Miscellaneous Puzzles 57
the cubical contents; (3) when the square of the surface area equals the
cubical contents?
- A COMMON DlVlSOR
Here is a puzzle that has been the subject of frequent inquiries by corre-
spondents, only, of course, the actual figures are varied considerably. A
country newspaper stated that many schoolmasters have suffered in health in
their attempts to master it! Perhaps this is merely a little journalistic exagger-
ation, for it is really a simple question if only you have the cunning to hit on
the method of attacking it.
This is the question: Find a common divisor for the three numbers, 480,608;
508,8\; and 723,217 so that the remainder shall be the same in every case.
- CURIOUS MULTIPLICATION
I have frequently been asked to explain the following, which will doubtless
interest many readers who have not seen it. If a person can add correctly but
is incapable of multiplying or dividing by a number higher than 2, it is
possible to obtain the product of any two numbers in this curious way.
Multiply 97 by 23.
97 23
48 (46)
24 (92)
12 (184)
6 (368)
3 736
1,472
2,231
In the first column we divide by 2, rejecting the remainders, until 1 is
reached. In the second column we multiply 23 by 2 the same number of
times. If we now strike out those products that are opposite to the even
numbers in the first column (we have enclosed these in parentheses for conven-
ience in printing) and add up the remaining numbers we get 2,231, which is
the correct answer. Why is this?