248 Answersthe top line, and so on until we come to a 7 in the bottom line without any
remainder. Then stop, for the correct number is found. A solution may
be found for any divisor and with any given figure at the beginning. A gen-
eral examination of the question is very interesting.
If we take the divisor 2 we get the number 2-1 0-5 2-6-3 1 5 7 8-9 4-7 3
6-8-4-.
This is a complete circuit. The hyphens are at the places where there is no
remainder when divided by 2. Note that the successive figures immediately
following a hyphen are 1,5,6,3,9, 7, 8, 4, 2, so ifI want my number to be-
gin with an 8, I start 8 4 2 1 0 5, etc., taking the 8 that follows a hyphen.
Where there is a complete circuit, as in this case, and in the case of the
divisors 3, 6, and 11, the number of figures in the number will always be 10
times the divisor less 2. If you try the divisor 4, there are five broken circuits.
Thus, 4-1 025 6-will give you numbers beginning with 4 or I; or 2 0-5 I 2-8-
with 2,5, or 8; or 7 1 7948-beginning with 7; or 3076-92-with 9 or 3;
or 6 I 5 3 8 4-for a number beginning with 6.
Some divisors, like 5 and 9, though producing broken circuits, require the
same total of figures as if they were complete circuits. Our divisor 7 gives
three broken circuits: the one shown above for the initial figures 7, 1, or 4,
another for 5, 8, or 2, and a third for 6, 9, or 3.
- A MISUNDERSTANDING
We may divide 8 5 7 I 4 2 by 3 by simply transferring the 2 from the end to
the beginning. Or 4 2 8 5 7 1, by transferring the 1.109. THE TWO FOURS
This is how 64 may be expressed by the use of only two fours with arith-
metical signs:
The process of simplification shown should make everything quite clear.j (fl4t= J( ~)24 = yI2i2 = y4096 = 64
[Interest in the "Four Fours" problem has had periodic revivals since its