by Dudeney, Henry Ernest, 1857-1930
Available in 215 free installments
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Twenty-one different squares may be selected. Of these nine will be of the size shown by the four A's in the diagram, four of the size shown by the B's, four of the size shown by the C's, two of the size shown by the D's, and two of the size indicated by the upper single A, the upper single E, the lower single C, and the EB. It is an interesting fact that you cannot form any one of these twenty-one squares without using at least one of the six circles marked E.
285.?THE FOUR POSTAGE STAMPS.? solution
Referring to the original diagram, the four stamps may be given in the shape 1,2,3, 4, in three ways; in the shape 1,2,5, 6, in six ways; in the shape 1, 2, 3, 5, or 1, 2, 3, 7, or 1, 5, 6, 7, or 3, 5, 6, 7, in twenty-eight ways; in shape 1, 2, 3, 6, or 2, 5, 6, 7, in fourteen ways; in shape 1, 2, 6, 7, or 2, 3, 5, 6, or 1, 5, 6, 10, or 2, 5, 6, 9, in fourteen ways. Thus there are sixty-five ways in all.
286.?PAINTING THE DIE.? solution
The 1 can be marked on any one of six different sides. For every side occupied by 1 we have a selection of four sides for the 2. For every situation of the 2 we have two places for the 3. (The 6, 5, and 4 need not be considered, as their positions are determined by the 1, 2, and 3.) Therefore 6, 4, and 2 multiplied together make 48 different ways?the correct answer.
287.?AN ACROSTIC PUZZLE.? solution
There are twenty-six letters in the alphabet, giving 325 different pairs. Every one of these pairs may be reversed, making 650 ways. But every initial letter may be repeated as the final, producing 26 other ways. The total is therefore 676 different pairs. In other words, the answer is the square of the number of letters in the alphabet.
288.?CHEQUERED BOARD DIVISIONS.? solution
There are 255 different ways of cutting the board into two pieces of exactly the same size and shape. Every Pg 211
way must involve one of the five cuts shown in Diagrams A, B, C, D, and E. To avoid repetitions by reversal
and reflection, we need only consider cuts that enter at the points a, b, and c. But the exit must always be at a
point in a straight line from the entry through the centre. This is the most important condition to remember. In
case B you cannot enter at a, or you will get the cut provided for in E. Similarly in C or D, you must not enter
the key-line in the same direction as itself, or you will get A or B. If you are working on A or C and entering at
