Amusements in Mathematics Part 26

Starting from A, the inspector need only travel 36 furlongs if he takes the following route: A to B, G, H, C, D, I, H, M, N, I, J, O, N, S, R, M, L, G, F, K, L, Q, R, S, T, O, J, E, D, C, B, A, F, K, P, Q. He thus pa.s.ses between A and B twice, between C and D twice, between F and K twice, between J and O twice, and between R and S twice--five repet.i.tions. Therefore 31 pa.s.sages plus 5 repeated equal 36 furlongs. The little pitfall in this puzzle lies in the fact that we start from an even node. Otherwise we need only travel 35 furlongs.

248.--THE CYCLIST'S TOUR.

When Mr. Maggs replied, "No way, I'm sure," he was not saying that the thing was impossible, but was really giving the actual route by which the problem can be solved. Starting from the star, if you visit the towns in the order, NO WAY, I'M SURE, you will visit every town once, and only once, and end at E. So both men were correct. This was the little joke of the puzzle, which is not by any means difficult.

249.--THE SAILOR'S PUZZLE.

[Ill.u.s.tration]

There are only four different routes (or eight, if we count the reverse ways) by which the sailor can start at the island marked A, visit all the islands once, and once only, and return again to A. Here they are:-- A I P T L O E H R Q D C F U G N S K M B A A I P T S N G L O E U F C D K M B Q R H A A B M K S N G L T P I O E U F C D Q R H A A I P T L O E U G N S K M B Q D C F R H A Now, if the sailor takes the first route he will make C his 12th island (counting A as 1); by the second route he will make C his 13th island; by the third route, his 16th island; and by the fourth route, his 17th island. If he goes the reverse way, C will be respectively his 10th, 9th, 6th, and 5th island. As these are the only possible routes, it is evident that if the sailor puts off his visit to C as long as possible, he must take the last route reading from left to right. This route I show by the dark lines in the diagram, and it is the correct answer to the puzzle.

The map may be greatly simplified by the "b.u.t.tons and string" method, explained in the solution to No. 341, "The Four Frogs."

250.--THE GRAND TOUR.

The first thing to do in trying to solve a puzzle like this is to attempt to simplify it. If you look at Fig. 1, you will see that it is a simplified version of the map. Imagine the circular towns to be b.u.t.tons and the railways to be connecting strings. (See solution to No. 341.) Then, it will be seen, we have simply "straightened out" the previous diagram without affecting the conditions. Now we can further simplify by converting Fig. 1 into Fig. 2, which is a portion of a chessboard. Here the directions of the railways will resemble the moves of a rook in chess--that is, we may move in any direction parallel to the sides of the diagram, but not diagonally. Therefore the first town (or square) visited must be a black one; the second must be a white; the third must be a black; and so on. Every odd square visited will thus be black and every even one white. Now, we have 23 squares to visit (an odd number), so the last square visited must be black. But Z happens to be white, so the puzzle would seem to be impossible of solution.

[Ill.u.s.tration: Fig. 1.]

[Ill.u.s.tration: Fig. 2.]

As we were told that the man "succeeded" in carrying put his plan, we must try to find some loophole in the conditions. He was to "enter every town once and only once," and we find no prohibition against his entering once the town A after leaving it, especially as he has never left it since he was born, and would thus be "entering" it for the first time in his life. But he must return at once from the first town he visits, and then he will have only 22 towns to visit, and as 22 is an even number, there is no reason why he should not end on the white square Z. A possible route for him is indicated by the dotted line from A to Z. This route is repeated by the dark lines in Fig. 1, and the reader will now have no difficulty in applying; it to the original map. We have thus proved that the puzzle can only be solved by a return to A immediately after leaving it.

251.--WATER, GAS, AND ELECTRICITY.

[Ill.u.s.tration]

According to the conditions, in the strict sense in which one at first understands them, there is no possible solution to this puzzle. In such a dilemma one always has to look for some verbal quibble or trick. If the owner of house A will allow the water company to run their pipe for house C through his property (and we are not bound to a.s.sume that he would object), then the difficulty is got over, as shown in our ill.u.s.tration. It will be seen that the dotted line from W to C pa.s.ses through house A, but no pipe ever crosses another pipe.

252.--A PUZZLE FOR MOTORISTS.

[Ill.u.s.tration]

The routes taken by the eight drivers are shown in the ill.u.s.tration, where the dotted line roads are omitted to make the paths clearer to the eye.

253.--A BANK HOLIDAY PUZZLE.

The simplest way is to write in the number of routes to all the towns in this manner. Put a 1 on all the towns in the top row and in the first column. Then the number of routes to any town will be the sum of the routes to the town immediately above and to the town immediately to the left. Thus the routes in the second row will be 1, 2, 3, 4, 5, 6, etc., in the third row, 1, 3, 6, 10, 15, 21, etc.; and so on with the other rows. It will then be seen that the only town to which there are exactly 1,365 different routes is the twelfth town in the fifth row--the one immediately over the letter E. This town was therefore the cyclist's destination.

The general formula for the number of routes from one corner to the corner diagonally opposite on any such rectangular reticulated arrangement, under the conditions as to direction, is (m+n)!/m!n!, where m is the number of towns on one side, less one, and n the number on the other side, less one. Our solution involves the case where there are 12 towns by 5. Therefore m = 11 and n = 4. Then the formula gives us the answer 1,365 as above.

254.-- THE MOTOR-CAR TOUR.

First of all I will ask the reader to compare the original square diagram with the circular one shown in Figs. 1, 2, and 3 below. If for the moment we ignore the shading (the purpose of which I shall proceed to explain), we find that the circular diagram in each case is merely a simplification of the original square one--that is, the roads from A lead to B, E, and M in both cases, the roads from L (London) lead to I, K, and S, and so on. The form below, being circular and symmetrical, answers my purpose better in applying a mechanical solution, and I therefore adopt it without altering in any way the conditions of the puzzle. If such a question as distances from town to town came into the problem, the new diagrams might require the addition of numbers to indicate these distances, or they might conceivably not be at all practicable.

[Ill.u.s.tration: Figs. 1, 2, and 3]

Now, I draw the three circular diagrams, as shown, on a sheet of paper and then cut out three pieces of cardboard of the forms indicated by the shaded parts of these diagrams. It can be shown that every route, if marked out with a red pencil, will form one or other of the designs indicated by the edges of the cards, or a reflection thereof. Let us direct our attention to Fig. 1. Here the card is so placed that the star is at the town T; it therefore gives us (by following the edge of the card) one of the circular routes from London: L, S, R, T, M, A, E, P, O, J, D, C, B, G, N, Q, K, H, F, I, L. If we went the other way, we should get L, I, F, H, K, Q, etc., but these reverse routes were not to be counted. When we have written out this first route we revolve the card until the star is at M, when we get another different route, at A a third route, at E a fourth route, and at P a fifth route. We have thus obtained five different routes by revolving the card as it lies. But it is evident that if we now take up the card and replace it with the other side uppermost, we shall in the same manner get five other routes by revolution.

We therefore see how, by using the revolving card in Fig. 1, we may, without any difficulty, at once write out ten routes. And if we employ the cards in Figs. 2 and 3, we similarly obtain in each case ten other routes. These thirty routes are all that are possible. I do not give the actual proof that the three cards exhaust all the possible cases, but leave the reader to reason that out for himself. If he works out any route at haphazard, he will certainly find that it falls into one or other of the three categories.

255.--THE LEVEL PUZZLE.

Let us confine our attention to the L in the top left-hand corner. Suppose we go by way of the E on the right: we must then go straight on to the V, from which letter the word may be completed in four ways, for there are four E's available through which we may reach an L. There are therefore four ways of reading through the right-hand E. It is also clear that there must be the same number of ways through the E that is immediately below our starting point. That makes eight. If, however, we take the third route through the E on the diagonal, we then have the option of any one of the three V's, by means of each of which we may complete the word in four ways. We can therefore spell LEVEL in twelve ways through the diagonal E. Twelve added to eight gives twenty readings, all emanating from the L in the top left-hand corner; and as the four corners are equal, the answer must be four times twenty, or eighty different ways.

256.--THE DIAMOND PUZZLE.

There are 252 different ways. The general formula is that, for words of n letters (not palindromes, as in the case of the next puzzle), when grouped in this manner, there are always 2^(n+1) - 4 different readings. This does not allow diagonal readings, such as you would get if you used instead such a word as DIGGING, where it would be possible to pa.s.s from one G to another G by a diagonal step.

257.--THE DEIFIED PUZZLE.

The correct answer is 1,992 different ways. Every F is either a corner F or a side F--standing next to a corner in its own square of F's. Now, FIED may be read from a corner F in 16 ways; therefore DEIF may be read into a corner F also in 16 ways; hence DEIFIED may be read through a corner F in 16 16 = 256 ways. Consequently, the four corner F's give 4 256 = 1,024 ways. Then FIED may be read from a side F in 11 ways, and DEIFIED therefore in 121 ways. But there are eight side F's; consequently these give together 8 121 = 968 ways. Add 968 to 1,024 and we get the answer, 1,992.

In this form the solution will depend on whether the number of letters in the palindrome be odd or even. For example, if you apply the word NUN in precisely the same manner, you will get 64 different readings; but if you use the word NOON, you will only get 56, because you cannot use the same letter twice in immediate succession (since you must "always pa.s.s from one letter to another") or diagonal readings, and every reading must involve the use of the central N.

The reader may like to find for himself the general formula in this case, which is complex and difficult. I will merely add that for such a case as MADAM, dealt with in the same way as DEIFIED, the number of readings is 400.

258.-- THE VOTERS' PUZZLE.

THE number of readings here is 63,504, as in the case of "WAS IT A RAT I SAW" (No. 30, _Canterbury Puzzles_). The general formula is that for palindromic sentences containing 2n + 1 letters there are (4(2^n -1)) readings.

259.-- HANNAH'S PUZZLE.

Starting from any one of the N's, there are 17 different readings of NAH, or 68 (4 times 17) for the 4 N's. Therefore there are also 68 ways of spelling HAN. If we were allowed to use the same N twice in a spelling, the answer would be 68 times 68, or 4,624 ways. But the conditions were, "always pa.s.sing from one letter to another." Therefore, for every one of the 17 ways of spelling HAN with a particular N, there would be 51 ways (3 times 17) of completing the NAH, or 867 (17 times 51) ways for the complete word. Hence, as there are four N's to use in HAN, the correct solution of the puzzle is 3,468 (4 times 867) different ways.

260.--THE HONEYCOMB PUZZLE.

The required proverb is, "There is many a slip 'twixt the cup and the lip." Start at the T on the outside at the bottom right-hand corner, pa.s.s to the H above it, and the rest is easy.

261.-- THE MONK AND THE BRIDGES.

[Ill.u.s.tration]

The problem of the Bridges may be reduced to the simple diagram shown in ill.u.s.tration. The point M represents the Monk, the point I the Island, and the point Y the Monastery. Now the only direct ways from M to I are by the bridges a and b; the only direct ways from I to Y are by the bridges c and d; and there is a direct way from M to Y by the bridge e. Now, what we have to do is to count all the routes that will lead from M to Y, pa.s.sing over all the bridges, a, b, c, d, and e once and once only. With the simple diagram under the eye it is quite easy, without any elaborate rule, to count these routes methodically. Thus, starting from a, b, we find there are only two ways of completing the route; with _a, c_, there are only two routes; with a, d, only two routes; and so on. It will be found that there are sixteen such routes in all, as in the following list:-- a b e c d b c d a e a b e d c b c e a d a c d b e b d c a e a c e b d b d e a c a d e b c e c a b d a d c b e e c b a d b a e c d e d a b c b a e d c e d b a c If the reader will transfer the letters indicating the bridges from the diagram to the corresponding bridges in the original ill.u.s.tration, everything will be quite obvious.

262.--THOSE FIFTEEN SHEEP.

If we read the exact words of the writer in the cyclopaedia, we find that we are not told that the pens were all necessarily empty! In fact, if the reader will refer back to the ill.u.s.tration, he will see that one sheep is already in one of the pens. It was just at this point that the wily farmer said to me, "Now I'm going to start placing the fifteen sheep." He thereupon proceeded to drive three from his flock into the already occupied pen, and then placed four sheep in each of the other three pens. "There," says he, "you have seen me place fifteen sheep in four pens so that there shall be the same number of sheep in every pen." I was, of course, forced to admit that he was perfectly correct, according to the exact wording of the question.

263.--KING ARTHUR'S KNIGHTS.

On the second evening King Arthur arranged the knights and himself in the following order round the table: A, F, B, D, G, E, C. On the third evening they sat thus, A, E, B, G, C, F, D. He thus had B next but one to him on both occasions (the nearest possible), and G was the third from him at both sittings (the furthest position possible). No other way of sitting the knights would have been so satisfactory.

264.--THE CITY LUNCHEONS.

The men may be grouped as follows, where each line represents a day and each column a table:-- AB CD EF GH IJ KL AE DL GK FI CB HJ AG LJ FH KC DE IB AF JB KI HD LG CE AK BE HC IL JF DG AH EG ID CJ BK LF AI GF CL DB EH JK AC FK DJ LE GI BH AD KH LB JG FC EI AL HI JE BF KD GC AJ IC BG EK HL FD Note that in every column (except in the case of the A's) all the letters descend cyclically in the same order, B, E, G, F, up to J, which is followed by B.

265.--A PUZZLE FOR CARD-PLAYERS.

In the following solution each of the eleven lines represents a sitting, each column a table, and each pair of letters a pair of partners.

A B -- I L | E J -- G K | F H -- C D A C -- J B | F K -- H L | G I -- D E A D -- K C | G L -- I B | H J -- E F A E -- L D | H B -- J C | I K -- F G A F -- B E | I C -- K D | J L -- G H A G -- C F | J D -- L E | K B -- H I A H -- D G | K E -- B F | L C -- I J A I -- E H | L F -- C G | B D -- J K A J -- F I | B G -- D H | C E -- K L A K -- G J | C H -- E I | D F -- L B A L -- H K | D I -- F J | E G -- B C It will be seen that the letters B, C, D ...L descend cyclically. The solution given above is absolutely perfect in all respects. It will be found that every player has every other player once as his partner and twice as his opponent.

266.--A TENNIS TOURNAMENT.

Call the men A, B, D, E, and their wives a, b, d, e. Then they may play as follows without any person ever playing twice with or against any other person:-- First Court. Second Court. 1st Day | A d against B e | D a against E b 2nd Day | A e " D b | E a " B d 3rd Day | A b " E d | B a " D e It will be seen that no man ever plays with or against his own wife--an ideal arrangement. If the reader wants a hard puzzle, let him try to arrange eight married couples (in four courts on seven days) under exactly similar conditions. It can be done, but I leave the reader in this case the pleasure of seeking the answer and the general solution.

267.--THE WRONG HATS.

The number of different ways in which eight persons, with eight hats, can each take the wrong hat, is 14,833.

Here are the successive solutions for any number of persons from one to eight:-- 1 = 0 2 = 1 3 = 2 4 = 9 5 = 44 6 = 265 7 = 1,854 8 = 14,833 To get these numbers, multiply successively by 2, 3, 4, 5, etc. When the multiplier is even, add 1; when odd, deduct 1. Thus, 3 1 - 1 = 2; 4 2 + 1 = 9; 5 9 - 1 = 44; and so on. Or you can multiply the sum of the number of ways for n - 1 and n - 2 persons by n - 1, and so get the solution for n persons. Thus, 4(2 + 9) = 44; 5(9 + 44) = 265; and so on.

268.--THE PEAL OF BELLS.

The bells should be rung as follows:-- 1 2 3 4 2 1 4 3 2 4 1 3 4 2 3 1 4 3 2 1 3 4 1 2 3 1 4 2 1 3 2 4 3 1 2 4 1 3 4 2 1 4 3 2 4 1 2 3 4 2 1 3 2 4 3 1 2 3 4 1 3 2 1 4 2 3 1 4 3 2 4 1 3 4 2 1 4 3 1 2 4 1 3 2 1 4 2 3 1 2 4 3 2 1 3 4 I have constructed peals for five and six bells respectively, and a solution is possible for any number of bells under the conditions previously stated.

269.--THREE MEN IN A BOAT.

If there were no conditions whatever, except that the men were all to go out together, in threes, they could row in an immense number of different ways. If the reader wishes to know how many, the number is 455^7. And with the condition that no two may ever be together more than once, there are no fewer than 15,567,552,000 different solutions--that is, different ways of arranging the men. With one solution before him, the reader will realize why this must be, for although, as an example, A must go out once with B and once with C, it does not necessarily follow that he must go out with C on the same occasion that he goes with B. He might take any other letter with him on that occasion, though the fact of his taking other than B would have its effect on the arrangement of the other triplets.

Of course only a certain number of all these arrangements are available when we have that other condition of using the smallest possible number of boats. As a matter of fact we need employ only ten different boats. Here is one the arrangements:-- 1 2 3 4 5 1st Day (ABC) (DBF) (GHI) (JKL) (MNO) 8 6 7 9 10 2nd Day (ADG) (BKN) (COL) (JEI) (MHF) 3 5 4 1 2 3rd Day (AJM) (BEH) (CFI) (DKO) (GNL) 7 6 8 9 1 4th Day (AEK) (CGM) (BOI) (DHL) (JNF) 4 5 3 10 2 5th Day (AHN) (CDJ) (BFL) (GEO) (MKI) 6 7 8 10 1 6th Day (AFO) (BGJ) (CKH) (DNI) (MEL) 5 4 3 9 2 7th Day (AIL) (BDM) (CEN) (GKF) (JHO) It will be found that no two men ever go out twice together, and that no man ever goes out twice in the same boat.

This is an extension of the well-known problem of the "Fifteen Schoolgirls," by Kirkman. The original conditions were simply that fifteen girls walked out on seven days in triplets without any girl ever walking twice in a triplet with another girl. Attempts at a general solution of this puzzle had exercised the ingenuity of mathematicians since 1850, when the question was first propounded, until recently. In 1908 and the two following years I indicated (see Educational Times Reprints, Vols. XIV., XV., and XVII.) that all our trouble had arisen from a failure to discover that 15 is a special case (too small to enter into the general law for all higher numbers of girls of the form 6n+3), and showed what that general law is and how the groups should be posed for any number of girls. I gave actual arrangements for numbers that had previously baffled all attempts to manipulate, and the problem may now be considered generally solved. Readers will find an excellent full account of the puzzle in W.W. Rouse Ball's Mathematical Recreations, 5th edition.

270.--THE GLa.s.s b.a.l.l.s.

There are, in all, sixteen b.a.l.l.s to be broken, or sixteen places in the order of breaking. Call the four strings A, B, C, and D--order is here of no importance. The breaking of the b.a.l.l.s on A may occupy any 4 out of these 16 places--that is, the combinations of 16 things, taken 4 together, will be 13 14 15 16 ----------------- = 1,820 1 2 3 4 ways for A. In every one of these cases B may occupy any 4 out of the remaining 12 places, making 9 10 11 12 ----------------- = 495 1 2 3 4 ways. Thus 1,820 495 = 900,900 different placings are open to A and B. But for every one of these cases C may occupy 5 6 7 8 ------------- = 70 1 2 3 4 different places; so that 900,900 70 = 63,063,000 different placings are open to A, B, and C. In every one of these cases, D has no choice but to take the four places that remain. Therefore the correct answer is that the b.a.l.l.s may be broken in 63,063,000 different ways under the conditions. Readers should compare this problem with No. 345, "The Two p.a.w.ns," which they will then know how to solve for cases where there are three, four, or more p.a.w.ns on the board.

271.--FIFTEEN LETTER PUZZLE.

The following will be found to comply with the conditions of grouping:-- ALE MET MOP BLM BAG CAP YOU CLT IRE OIL LUG LNR NAY BIT BUN BPR AIM BEY RUM GMY OAR GIN PLY CGR PEG ICY TRY CMN CUE COB TAU PNT ONE GOT PIU The fifteen letters used are A, E, I, O, U, Y, and B, C, G, L, M, N, P, R, T. The number of words is 27, and these are all shown in the first three columns. The last word, PIU, is a musical term in common use; but although it has crept into some of our dictionaries, it is Italian, meaning "a little; slightly." The remaining twenty-six are good words. Of course a TAU-cross is a T-shaped cross, also called the cross of St. Anthony, and borne on a badge in the Bishop's Palace at Exeter. It is also a name for the toad-fish.

We thus have twenty-six good words and one doubtful, obtained under the required conditions, and I do not think it will be easy to improve on this answer. Of course we are not bound by dictionaries but by common usage. If we went by the dictionary only in a case of this kind, we should find ourselves involved in prefixes, contractions, and such absurdities as I.O.U., which Nuttall actually gives as a word.

272.--THE NINE SCHOOLBOYS.

The boys can walk out as follows:-- 1st Day. 2nd Day. 3rd Day. A B C B F H F A G D E F E I A I D B G H I C G D H C E 4th Day. 5th Day. 6th Day. A D H G B I D C A B E G C F D E H B F I C H A E I G F Every boy will then have walked by the side of every other boy once and once only.

Dealing with the problem generally, 12n+9 boys may walk out in triplets under the conditions on 9n+6 days, where n may be nought or any integer. Every possible pair will occur once. Call the number of boys m. Then every boy will pair m-1 times, of which (m-1)/4 times he will be in the middle of a triplet and (m-1)/2 times on the outside. Thus, if we refer to the solution above, we find that every boy is in the middle twice (making 4 pairs) and four times on the outside (making the remaining 4 pairs of his 8). The reader may now like to try his hand at solving the two next cases of 21 boys on 15 days, and 33 boys on 24 days. It is, perhaps, interesting to note that a school of 489 boys could thus walk out daily in one leap year, but it would take 731 girls (referred to in the solution to No. 269) to perform their particular feat by a daily walk in a year of 365 days.

273.--THE ROUND TABLE.

The history of this problem will be found in The Canterbury Puzzles (No. 90). Since the publication of that book in 1907, so far as I know, n.o.body has succeeded in solving the case for that unlucky number of persons, 13, seated at a table on 66 occasions. A solution is possible for any number of persons, and I have recorded schedules for every number up to 25 persons inclusive and for 33. But as I know a good many mathematicians are still considering the case of 13, I will not at this stage rob them of the pleasure of solving it by showing the answer. But I will now display the solutions for all the cases up to 12 persons inclusive. Some of these solutions are now published for the first time, and they may afford useful clues to investigators.

The solution for the case of 3 persons seated on 1 occasion needs no remark.

A solution for the case of 4 persons on 3 occasions is as follows:-- 1 2 3 4 1 3 4 2 1 4 2 3 Each line represents the order for a sitting, and the person represented by the last number in a line must, of course, be regarded as sitting next to the first person in the same line, when placed at the round table.

The case of 5 persons on 6 occasions may be solved as follows:-- 1 2 3 4 5 1 2 4 5 3 1 2 5 3 4 --------- 1 3 2 5 4 1 4 2 3 5 1 5 2 4 3 The case for 6 persons on 10 occasions is solved thus:-- 1 2 3 6 4 5 1 3 4 2 5 6 1 4 5 3 6 2 1 5 6 4 2 3 1 6 2 5 3 4 ----------- 1 2 4 5 6 3 1 3 5 6 2 4 1 4 6 2 3 5 1 5 2 3 4 6 1 6 3 4 5 2 It will now no longer be necessary to give the solutions in full, for reasons that I will explain. It will be seen in the examples above that the 1 (and, in the case of 5 persons, also the 2) is repeated down the column. Such a number I call a "repeater." The other numbers descend in cyclical order. Thus, for 6 persons we get the cycle, 2, 3, 4, 5, 6, 2, and so on, in every column. So it is only necessary to give the two lines 1 2 3 6 4 5 and 1 2 4 5 6 3, and denote the cycle and repeaters, to enable any one to write out the full solution straight away. The reader may wonder why I do not start the last solution with the numbers in their natural order, 1 2 3 4 5 6. If I did so the numbers in the descending cycle would not be in their natural order, and it is more convenient to have a regular cycle than to consider the order in the first line.

The difficult case of 7 persons on 15 occasions is solved as follows, and was given by me in _The Canterbury Puzzles_:-- 1 2 3 4 5 7 6 1 6 2 7 5 3 4 1 3 5 2 6 7 4 1 5 7 4 3 6 2 1 5 2 7 3 4 6 In this case the 1 is a repeater, and there are two separate cycles, 2, 3, 4, 2, and 5, 6, 7, 5. We thus get five groups of three lines each, for a fourth line in any group will merely repeat the first line.

A solution for 8 persons on 21 occasions is as follows:-- 1 8 6 3 4 5 2 7 1 8 4 5 7 2 3 6 1 8 2 7 3 6 4 5 The 1 is here a repeater, and the cycle 2, 3, 4, 5, 6, 7, 8. Every one of the 3 groups will give 7 lines.

Here is my solution for 9 persons on 28 occasions:-- 2 1 9 7 4 5 6 3 8 2 9 5 1 6 8 3 4 7 2 9 3 1 8 4 7 5 6 2 9 1 5 6 4 7 8 3 There are here two repeaters, 1 and 2, and the cycle is 3, 4, 5, 6, 7, 8, 9. We thus get 4 groups of 7 lines each.

The case of 10 persons on 36 occasions is solved as follows:-- 1 10 8 3 6 5 4 7 2 9 1 10 6 5 2 9 7 4 3 8 1 10 2 9 3 8 6 5 7 4 1 10 7 4 8 3 2 9 5 6 The repeater is 1, and the cycle, 2, 3, 4, 5, 6, 7, 8, 9, 10. We here have 4 groups of 9 lines each.

My solution for 11 persons on 45 occasions is as follows:-- 2 11 9 4 7 6 5 1 8 3 10 2 1 11 7 6 3 10 8 5 4 9 2 11 10 3 9 4 8 5 1 7 6 2 11 5 8 1 3 10 6 7 9 4 2 11 1 10 3 4 9 6 7 5 8 There are two repeaters, 1 and 2, and the cycle is, 3, 4, 5,... 11. We thus get 5 groups of 9 lines each.

The case of 12 persons on 55 occasions is solved thus:-- 1 2 3 12 4 11 5 10 6 9 7 8 1 2 4 11 6 9 8 7 10 5 12 3 1 2 5 10 8 7 11 4 3 12 6 9 1 2 6 9 10 5 3 12 7 8 11 4 1 2 7 8 12 3 6 9 11 4 5 10 Here 1 is a repeater, and the cycle is 2, 3, 4, 5,... 12. We thus get 5 groups of 11 lines each.

274.--THE MOUSE-TRAP PUZZLE.

If we interchange cards 6 and 13 and begin our count at 14, we may take up all the twenty-one cards--that is, make twenty-one "catches"--in the following order: 6, 8, 13, 2, 10, 1, 11, 4, 14, 3, 5, 7, 21, 12, 15, 20, 9, 16, 18, 17, 19. We may also exchange 10 and 14 and start at 16, or exchange 6 and 8 and start at 19.

275.--THE SIXTEEN SHEEP.

The six diagrams on next page show solutions for the cases where we replace 2, 3, 4, 5, 6, and 7 hurdles. The dark lines indicate the hurdles that have been replaced. There are, of course, other ways of making the removals.

276.--THE EIGHT VILLAS.

There are several ways of solving the puzzle, but there is very little difference between them. The solver should, however, first of all bear in mind that in making his calculations he need only consider the four villas that stand at the corners, because the intermediate villas can never vary when the corners are known. One way is to place the numbers nought to 9 one at a time in the top left-hand corner, and then consider each case in turn.

Now, if we place 9 in the corner as shown in the Diagram A, two of the corners cannot be occupied, while the corner that is diagonally opposite may be filled by 0, 1, 2, 3, 4, 5, 6, 7, 8, or 9 persons. We thus see that there are 10 [Ill.u.s.tration: +---+---+ +-+-----+ +---+---+ |O OHO O| |OHO O O| |O OHO O| | H | | + | | +=+ | |O OHO O| |OHO O O| |O OHOHO| +-+ +-+-+ +-+-----+ +---+ + | |O|O O|O| |O|O O O| |O O O|O| | +---+ | | +-+-+ | | +-+ | |O O O O| |O O OHO| |O O|O O| +-------+ +-------+ +-------+ 2 3 4 +-----+-+ +-+-----+ +-------+ |O O OHO| |OHO O O| |O O O O| | +=+ | | +=+ | | +=+=+=+ |O OHO O| |OHOHO O| |OHOHO O| | +-+-+ + | + +-+ | + + + | |O|O O|O| |O|O O|O| |O|OHO O| +=+ +=+ | + +=+ +=+ + | |O O O O| |OHO O O| |O O|O O| +-------+ +-+-----+ +---+---+ 5 6 7 THE SIXTEEN SHEEP ]

solutions with a 9 in the corner. If, however, we subst.i.tute 8, the two corners in the same row and column may contain 0, 0, or 1, 1, or 0, 1, or 1, 0. In the case of B, ten different selections may be made for the fourth corner; but in each of the cases C, D, and E, only nine selections are possible, because we cannot use the 9. Therefore with 8 in the top left-hand corner there are 10 + (3 9) = 37 different solutions. If we then try 7 in the corner, the result will be 10 + 27 + 40, or 77 solutions. With 6 we get 10 + 27 + 40 + 49 = 126; with 5, 10 + 27 + 40 + 49 + 54 = 180; with 4, the same as with 5, + 55 = 235 ; with 3, the same as with 4, + 52 = 287; with 2, the same as with 3, + 45 = 332; with 1, the same as with 2, + 34 = 366, and with nought in the top left-hand corner the number of solutions will be found to be 10 + 27 + 40 + 49 + 54 + 55 + 52 + 45 + 34 + 19 = 385. As there is no other number to be placed in the top left-hand corner, we have now only to add these totals together thus, 10 + 37 + 77 + 126 + 180 + 235 + 287 + 332 + 366 + 385 = 2,035. We therefore find that the total number of ways in which tenants may occupy some or all of the eight villas so that there shall be always nine persons living along each side of the square is 2,035. Of course, this method must obviously cover all the reversals and reflections, since each corner in turn is occupied by every number in all possible combinations with the other two corners that are in line with it.

[Ill.u.s.tration: A B C D E +-+-+-+ +-+-+-+ +-+-+-+ +-+-+-+ +-+-+-+ |9| |0| |8| |0| |8| |1| |8| |0| |8| |1| +-+-+-+ +-+-+-+ +-+-+-+ +-+-+-+ +-+-+-+ | |*| | | |*| | | |*| | | |*| | | |*| | +-+-+-+ +-+-+-+ +-+-+-+ +-+-+-+ +-+-+-+ |0| | | |0| | | |1| | | |1| | | |0| | | +-+-+-+ +-+-+-+ +-+-+-+ +-+-+-+ +-+-+-+ ]

Here is a general formula for solving the puzzle: (n + 3n + 2)(n + 3n + 3)/6. Whatever may be the stipulated number of residents along each of the sides (which number is represented by n), the total number of different arrangements may be thus ascertained. In our particular case the number of residents was nine. Therefore (81 + 27 + 2) (81 + 27 + 3) and the product, divided by 6, gives 2,035. If the number of residents had been 0, 1, 2, 3, 4, 5, 6, 7, or 8, the total arrangements would be 1, 7, 26, 70, 155, 301, 532, 876, or 1,365 respectively.

277.--COUNTER CROSSES.

Let us first deal with the Greek Cross. There are just eighteen forms in which the numbers may be paired for the two arms. Here they are:-- 12978 13968 14958 34956 24957 23967 23958 13769 14759 14967 24758 23768 12589 23759 13579 34567 14768 24568 14569 23569 14379 23578 14578 25368 15369 24369 23189 24378 15378 45167 24179 25169 34169 35168 34178 25178 Of course, the number in the middle is common to both arms. The first pair is the one I gave as an example. I will suppose that we have written out all these crosses, always placing the first row of a pair in the upright and the second row in the horizontal arm. Now, if we leave the central figure fixed, there are 24 ways in which the numbers in the upright may be varied, for the four counters may be changed in 1 2 3 4 = 24 ways. And as the four in the horizontal may also be changed in 24 ways for every arrangement on the other arm, we find that there are 24 24 = 576 variations for every form; therefore, as there are 18 forms, we get 18 576 = 10,368 ways. But this will include half the four reversals and half the four reflections that we barred, so we must divide this by 4 to obtain the correct answer to the Greek Cross, which is thus 2,592 different ways. The division is by 4 and not by 8, because we provided against half the reversals and reflections by always reserving one number for the upright and the other for the horizontal.

In the case of the Latin Cross, it is obvious that we have to deal with the same 18 forms of pairing. The total number of different ways in this case is the full number, 18 576. Owing to the fact that the upper and lower arms are unequal in length, permutations will repeat by reflection, but not by reversal, for we cannot reverse. Therefore this fact only entails division by 2. But in every pair we may exchange the figures in the upright with those in the horizontal (which we could not do in the case of the Greek Cross, as the arms are there all alike); consequently we must multiply by 2. This multiplication by 2 and division by 2 cancel one another. Hence 10,368 is here the correct answer.

278.--A DORMITORY PUZZLE.

[Ill.u.s.tration: MON. TUES. WED. +---+---+---+ +---+---+---+ +---+---+---+ | 1 | 2 | 1 | | 1 | 3 | 1 | | 1 | 4 | 1 | +---+---+---+ +---+---+---+ +---+---+---+ | 2 | | 2 | | 1 | | 1 | | 1 | | 1 | +---+---+---+ +---+---+---+ +---+---+---+ | 1 | 22| 1 | | 3 | 19| 3 | | 4 | 16| 4 | +---+---+---+ +---+---+---+ +---+---+---+ THURS. FRI. SAT. +---+---+---+ +---+---+---+ +---+---+---+ | 1 | 5 | 1 | | 2 | 6 | 2 | | 4 | 4 | 4 | +---+---+---+ +---+---+---+ +---+---+---+ | 2 | | 2 | | 1 | | 1 | | 4 | | 4 | +---+---+---+ +---+---+---+ +---+---+---+ | 4 | 13| 4 | | 7 | 6 | 7 | | 4 | 4 | 4 | +---+---+---+ +---+---+---+ +---+---+---+ ]

Arrange the nuns from day to day as shown in the six diagrams. The smallest possible number of nuns would be thirty-two, and the arrangements on the last three days admit of variation.

279.--THE BARRELS OF BALSAM.

This is quite easy to solve for any number of barrels--if you know how. This is the way to do it. There are five barrels in each row Multiply the numbers 1, 2, 3, 4, 5 together; and also multiply 6, 7, 8, 9, 10 together. Divide one result by the other, and we get the number of different combinations or selections of ten things taken five at a time. This is here 252. Now, if we divide this by 6 (1 more than the number in the row) we get 42, which is the correct answer to the puzzle, for there are 42 different ways of arranging the barrels. Try this method of solution in the case of six barrels, three in each row, and you will find the answer is 5 ways. If you check this by trial, you will discover the five arrangements with 123, 124, 125, 134, 135 respectively in the top row, and you will find no others.