is thus rendered ridiculously easy by this method. See also notes on
solutions to Nos. 13 and 85.

95.--_Robinson Crusoe's Table._

The diagram shows how the piece of wood should be cut in two pieces to
form the square table-top. A, B, C, D are the corners of the table. The
way in which the piece E fits into the piece F will be obvious to the eye
of the reader. The shaded part is the wood that is discarded.

96.--_The Fifteen Orchards._

The number must be the least common multiple of 1, 2, 3, etc., up to 15,
that, when divided by 7, leaves the remainder 1, by 9 leaves 3, by 11
leaves 10, by 13 leaves 3, and by 14 leaves 8. Such a number is 120. The
next number is 360,480, but as we have no record of a tree--especially a
very young one--bearing anything like such a large number of apples, we
may take 120 to be the only answer that is acceptable.

97.--_The Perplexed Plumber._

The rectangular closed cistern that shall hold a given quantity of water
and yet have the smallest possible surface of metal must be a perfect
cube--that is, a cistern every side of which is a square. For 1,000 cubic
feet of water the internal dimensions will be 10 ft. x 10 ft. x 10 ft.,
and the zinc required will be 600 square feet. In the case of a cistern
without a top the proportions will be exactly half a cube. These are the
"exact proportions" asked for in the second case. The exact dimensions
cannot be given, but 12.6 ft. x 12.6 ft. x 6.3 ft. is a close
approximation. The cistern will hold a little too much water, at which
the buyer will not complain, and it will involve the plumber in a
trifling loss not worth considering.

98.--_The Nelson Column._

If you take a sheet of paper and mark it with a diagonal line, as in
Figure A, you will find that when you roll it into cylindrical form, with
the line outside, it will appear as in Figure B.

[Illustration]

It will be seen that the spiral (in one complete turn) is merely the
hypotenuse of a right-angled triangle, of which the length and width of
the paper are the other two sides. In the puzzle given, the lengths of
the two sides of the triangle are 40 ft. (one-fifth of 200 ft.) and 16
ft. 8 in. Therefore the hypotenuse is 43 ft. 4 in. The length of the
garland is therefore five times as long--216 ft. 8 in. A curious feature
of the puzzle is the fact that with the dimensions given the result is
exactly the sum of the height and the circumference.

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