of the age of the former. Two years ago Jack was four times older than his brother, but now he is only three times as old. What are the ages of the two respectively? Although there is no catch in this question it has to be carefully read and understood before the necessary calculation is attempted,, otherwise the would-be solver’s mathematics may be astray. THIRTY-FIVE FROM FORTY-FIVE. Forty-five counters are placed on a square board of 49 cells, seven by seven, and from them it is required to remove 35 so that the 10 that are left will form five straight lines in any direction with four counters in each of those lines or rows. If the 49 cells comprising the square be numbered consecutively from one, commencing from the top left hand corner and reading horizontally, it will be easy to identify the cells which are vacant originally. They are the fourth and the seventh cells in the top row and the first and the seventh in the fourth row, their numbers being 4,7, 22, and 28 respectively. It may require some patience, in addition to the skill which must be brought into action, to find what cells will be occupied by the 10 counters, one on each square, when they form five straight rows of four counters each in the manner described. Can the reader arrange these? FOUR RUNNERS. In a foot race over a distance of one mile there were four runners. The result was a dead heat between two of them for first place, and the other two breasted the tape together, two seconds afterwards. Let us have a little problem concerning the winning two, whose sectional times were taken, and for the purpose of finding the exact time they took to run the mile we will assume that they ran together the whole way. The first half

mile took exactly the same time as the remaining part of the race, but the distance between the pegs at the 40 chains and the 60 chains was run in exactly two seconds slower time than was taken in covering the length between the mark at the starting point and that at the point indicating the quarter mile. If the two winners started together off scratch, and ran the first 60 chains in exactly three minutes and two seconds, and if the last 20 chains took four seconds less than the third quarter of the mile, can the reader say in what time the winning deadheaters ran the full distance? SOLUTIONS OF LAST WEEK’S PROBLEMS. AN ALPHABETICAL SUM. 417689325 are the digital equivalents of the letters R S T U V W X Y Z respectively. A PECULIAR NUMBER. The number is 12345679. If 5 be the digit selected as the first multiplier and the product multiplied by the key digit 9, the peculiarity described will be seen. TWO AT A TABLE. The party must have stayed until after the third meal on the following Friday. TESTING TWO CANDLES. They burnt for two and a-quarter hours, the portion left of the larger being six inches, and five of the other. A QUESTION OF AREA. As the south boundary of the farmer’s block when extended to the river bank is exactly one mile in length, the section in dispute contained 50 acres and cost the farmer £lO5O.

ANSWERS TO CORRESPONDENTS. “ Alphabetical.”—The crux of the prob-lem-puzzle is not to try to place the highest value in every square available under the conditions, even though the object is to place the maximum value on the whole board. If that method is followed some of the squares must be vacant if the stipulations are strictly observed, resulting in a lessened total. “ Zeno.”—(l) The fallacy of Achilles and the tortoise lies in the use of the word “ never.” It is impossible to think of a minimum measure of time as it is also its infinite divisibility. (2) Thauks for your interest. “ Colenso.”—Explanation appeared on the day you wrote.

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Permanent link to this item

https://paperspast.natlib.govt.nz/newspapers/OW19310901.2.33

Bibliographic details

Otago Witness, Issue 4042, 1 September 1931, Page 8

Word Count
1,256

NUTS TO CRACK Otago Witness, Issue 4042, 1 September 1931, Page 8

NUTS TO CRACK Otago Witness, Issue 4042, 1 September 1931, Page 8