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
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:
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
The sum of the digits in the total is always governed by the digit
omitted. 9/9 - 7/10 - 5/11
- 3/12 - 1/13 - 8/14 - 6/15 - 4/16 - 2/17 - 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.