We next write out in a column these 28 sets of five figures, and proceed to tabulate the possible factors, or multipliers, into which they may be split.  Roughly speaking, there would now appear to be about 2,000 possible cases to be tried, instead of the 30,240 mentioned above; but the process of elimination now begins, and if the reader has a quick eye and a clear head he can rapidly dispose of the large bulk of these cases, and there will be comparatively few test multiplications necessary.  It would take far too much space to explain my own method in detail, but I will take the first set of figures in my table and show how easily it is done by the aid of little tricks and dodges that should occur to everybody as he goes along.

My first product group of five figures is 84,321.  Here, as we have seen, the root of each factor must be 3 or a multiple of 3.  As there is no 6 or 9, the only single multiplier is 3.  Now, the remaining four figures can be arranged in 24 different ways, but there is no need to make 24 multiplications.  We see at a glance that, in order to get a five-figure product, either the 8 or the 4 must be the first figure to the left.  But unless the 2 is preceded on the right by the 8, it will produce when multiplied either a 6 or a 7, which must not occur.  We are, therefore, reduced at once to the two cases, 3 x 4,128 and 3 x 4,281, both of which give correct solutions.  Suppose next that we are trying the two-figure factor, 21.  Here we see that if the number to be multiplied is under 500 the product will either have only four figures or begin with 10.  Therefore we have only to examine the cases 21 x 843 and 21 x 834.  But we know that the first figure will be repeated, and that the second figure will be twice the first figure added to the second.  Consequently, as twice 3 added to 4 produces a nought in our product, the first case is at once rejected.  It only remains to try the remaining case by multiplication, when we find it does not give a correct answer.  If we are next trying the factor 12, we see at the start that neither the 8 nor the 3 can be in the units place, because they would produce a 6, and so on.  A sharp eye and an alert judgment will enable us thus to run through our table in a much shorter time than would be expected.  The process took me a little more than three hours.

I have not attempted to enumerate the solutions in the cases of six, seven, eight, and nine digits, but I have recorded nearly fifty examples with nine digits alone.

86.—­QUEER MULTIPLICATION.

If we multiply 32547891 by 6, we get the product, 195287346.  In both cases all the nine digits are used once and once only.

87.—­THE NUMBER CHECKS PUZZLE.

Divide the ten checks into the following three groups:  7 1 5—­4 6—­3 2 8 9 0, and the first multiplied by the second produces the third.

88.—­DIGITAL DIVISION.

It is convenient to consider the digits as arranged to form fractions of the respective values, one-half, one-third, one-fourth, one-fifth, one-sixth, one-seventh, one-eighth, and one-ninth.  I will first give the eight answers, as follows:—­