The Canterbury Puzzles Part 29

7 5 2 4 6 3 8 .[.9]

Every column, every row, and each of the two diagonals now add up to 12.

This is the correct solution to the puzzle.

82.--_The Wizard's Arithmetic._

This puzzle is both easy and difficult, for it is a very simple matter to find one of the multipliers, which is 86. If we multiply 8 by 86, all we need do is to place the 6 in front and the 8 behind in order to get the correct answer, 688. But the second number is not to be found by mere trial. It is 71, and the number to be multiplied is no less than 1639344262295081967213114754098360655737704918032787. If you want to multiply this by 71, all you have to do is to place another 1 at the beginning and another 7 at the end--a considerable saving of labour!

These two, and the example shown by the wizard, are the only two-figure multipliers, but the number to be multiplied may always be increased.

Thus, if you prefix to 41096 the number 41095890, repeated any number of times, the result may always be multiplied by 83 in the wizard's peculiar manner.

If we add the figures of any number together and then, if necessary, again add, we at last get a single-figure number. This I call the "digital root." Thus, the digital root of 521 is 8, and of 697 it is 4.

This digital a.n.a.lysis is extensively dealt with in _A. in M._ Now, it is evident that the digital roots of the two numbers required by the puzzle must produce the same root in sum and product. This can only happen when the roots of the two numbers are 2 and 2, or 9 and 9, or 3 and 6, or 5 and 8. Therefore the two-figure multiplier must have a digital root of 2, 3, 5, 6, 8, or 9. There are ten such numbers in each case. I write out all the sixty, then I strike out all those numbers where the second figure is higher than the first, and where the two figures are alike (thirty-six numbers in all); also all remaining numbers where the first figure is odd and the second figure even (seven numbers); also all multiples of 5 (three more numbers). The numbers 21 and 62 I reject on inspection, for reasons that I will not enter into. I then have left, out of the original sixty, only the following twelve numbers: 83, 63, 81, 84, 93, 42, 51, 87, 41, 86, 53, and 71. These are the only possible multipliers that I have really to examine.

My process is now as curious as it is simple in working. First trying 83, I deduct 10 and call it 73. Adding 0's to the second figure, I say if 30000, etc., ever has a remainder 43 when divided by 73, the dividend will be the required multiplier for 83. I get the 43 in this way. The only multiplier of 3 that produces an 8 in the digits place is 6. I therefore multiply 73 by 6 and get 438, or 43 after rejecting the 8. Now, 300,000 divided by 73 leaves the remainder 43, and the dividend is 4,109.

To this 1 add the 6 mentioned above and get 41,096 x 83, the example given on page 129.

In trying the even numbers there are two cases to be considered. Thus, taking 86, we may say that if 60000, etc., when divided by 76 leaves either 22 or 60 (because 36 and 86 both produce 8), we get a solution.

But I reject the former on inspection, and see that 60 divided by 76 is 0, leaving a remainder 60. Therefore 8 x 86 = 688, the other example. It will be found in the case of 71 that 100000, etc., divided by 61 gives a remainder 42, (7 61 = 427) after producing the long dividend at the beginning of this article, with the 7 added.

The other multipliers fail to produce a solution, so 83, 86, and 71 are the only three possible multipliers. Those who are familiar with the principle of recurring decimals (as somewhat explained in my next note on No. 83, "The Ribbon Problem") will understand the conditions under which the remainders repeat themselves after certain periods, and will only find it necessary in two or three cases to make any lengthy divisions. It clearly follows that there is an unlimited number of multiplicands for each multiplier.

83.--_The Ribbon Problem._

The solution is as follows: Place this rather lengthy number on the ribbon, 0212765957446808510638297872340425531914393617. It may be multiplied by any number up to 46 inclusive to give the same order of figures in the ring. The number previously given can be multiplied by any number up to 16. I made the limit 9 in order to put readers off the scent. The fact is these two numbers are simply the recurring decimals that equal 1/17 and 1/47 respectively. Multiply the one by seventeen and the other by forty-seven, and you will get all nines in each case.

In transforming a vulgar fraction, say 1/17, to a decimal fraction, we proceed as below, adding as many noughts to the dividend as we like until there is no remainder, or until we get a recurring series of figures, or until we have carried it as far as we require, since every additional figure in a never-ending decimal carries us nearer and nearer to exact.i.tude.

17) 100 (.058823 85 ---- 150 136 ---- 140 136 ---- 40 34 ---- 60 51 ---- 9

Now, since all powers of 10 can only contain factors of the powers of 2 and 5, it clearly follows that your decimal never will come to an end if any other factor than these occurs in the denominator of your vulgar fraction. Thus, 1/2, 1/4, and 1/8 give us the exact decimals, .5, .25, and .125; 1/5 and 1/25 give us .2 and .04; 1/10 and 1/20 give us .1 and .05: because the denominators are all composed of 2 and 5 factors. But if you wish to convert 1/3, 1/6, or 1/7, your division sum will never end, but you will get these decimals, .33333, etc., .166666, etc., and .142857142857142857, etc., where, in the first case, the 3 keeps on repeating for ever and ever; in the second case the 6 is the repeater, and in the last case we get the recurring period of 142857. In the case of 1/17 (in "The Ribbon Problem") we find the circulating period to be .0588235294117647.

Now, in the division sum above, the successive remainders are 1, 10, 15, 14, 4, 6, 9, etc., and these numbers I have inserted around the inner ring of the diagram. It will be seen that every number from 1 to 16 occurs once, and that if we multiply our ribbon number by any one of the numbers in the inner ring its position indicates exactly the point at which the product will begin. Thus, if we multiply by 4, the product will be 235, etc.; if we multiply by 6, 352, etc. We can therefore multiply by any number from 1 to 16 and get the desired result.

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The kernel of the puzzle is this: Any prime number, with the exception of 2 and 5, which are the factors of 10, will exactly divide without remainder a number consisting of as many nines as the number itself, less one. Thus 999999 (six 9's) is divisible by 7, sixteen 9's are divisible by 17, eighteen 9's by 19, and so on. This is always the case, though frequently fewer 9's will suffice; for one 9 is divisible by 3, two by 11, six by 13, when our ribbon rule for consecutive multipliers breaks down and another law comes in. Therefore, since the 0 and 7 at the ends of the ribbon may not be removed, we must seek a fraction with a prime denominator ending in 7 that gives a full period circulator. We try 37, and find that it gives a short period decimal, .027, because 37 exactly divides 999; it, therefore, will not do. We next examine 47, and find that it gives us the full period circulator, in 46 figures, at the beginning of this article.

If you cut any of these full period circulators in half and place one half under the other, you will find that they will add up all 9's; so you need only work out one half and then write down the complements. Thus, in the ribbon above, if you add 05882352 to 94117647 the result is 99999999, and so with our long solution number. Note also in the diagram above that not only are the opposite numbers on the outer ring complementary, always making 9 when added, but that opposite numbers in the inner ring, our remainders, are also complementary, adding to 17 in every case. I ought perhaps to point out that in limiting our multipliers to the first nine numbers it seems just possible that a short period circulator might give a solution in fewer figures, but there are reasons for thinking it improbable.

84.--_The j.a.panese Ladies and the Carpet._

If the squares had not to be all the same size, the carpet could be cut in four pieces in any one of the three manners shown. In each case the two pieces marked A will fit together and form one of the three squares, the other two squares being entire. But in order to have the squares exactly equal in size, we shall require six pieces, as shown in the larger diagram. No. 1 is a complete square, pieces 4 and 5 will form a second square, and pieces 2, 3, and 6 will form the third--all of exactly the same size.

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If with the three equal squares we form the rectangle IDBA, then the mean proportional of the two sides of the rectangle will be the side of a square of equal area. Produce AB to C, making BC equal to BD. Then place the point of the compa.s.ses at E (midway between A and C) and describe the arc AC. I am showing the quite general method for converting rectangles to squares, but in this particular case we may, of course, at once place our compa.s.ses at E, which requires no finding. Produce the line BD, cutting the arc in F, and BF will be the required side of the square. Now mark off AG and DH, each equal to BF, and make the cut IG, and also the cut HK from H, perpendicular to ID. The six pieces produced are numbered as in the diagram on last page.

It will be seen that I have here given the reverse method first: to cut the three small squares into six pieces to form a large square. In the case of our puzzle we can proceed as follows:--

Make LM equal to half the diagonal ON. Draw the line NM and drop from L a perpendicular on NM. Then LP will be the side of all the three squares of combined area equal to the large square QNLO. The reader can now cut out without difficulty the six pieces, as shown in the numbered square on the last page.

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85.--_Captain Longbow and the Bears._

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It might have struck the reader that the story of the bear impaled on the North Pole had no connection with the problem that followed. As a matter of fact it is essential to a solution. Eleven bears cannot possibly be arranged to form of themselves seven rows of bears with four bears in every row. But it is a different matter when Captain Longbow informs us that "they had so placed themselves that _there were_" seven rows of four bears. For if they were grouped as shown in the diagram, so that three of the bears, as indicated, were in line with the North Pole, that impaled animal would complete the seventh row of four, which cannot be obtained in any other way. It obviously does not affect the problem whether this seventh row is a hundred miles long or a hundred feet, so long as they were really in a straight line--a point that might perhaps be settled by the captain's pocket compa.s.s.

86.--_The English Tour._

It was required to show how a resident at the town marked A might visit every one of the towns once, and only once, and finish up his tour at Z.

This puzzle conceals a little trick. After the solver has demonstrated to his satisfaction that it cannot be done in accordance with the conditions as he at first understood them, he should carefully examine the wording in order to find some flaw. It was said, "This would be easy enough if he were able to cut across country by road, as well as by rail, but he is not."

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Now, although he is prohibited from cutting across country by road, nothing is said about his going by sea! If, therefore, we carefully look again at the map, we shall find that two towns, and two only, lie on the sea coast. When he reaches one of these towns he takes his departure on board a coasting vessel and sails to the other port. The annexed ill.u.s.tration shows, by a dark line, the complete route.

This problem should be compared with No. 250, "The Grand Tour," in _A. in M._ It can be simplified in practically an identical manner, but as there is here no choice on the first stage from A, the solutions are necessarily quite different. See also solution to No. 94.

87.--_The Chifu-Chemulpo Puzzle._

The solution is as follows. You may accept the invitation to "try to do it in twenty moves," but you will never succeed in performing the feat.

The fewest possible moves are twenty-six. Play the cars so as to reach the following positions:--

E5678 -------- = 10 moves.

1234

E56 -------- = 2 moves.