By Schuh, Frederick; Schuh, Fred
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From the outlines of the pieces and the position of the doubles it follows that: b # 0; c # 6, e # 0, e # I; g # 5, g # 6; i # 2, i # 3;j # 3,j # 4. Since a piece does not occur twice, we further have: b # l,b # 6;c # O,c # 5;e # 2,e # 3;1# 0,1# 2,1# 3, f # 4,1 # 6; g # 3, g # 4; i # 0, i # l;j # 5,j # 6. The fact that e # 3, for example, may be seen from the occurrence of the pieces e-i and i-3, which cannot be the same. The fact that f # can be seen from the pieces O-b and b-f, similarly elsewhere.
Of these four ways, the last one seems simplest. If we adapt the order of the numbers and the letters to the original puzzle, then the solution in question is the one shown at the left. IA2B3 2 B 3 .. I A 3 .. IA2B 3A21 .. B 2 I3AB IA2B3 2B 3 IA 31A2B 3 .. 2 B I A 3B12 .. A 123 B A IA2B 3 IA .. 32B 32BIA 3 .. IA2B 3A21 .. ced in a completely detached position. In these solutions, two moves are made with the same pair (1 A or 2B), with one move in between (with 2B and 1 A, respectively). The two moves with the same pair are equivalent to a single move; but splitting up this move into two moves can here be regarded as a means to make the correct preparation for another move.
In a number that satisfies the condition a bold digit is followed by a high digit, and an ordinary digit by nothing or by a low digit. 1 The number begins with a low digit and ends with an ordinary digit. 1 Hence, in a number that satisfies the condition, the number of bold digits is equal to the number of high digits, and the number of ordinary digits to that of SOME PUZZLES WITH MULTIPLES 33 A combination of sequences in which all bold digits are high (and hence all ordinary digits are low) cannot lead to a solution, because a bold digit would always have to be followed by a bold digit again.
The master book of mathematical recreations by Schuh, Frederick; Schuh, Fred