The Pierrot in the illustration is standing in a posture that represents the sign of multiplication.  He is indicating the peculiar fact that 15 multiplied by 93 produces exactly the same figures (1,395), differently arranged.  The puzzle is to take any four digits you like (all different) and similarly arrange them so that the number formed on one side of the Pierrot when multiplied by the number on the other side shall produce the same figures.  There are very few ways of doing it, and I shall give all the cases possible.  Can you find them all?  You are allowed to put two figures on each side of the Pierrot as in the example shown, or to place a single figure on one side and three figures on the other.  If we only used three digits instead of four, the only possible ways are these:  3 multiplied by 51 equals 153, and 6 multiplied by 21 equals 126.

85.—­THE CAB NUMBERS.

A London policeman one night saw two cabs drive off in opposite directions under suspicious circumstances.  This officer was a particularly careful and wide-awake man, and he took out his pocket-book to make an entry of the numbers of the cabs, but discovered that he had lost his pencil.  Luckily, however, he found a small piece of chalk, with which he marked the two numbers on the gateway of a wharf close by.  When he returned to the same spot on his beat he stood and looked again at the numbers, and noticed this peculiarity, that all the nine digits (no nought) were used and that no figure was repeated, but that if he multiplied the two numbers together they again produced the nine digits, all once, and once only.  When one of the clerks arrived at the wharf in the early morning, he observed the chalk marks and carefully rubbed them out.  As the policeman could not remember them, certain mathematicians were then consulted as to whether there was any known method for discovering all the pairs of numbers that have the peculiarity that the officer had noticed; but they knew of none.  The investigation, however, was interesting, and the following question out of many was proposed:  What two numbers, containing together all the nine digits, will, when multiplied together, produce another number (the highest possible) containing also all the nine digits?  The nought is not allowed anywhere.

86.—­QUEER MULTIPLICATION.

If I multiply 51,249,876 by 3 (thus using all the nine digits once, and once only), I get 153,749,628 (which again contains all the nine digits once).  Similarly, if I multiply 16,583,742 by 9 the result is 149,253,678, where in each case all the nine digits are used.  Now, take 6 as your multiplier and try to arrange the remaining eight digits so as to produce by multiplication a number containing all nine once, and once only.  You will find it far from easy, but it can be done.

87.—­THE NUMBER-CHECKS PUZZLE.

[Illustration]