by Dudeney, Henry Ernest, 1857-1930
Available in 215 free installments
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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? solution
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.
1+2 + 3+4 + 5 + 6 + 7 + (8x9) =100
(1 x 2)-3-4-5+ (6x7) + (8x9) =100
1 + (2 x 3) + (4 x 5) - 6 + 7 + (8 x 9) = 100
(1 +2-3-4)(5-6-7-8-9)=100
1 +(2x3) + 4 + 5 + 67 + 8 + 9 = 100
(1 x 2)+ 34+ 56+ 7-8 + 9 = 100
12 + 3-4 + 5 + 67 + 8 + 9 = 100
123-4-5-6-7 + 8-9 = 100
123+4-5 + 67-8-9 = 100
123 + 45-67 + 8-9 = 100
123-45-67 + 89 = 100
It will be noticed that in the above I have counted the bracket as one sign and two strokes. The last solution is p 9 160 singularly simple, and I do not think it will ever be beaten.
95.?THE FOUR SEVENS.? solution
The way to write four sevens with simple arithmetical signs so that they represent 100 is as follows:?
7 /. 7 x 7 / 7 =100.
Of course the fraction, 7 over decimal 7, equals 7 divided by 7 A|q, 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.? solution
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
