NUTS TO CRACK

Otago Witness, Issue 4033, 30 June 1931, Page 61

NUTS TO CRACK

By

T. L. Briton

(Fob 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 lh~ “ nuts " may appear harder than others. It will be found that none will require a sledge-hammer to crack them Solutions will appear in 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 SEVEN SHARED. Seven-brothers shared in their father’s will, but they aid not participate equally. Two-thirds of the sum received by C and D together were equivalent to ’tie joint amount to which A and B were eil- 1 titled, and the combined shares of the I two latter brothers exactly equalled the | sum received by each of the older brothers E and F. To G was given the residue of the legacy, which represented exactly £5 more than the sum which the two youngest sons A and B received jointly, and £9O more than G would have been entitled to had the will provided for an equal distribution between the seven beneficiaries. As A, C, and E together received £9OO, can the reader find what was the total amount left to the seven brothers? THREE BANK ACCOUNTS. A trading firm which was accustomed to hold money in trust had three bank accounts, No. I and No. 2 being ordinary current accounts, while No. 3 was the trust account. On March 31 last No. 1 was overdrawn. No. 2 was in credit, while No. 3 was also in credit to the amount of £l2BO. On April 1 the sum to the credit of No. 2 account was transferred to the overdrawn account No. 1. which reduced the debit balance of that particular account to f 44. The figures mentioned have a curious feature when taken relatively, for if tlie state of the two current ordinary accounts, before the transfer from one to the other was made, is examined, it will be seen that if the amount to the credit of No. 2 account,then be multiplied by the num her representing the sum to the debit of No. 1. the result will be found to he the amount to the credit of the trust account .mentioned, viz., £l2BO. With only this information to work upon, can the reader find how the two smaller ac--1 counts stood before the transfer on April ’ 1?

ON SIX SIDES. If a cube be marked on its six respective sides 1,2, 3,4, .5, 6, there are many different arrangements for placing the set of figures, one of which will be used in this problem —viz., that the figures on the two opposite sides will add up seven. Thus in marking a cube in this way and a side is chosen for the figure 1, there are five other sides which would do equally as well. But when the second figure, 2, comes to be placed, there are obviously only four sides available for it, and consequently for the third figure, 3, there are only two sides from which to select. It is a very simple calculation, though liable perhaps to puzzle the reader for a little while, to find in how many different ways it is possible to mark a cube, in the manner described, so that every two opposite sides will add up. seven. The explanation given above will probably lessen whatever difficulty there may be in finding the correct number. zl CATERER’S ACCOUNT. A difficulty arose at a meeting of the committee of a public function w’hen accounts were being pas®—! for payment, the caterer’s bill being the particular trouble. His bill was for the sum of £l6 4s, and he had sent a memo with the account, stating the number who sat down to the dinner for which the account was rendered. It appears that the secretary of the function had taken the number of diners as supplied by the caterer and divided it into the amount of the bill for the purpose of determining the quota each had to pay. But inadvertently he had overlooked the fact that nine of the diners were guests of the committee, so that his divisor when he made the calculation was that number too many. As the’ sum of the caterer’s bill had to be shared equally with the others who sat down to the repast, can the reader nay how many diners there were if the secretary’s miscalculation made the quota for each subscriber three shillings less than it should have been?

AT THE WOOL SALES. A firm of wool brokers submitted a quantity of bales for sale on every day of the week except Saturday, the five days in succession being specially arranged by way of experiment. On Monday a good percentage of that offered was’ sold,' and Tuesday’s lot submitted consisted of the bales left over from Monday plus as many more. Tuesday’s sales showed that the same number of bales were sold as on Monday, and those not sold were offered at Wednesday’s auction together with twice as many more. Again the same number were disposed of on the third day as on the tw’o previous days, and again there were some left over. These unsold bales were sent to auction on Thursday, together with three times

as many more new ones, and once more the number sold was the same as on the other sale days. The wool offered on the last day of the series consisted of the bales- left over from Thursday’s sale plus four times as many more new bales that had just arrived from the country. The result of Friday’s auction was that every bale was sold, the total number being exactly the same as the quantity sold on each of the previous four days. Can the reader find the number of bales offered on the first day, it being the smallest possible number under these conditions?

SOLUTIONS OF LAST WEEK’S PROBLEMS. A SMART NEWSBOY. Eighteen pence per dozen. BOAT ROWING. Ten boats are all that would be necessary for the 15 rowers to make use of under the conditions set out. PRACTICAL, DISSECTION. From F, the centre of the line A B. draw a line F H to the centre of the square and parallel to A D. Join D H, and from H draw a line to the centre of line B E, which will be parallel to D E, the centre of B E being at a point G. Then if a line be drawn from C at right angles to D E to meet the line H G, the original figure will be divided into four parts of equal shape and size. IN ANOTHER WAY. To convert it into a magic square in the manner described, 37 moves are necessary. A MOVING STAIRCASE. Forty-six steps, descending at the rate of one' step in one and a-half seconds, the. speed at which the person walks not affecting the question. ANSWERS TO CORRESPONDENTS. C. M. P.— (1) Interesting, but if the digits in the diagonal are of one colour, the arrangement falls short of the requirements, the conditions being that no colour or digit shal l be repeated in the same line in any of three directions. (2) The other item ’ appeared on December 13. 1930. Thqnks for interest. ' J. T.—The plan suggested satisfies only one side, the other person’s vessel *- containing one gallon more than required. “ Colenso.” —Not without stipulating the other condition, which was omitted with a purpose. It exemplifies the good old maxim in mathematics to take nothing for granted. Thanks.