by Dudeney, Henry Ernest, 1857-1930
Available in 215 free installments
Owner:
The constable took thirty steps. In the same time the thief would take forty-eight, which, added to his start of twenty-seven, carried him seventy-five steps. This distance would be exactly equal to thirty steps of the constable.
105.?THE PARISH COUNCIL ELECTION.? solution
The voter can vote for one candidate in 23 ways, for two in 253 ways, for three in 1,771, for four in 8,855, for five in 33,649, for six in 100,947, for seven in 245,157, for eight in 490,314, and for nine candidates in 817,190 different ways. Add these together, and we get the total of 1,698,159 ways of voting.
106.?THE MUDDLETOWN ELECTION.? solution
The numbers of votes polled respectively by the Liberal, the Conservative, the Independent, and the Socialist were 1,553,1,535, 1,407, and 978 All that was necessary was to add the sum of the three majorities (739) to the total poll of 5,473 (making 6,212) and divide by 4, which gives us 1,553 as the poll of the Liberal. Then the polls of the other three candidates can, of course, be found by deducting the successive majorities from the last-mentioned number.
107._THE SUFFRAGISTS' MEETING.? solution
Eighteen were present at the meeting and eleven left. If twelve had gone, two-thirds would have retired. If only nine had gone, the meeting would have lost half its members.
108.?THE LEAP-YEAR LADIES.? solution
The correct and only answer is that 11,616 ladies made proposals of marriage. Here are all the details, which
the reader can check for himself with the original statements. Of 10,164 spinsters, 8,085 married bachelors, 627 married widowers, 1,221 were declined by bachelors, and 231 declined by widowers. Of the 1,452 widows, 1,155 married bachelors, and 297 married widowers. No widows were declined. The problem is not difficult, by algebra, when once we have succeeded in correctly stating it.
109.?THE GREAT SCRAMBLE.? solution
The smallest number of sugar plums that will fulfil the conditions is 26,880. The five boys obtained respectively: Andrew, 2,863; Bob, 6,335; Charlie, 2,438; David, 10,294; Edgar, 4,950. There is a little trap concealed in the words near the end, "one-fifth of the same," that seems at first sight to upset the whole account of the affair. But a little thought will show that the words could only mean "one-fifth of five-eighths", the fraction last mentioned?that is, one-eighth of the three-quarters that Bob and Andrew had last acquired.
110.?THE ABBOT'S PUZZLE.? solution
The only answer is that there were 5 men, 25 women, and 70 children. There were thus 100 persons in all, 5 times as many women as men, and as the men would together receive 15 bushels, the women 50 bushels, and the children 35 bushels, exactly 100 bushels would be distributed.
111.?REAPING THE CORN.? solution
The whole field must have contained 46.626 square rods. The side of the central square, left by the farmer, is 4.8284 rods, so it contains 23.313 square rods. The area of the field was thus something more than a quarter of an acre and less than one-third; to be more precise, .2914 of an acre.
112.?A PUZZLING LEGACY? solution
As the share of Charles falls in through his death, we have merely to divide the whole hundred acres between Alfred and Benjamin in the proportion of one-third to one-fourth?that is in the proportion of four-twelfths to three-twelfths, which is the same as four to three. Therefore Alfred takes four-sevenths of the hundred acres p 9 162 and Benjamin three-sevenths.
113.?THE TORN NUMBER.? solution
The other number that answers all the requirements of the puzzle is 9,801. If we divide this in the middle into two numbers and add them together we get 99, which, multiplied by itself, produces 9,801. It is true that 2,025 may be treated in the same way, only this number is excluded by the condition which requires that no two figures should be alike.
