(related to Problem: The Millionaire's Perplexity)

The answer to this quite easy puzzle may, of course, be readily obtained by trial, deducting the largest power of $7$ that is contained in one million dollars, then the next largest power from the remainder, and so on. But the little problem is intended to illustrate a simple direct method. The answer is given at once by converting $1,000,000$ to the septenary scale, and it is on this subject of scales of notation that I propose to write a few words for the benefit of those who have never sufficiently considered the matter.

Our manner of figuring is a sort of perfected arithmetical shorthand, a system devised to enable us to manipulate numbers as rapidly and correctly as possible by means of symbols. If we write the number $2,341$ to represent two thousand three hundred and forty-one dollars, we wish to imply $1$ dollar, added to four times $10$ dollars, added to three times $100$ dollars, added to two times $1,000$ dollars. From the number in the units place on the right, every figure to the left is understood to represent a multiple of the particular power of $10$ that its position indicates, while a cipher ($0$) must be inserted where necessary in order to prevent confusion, for if instead of $207$ we wrote $27$ it would be obviously misleading. We thus only require ten figures, because directly a number exceeds $9$ we put the second figure to the left, directly it exceeds $99$ we put the third figure to the left, and so on. It will be seen that this is a purely arbitrary method. It is working in the denary (or ten) scale of notation, a system undoubtedly derived from the fact that our forefathers who devised it had ten fingers upon which they were accustomed to counting, like our children of today. It is unnecessary for us ordinarily to state that we are using the denary scale because this is always understood in the common affairs of life.

But if a man said that he had $6,553$ dollars in the septenary (or seven) scale of notation, you will find that this is precisely the same amount as $2,341$ in our ordinary denary scale. Instead of using powers of ten, he uses powers of $7,$ so that he never needs any figure higher than $6,$ and $6,553$ really stands for $3,$ added to five times $7,$ added to five times $49,$ added to six times $343$ (in the ordinary notation), or $2,341.$ To reverse the operation, and convert $2,341$ from the denary to the septenary scale, we divide it by $7,$ and get $334$ and remainder $3;$ divide $334$ by $7,$ and get $47$ and remainder $5;$ and so keep on dividing by $7$ as long as there is anything to divide. The remainders, read backward, $6, 5, 5, 3,$ give us the answer, $6,553.$

Now, as I have said, our puzzle may be solved at once by merely converting $1,000,000$ dollars to the septenary scale. Keep on dividing this number by $7$ until there is nothing more left to divide, and the remainders will be found to be $11333311$ which is $1,000,000$ expressed in the septenary scale. Therefore, $1$ gift of $1$ dollar, $1$ gift of $7$ dollars, $3$ gifts of $49$ dollars, $3$ gifts of $343$ dollars, $3$ gifts of $2,401$ dollars, $3$ gifts of $16,807$ dollars, $1$ gift of $117,649$ dollars, and one substantial gift of $823,543$ dollars, satisfactorily solves our problem. And it is the only possible solution. It is thus seen that no "trials" are necessary; by converting to the septenary scale of notation we go directly to the answer.

**Dudeney, H. E.**: "Amusements in Mathematics", The Authors' Club, 1917

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