Part 72

Most people know that if the sum of the digits in the odd places of any number is the same as the sum of the digits in the even places, then the number is divisible by 11 without remainder. Thus in 896743012 the odd digits, 20468, add up 20, and the even digits, 1379, also add up 20. Therefore the number may be divided by 11. But few seem to know that if the difference between the sum of the odd and the even digits is 11, or a multiple of 11, the rule equally applies. This law enables us to find, with a very little trial, that the smallest number containing nine of the ten digits (calling nought a digit) that is divisible by 11 is 102,347,586, and the highest number possible, 987,652,413.

94.--THE DIGITAL CENTURY.

There is a very large number of different ways in which arithmetical signs may be placed between the nine digits, arranged in numerical order, so as to give an expression equal to 100. In fact, unless the reader investigated the matter very closely, he might not suspect that so many ways are possible. It was for this reason that I added the condition that not only must the fewest possible signs be used, but also the fewest possible strokes. In this way we limit the problem to a single solution, and arrive at the simplest and therefore (in this case) the best result.

Just as in the case of magic squares there are methods by which we may write down with the greatest ease a large number of solutions, but not all the solutions, so there are several ways in which we may quickly arrive at dozens of arrangements of the "Digital Century," without finding all the possible arrangements. There is, in fact, very little principle in the thing, and there is no certain way of demonstrating that we have got the best possible solution. All I can say is that the arrangement I shall give as the best is the best I have up to the present succeeded in discovering. I will give the reader a few interesting specimens, the first being the solution usually published, and the last the best solution that I know.

Signs. Strokes. 1 + 2 + 3 + 4 + 5 + 6 + 7 + (8 x 9) = 100 ( 9 18)

- (1 x 2) - 3 - 4 - 5 + (6 x 7) + (8 x 9) = 100 (12 20)

1 + (2 x 3) + (4 x 5) - 6 + 7 + (8 x 9) = 100 (11 21)

(1 + 2 - 3 - 4)(5 - 6 - 7 - 8 - 9) = 100 ( 9 12)

1 + (2 x 3) + 4 + 5 + 67 + 8 + 9 =100 (8 16)

(1 x 2) + 34 + 56 + 7 - 8 + 9 = 100 (7 13)

12 + 3 - 4 + 5 + 67 + 8 + 9 = 100 (6 11)

123 - 4 - 5 - 6 - 7 + 8 - 9 = 100 (6 7)

123 + 4 - 5 + 67 - 8 - 9 = 100 (4 6)

123 + 45 - 67 + 8 - 9 = 100 (4 6)

123 - 45 - 67 + 89 = 100 (3 4)

It will be noticed that in the above I have counted the bracket as one sign and two strokes. The last solution is singularly simple, and I do not think it will ever be beaten.

95.--THE FOUR SEVENS.

The way to write four sevens with simple arithmetical signs so that they represent 100 is as follows:--

7 7 -- x -- = 100. .7 .7

Of course the fraction, 7 over decimal 7, equals 7 divided by 7/10, which is the same as 70 divided by 7, or 10. Then 10 multiplied by 10 is 100, and there you are! It will be seen that this solution applies equally to any number whatever that you may substitute for 7.

96.--THE DICE NUMBERS.

The sum of all the numbers that can be formed with any given set of four different figures is always 6,666 multiplied by the sum of the four figures. Thus, 1, 2, 3, 4 add up 10, and ten times 6,666 is 66,660. Now, there are thirty-five different ways of selecting four figures from the seven on the dice--remembering the 6 and 9 trick. The figures of all these thirty-five groups add up to 600. Therefore 6,666 multiplied by 600 gives us 3,999,600 as the correct answer.

Let us discard the dice and deal with the problem generally, using the nine digits, but excluding nought. Now, if you were given simply the sum of the digits--that is, if the condition were that you could use any four figures so long as they summed to a given amount--then we have to remember that several combinations of four digits will, in many cases, make the same sum.

10 11 12 13 14 15 16 17 18 19 20 1 1 2 3 5 6 8 9 11 11 12

21 22 23 24 25 26 27 28 29 30 11 11 9 8 6 5 3 2 1 1

Here the top row of numbers gives all the possible sums of four different figures, and the bottom row the number of different ways in which each sum may be made. For example 13 may be made in three ways: 1237, 1246, and 1345. It will be found that the numbers in the bottom row add up to 126, which is the number of combinations of nine figures taken four at a time. From this table we may at once calculate the answer to such a question as this: What is the sum of all the numbers composed of our different digits (nought excluded) that add up to 14? Multiply 14 by the number beneath t in the table, 5, and multiply the result by 6,666, and you will have the answer. It follows that, to know the sum of all the numbers composed of four different digits, if you multiply all the pairs in the two rows and then add the results together, you will get 2,520, which, multiplied by 6,666, gives the answer 16,798,320.

The following general solution for any number of digits will doubtless interest readers. Let n represent number of digits, then 5 (10^n - 1) 8! divided by (9 - n)! equals the required sum. Note that 0! equals 1. This may be reduced to the following practical rule: Multiply together 4 x 7 x 6 x 5 ... to (n - 1) factors; now add (n + 1) ciphers to the right, and from this result subtract the same set of figures with a single cipher to the right. Thus for n = 4 (as in the case last mentioned), 4 x 7 x 6 = 168. Therefore 16,800,000 less 1,680 gives us 16,798,320 in another way.

97.--THE SPOT ON THE TABLE.