THE DICE NUMBERS.
(
Money Puzzles)
I have a set of four dice, not marked with spots in the ordinary way,
but with Arabic figures, as shown in the illustration. Each die, of
course, bears the numbers 1 to 6. When put together they will form a
good many, different numbers. As represented they make the number 1246.
Now, if I make all the different four-figure numbers that are possible
with these dice (never putting the same figure more than once in any
number), what will they all add up to? You are allowed to turn the 6
upside down, so as to represent a 9. I do not ask, or expect, the reader
to go to all the labour of writing out the full list of numbers and then
adding them up. Life is not long enough for such wasted energy. Can you
get at the answer in any other way?
VARIOUS ARITHMETICAL AND ALGEBRAICAL PROBLEMS.
"Variety's the very spice of life,
That gives it all its flavour."
COWPER: _The Task._
Answer:
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.
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