(related to Problem: The Lockers Puzzle)

The smallest possible total is $356 = 107 + 249$ and the largest sum possible is $981 = 235 + 746,$ or $657 + 324.$ The middle sum may be either $720 =134+586,$ or $702 = 134 + 568,$ or $407 = 138 + 269.$ The total in this case must be made up of three of the figures $0, 2, 4, 7,$ but no sum other than the three given can possibly be obtained. We have therefore no choice in the case of the first locker, an alternative in the case of the third, and any one of three arrangements in the case of the middle locker. Here is one solution: —

$$\begin{array}{ccc} 107&134&235\\ 249&586&746\\ \hline 356&720&981\\ \end{array}$$ Of course, in each case figures in the first two lines may be exchanged vertically without altering the total, and as a result, there are just $3,072$ different ways in which the figures might be actually placed on the locker doors. I must content myself with showing one little principle involved in this puzzle. The sum of the digits in the total is always governed by the digit omitted. $$\frac{9}{9} - \frac{7}{10} - \frac{5}{11} - \frac{3}{12} - \frac{1}{13} - \frac{8}{14} - \frac{6}{15} - \frac{4}{16} - \frac{2}{17} - \frac 0{18}.$$ Whichever digit shown here in the upper line we omit, the sum of the digits in the total will be found beneath it. Thus in the case of locker A we omitted $8,$ and the figures in the total sum up to $14.$ If, therefore, we wanted to get $356,$ we may know at once to a certainty that it can only be obtained (if at all) by dropping the $8.$

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Project Gutenberg

  1. Dudeney, H. E.: "Amusements in Mathematics", The Authors' Club, 1917

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