36 Arithmetic & Algebraic Problems- ANOTHER 37 DIVISION
Here is an interesting extension of the last puzzle. If the nine digits are
written at haphazard in any order, for example, 4 1 2539768, what are the
chances that the number that happens to be produced will be divisible by 37
without remainder?- A DIGITAL DIFFICULTY
Arrange the ten digits, 1 2 3 4 5 678 9 0, in such order that they shall form
a number that may be divided by every number from 2 to 18 without in any
case a remainder. As an example, if I arrange them thus, 1 2 7 4 9 5 3 6 8 0,
this number can be divided by 2, 3,4, 5, and so on up to 16, without any re-
mainder, but it breaks down at 17.
- THREES AND SEVENS
What is the smallest number composed only of the digits 3 and 7 that may
be divided by 3 and by 7, and also the sum of its digits by 3 and by 7, with-
out any remainder? Thus, for example, 7733733 is divisible by 3 and by 7
without remainder, but the sum of its digits (33), while divisible by 3, is not
divisible by 7 without remainder. Therefore it is not a solution.
- ROOT EXTRACTION
In a conversation I had with Professor Simon Greathead, the eminent
mathematician, now living in retirement at Colney Hatch, * I had occasion to
refer to the extraction of the cube root.
"Ah," said the professor, "it is astounding what ignorance prevails on that
elementary matter! The world seems to have made little advance in the
process of the extraction of roots since the primitive method of employing
spades, forks, and trowels for the purpose. For example, nobody but myself
has ever discovered the simple fact that, to extract the cube root of a number,
all you have to do is to add together the digits. Thus, ignoring the obvious case
of the number 1, if we want the cube root of 512, add the digits-8, and there
you are!"
I suggested that that was a special case.
°A large mental hospital in Middlesex, near London.-M. G.