Pat must have painted six more posts than Tim, no matter how many lamp-posts there were.  For example, suppose twelve on each side; then Pat painted fifteen and Tim nine.  If a hundred on each side, Pat painted one hundred and three, and Tim only ninety-seven

104.—­CATCHING THE THIEF.

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,

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.

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.

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.

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.

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.

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.