by Dudeney, Henry Ernest, 1857-1930
Available in 215 free installments
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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. %7 /io5 /n - 3 /i2 - 1 /i3 - 8 ^14 - 6 ^15 - 4 ^16" 2 ^17" °^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 certaintythatitcanonlybe obtained (if at all) by dropping the 8.
80.?THE THREE GROUPS.? solution
There are nine solutions to this puzzle, as follows, and no more:?
12x483 = 5,796 27x198 = 5,346 42x138 = 5,796 39x186 = 7,254 18x297 = 5,346 48x159 = 7,632 28x157 = 4,396 4x1,738 = 6,952 4x1,963 = 7,852
The seventh answer is the one that is most likely to be overlooked by solvers of the puzzle.
81.?THE NINE COUNTERS.? solution
In this case a certain amount of mere "trial" is unavoidable. But there are two kinds of "trials"?those that are purely haphazard, and those that are methodical. The true puzzle lover is never satisfied with mere haphazard trials. The reader will find that by just reversing the figures in 23 and 46 (making the multipliers 32 and 64) both products will be 5,056. This is an improvement, but it is not the correct answer. We can get as large a product as 5,568 if we multiply 174 by 32 and 96 by 58, but this solution is not to be found without the exercise of some judgment and patience.
82.?THE TEN COUNTERS.? solution
As I pointed out, it is quite easy so to arrange the counters that they shall form a pair of simple multiplication sums, each of which will give the same product?in fact, this can be done by anybody in five minutes with a little patience. But it is quite another matter to find that pair which gives the largest product and that which gives the smallest product.
Now, in order to get the smallest product, it is necessary to select as multipliers the two smallest possible numbers. If, therefore, we place 1 and 2 as multipliers, all we have to do is to arrange the remaining eight counters in such a way that they shall form two numbers, one of which is just double the other; and in doing this we must, of course, try to make the smaller number as low as possible. Of course the lowest number we could get would be 3,045; but this will not work, neither will 3,405, 3,450, etc., and it may be ascertained that 3,485 is the lowest possible. One of the required answers is 3,485 x 2 = 6,970, and 6,970 x 1 = 6,970.
The other part of the puzzle (finding the pair with the highest product) is, however, the real knotty point, for it is not at all easy to discover whether we should let the multiplier consist of one or of two figures, though it is clear that we must keep, so far as we can, the largest figures to the left in both multiplier and multiplicand. It will be seen that by the following arrangement so high a number as 58,560 may be obtained. Thus, 915 x 64 = 58,560, and 732 x 80 = 58,560.
83.?DIGITAL MULTIPLICATION.? solution
The solution that gives the smallest possible sum of digits in the common product is 23 x 174 = 58 x 69 = 4,002, and the solution that gives the largest possible sum of digits, 9 x 654 = 18 x 327 = 5,886. In the first case the digits sum to 6 and in the second case to 27. There is no
way of obtaining the solution but by actual trial.
84.?THE PIERROT'S PUZZLE.? solution
There are just six different solutions to this puzzle, as follows:?
8 multiplied by 473 equals 3784
It will be seen that in every case the two multipliers contain exactly the same figures as the product.
85.?THE CAB NUMBERS.? solution Pg 157
The highest product is, I think, obtained by multiplying 8,745,231 by 96?namely, 839,542,176.
Dealing here with the problem generally, I have shown in the last puzzle that with three digits there are only two possible solutions, and with four digits only six different solutions.
These cases have all been given. With five digits there are just twenty-two solutions, as follows:?
3 x 4128 = 12384
3 x 4281 = 12843
3 x 7125 = 21375
3 x 7251 = 21753
2541 x 6 = 15246
651 x 24 = 15624
678 x 42 = 28476
246 x 51 = 12546
57 x 834 = 47538
75 x 231 = 17325
