A B C ================== ================== ================== | +--+ +--+ +--+ | | +--+ +--+ +--+ | | +--+ +--+ +--+ | | | | | | | | | | | | | | | | | | | | | | | | | | +--+ +--+ +--+ | | +--+ +--+ +--+ | | +--+ +--+ +--+ | | | | | | | | +--+ +--+ +--+ | | +--+ +--+ +--+ | | +--+ +--+ +--+ | | | | | | | | | | | | | | | | | | | | | | | | | | +--+ +--+ +--+ | | +--+ +--+ +--+ | | +--+ +--+ +--+ | | | | | | | ================== ================== ================== | +--+ +--+ +--+ | | +--+ +--+ +--+ | | +--+ +--+ +--+ | | | | | | | | | | | | | | | | | | | | | | | | | | +--+ +--+ +--+ | | +--+ +--+ +--+ | | +--+ +--+ +--+ | ------------------ ------------------ ------------------ ] A man had in his office three cupboards, each containing nine lockers, as shown in the diagram. He told his clerk to place a different one-figure number on each locker of cupboard A, and to do the same in the case of B, and of C. As we are here allowed to call nought a digit, and he was not prohibited from using nought as a number, he clearly had the option of omitting any one of ten digits from each cupboard. Now, the employer did not say the lockers were to be numbered in any numerical order, and he was surprised to find, when the work was done, that the figures had apparently been mixed up indiscriminately. Calling upon his clerk for an explanation, the eccentric lad stated that the notion had occurred to him so to arrange the figures that in each case they formed a simple addition sum, the two upper rows of figures producing the sum in the lowest row. But the most surprising point was this: that he had so arranged them that the addition in A gave the smallest possible sum, that the addition in C gave the largest possible sum, and that all the nine digits in the three totals were different. The puzzle is to show how this could be done. No decimals are allowed and the nought may not appear in the hundreds place. 80.--THE THREE GROUPS. There appeared in "Nouvelles Annales de Mathematiques" the following puzzle as a modification of one of my "Canterbury Puzzles." Arrange the nine digits in three groups of two, three, and four digits, so that the first two numbers when multiplied together make the third. Thus, 12 x 483 = 5,796. I now also propose to include the cases where there are one, four, and four digits, such as 4 x 1,738 = 6,952. Can you find all the possible solutions in both cases? 81.--THE NINE COUNTERS. [Illustration: (1)(5)(8) (7)(9) (2)(3) (4)(6) ]