Amusements in Mathematics Part 20

47.--ROVER'S AGE.

Rover's present age is ten years and Mildred's thirty years. Five years ago their respective ages were five and twenty-five. Remember that we said "four times older than the dog," which is the same as "five times as old." (See answer to No. 44.) 48.--CONCERNING TOMMY'S AGE.

Tommy Smart's age must have been nine years and three-fifths. Ann's age was sixteen and four-fifths, the mother's thirty-eight and two-fifths, and the father's fifty and two-fifths.

49.--NEXT-DOOR NEIGHBOURS.

Mr. Jupp 39, Mrs. Jupp 34, Julia 14, and Joe 13; Mr. Simkin 42; Mrs. Simkin 40; Sophy 10; and Sammy 8.

50.--THE BAG OF NUTS.

It will be found that when Herbert takes twelve, Robert and Christopher will take nine and fourteen respectively, and that they will have together taken thirty-five nuts. As 35 is contained in 770 twenty-two times, we have merely to multiply 12, 9, and 14 by 22 to discover that Herbert's share was 264, Robert's 198, and Christopher's 308. Then, as the total of their ages is 17 years or half the sum of 12, 9, and 14, their respective ages must be 6, 4, and 7 years.

51.--HOW OLD WAS MARY?

The age of Mary to that of Ann must be as 5 to 3. And as the sum of their ages was 44, Mary was 27 and Ann 16. One is exactly 11 years older than the other. I will now insert in brackets in the original statement the various ages specified: "Mary is (27) twice as old as Ann was (13) when Mary was half as old (24) as Ann will be (49) when Ann is three times as old (49) as Mary was (16) when Mary was (16) three times as old as Ann (5)." Now, check this backwards. When Mary was three times as old as Ann, Mary was 16 and Ann 5 (11 years younger). Then we get 49 for the age Ann will be when she is three times as old as Mary was then. When Mary was half this she was 24. And at that time Ann must have been 13 (11 years younger). Therefore Mary is now twice as old--27, and Ann 11 years younger--16.

52.--QUEER RELATIONs.h.i.+PS.

If a man marries a woman, who dies, and he then marries his deceased wife's sister and himself dies, it may be correctly said that he had (previously) married the sister of his widow.

The youth was not the nephew of Jane Brown, because he happened to be her son. Her surname was the same as that of her brother, because she had married a man of the same name as herself.

53.--HEARD ON THE TUBE RAILWAY.

The gentleman was the second lady's uncle.

54.--A FAMILY PARTY.

The party consisted of two little girls and a boy, their father and mother, and their father's father and mother.

55.--A MIXED PEDIGREE.

[Ill.u.s.tration: Thos. Bloggs m ..... | +------------------------+------------+ | | | | | | | W. Snoggs m Kate Bloggs. | | | | | | | . . m Henry Bloggs. | Joseph Bloggs m | | | | +--------+-------------+ | | | | | | | | | Jane John Alf. Mary Bloggs m Snoggs Snoggs m Bloggs ]

The letter m stands for "married." It will be seen that John Snoggs can say to Joseph Bloggs, "You are my _father's brother-in-law_, because my father married your sister Kate; you are my _brother's father-in-law_, because my brother Alfred married your daughter Mary; and you are my _father-in-law's brother_, because my wife Jane was your brother Henry's daughter."

56.--WILSON'S POSER.

If there are two men, each of whom marries the mother of the other, and there is a son of each marriage, then each of such sons will be at the same time uncle and nephew of the other. There are other ways in which the relations.h.i.+p may be brought about, but this is the simplest.

57.--WHAT WAS THE TIME?

The time must have been 9.36 p.m. A quarter of the time since noon is 2 hr. 24 min., and a half of the time till noon next day is 7 hr. 12 min. These added together make 9 hr. 36 min.

58.--A TIME PUZZLE.

Twenty-six minutes.

59.--A PUZZLING WATCH.

If the 65 minutes be counted on the face of the same watch, then the problem would be impossible: for the hands must coincide every 65+5/11 minutes as shown by its face, and it matters not whether it runs fast or slow; but if it is measured by true time, it gains 5/11 of a minute in 65 minutes, or 60/143 of a minute per hour.

60.--THE WAPSHAW'S WHARF MYSTERY.

There are eleven different times in twelve hours when the hour and minute hands of a clock are exactly one above the other. If we divide 12 hours by 11 we get 1 hr. 5 min. 27+3/11 sec., and this is the time after twelve o'clock when they are first together, and also the time that elapses between one occasion of the hands being together and the next. They are together for the second time at 2 hr. 10 min. 54+6/11 sec. (twice the above time); next at 3 hr. 16 min. 21+9/11 sec.; next at 4 hr. 21 min. 49+1/11 sec. This last is the only occasion on which the two hands are together with the second hand "just past the forty-ninth second." This, then, is the time at which the watch must have stopped. Guy Boothby, in the opening sentence of his Across the World for a Wife, says, "It was a cold, dreary winter's afternoon, and by the time the hands of the clock on my mantelpiece joined forces and stood at twenty minutes past four, my chambers were well-nigh as dark as midnight." It is evident that the author here made a slip, for, as we have seen above, he is 1 min. 49+1/11 sec. out in his reckoning.

61.--CHANGING PLACES.

There are thirty-six pairs of times when the hands exactly change places between three p.m. and midnight. The number of pairs of times from any hour (n) to midnight is the sum of 12 - (n + 1) natural numbers. In the case of the puzzle n = 3; therefore 12 - (3 + 1) = 8 and 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 = 36, the required answer.

The first pair of times is 3 hr. 21+57/143 min. and 4 hr. 16+112/143 min., and the last pair is 10 hr. 59+83/143 min. and 11 hr. 54+138/143 min. I will not give all the remainder of the thirty-six pairs of times, but supply a formula by which any of the sixty-six pairs that occur from midday to midnight may be at once found:-- 720b + 60a 720a + 60b min. a hr ---------- min. and b hr. --------------- 143 143 For the letter a may be subst.i.tuted any hour from 0, 1, 2, 3 up to 10 (where nought stands for 12 o'clock midday); and b may represent any hour, later than a, up to 11.

By the aid of this formula there is no difficulty in discovering the answer to the second question: a = 8 and b = 11 will give the pair 8 hr. 58+106/143 min. and 11 hr. 44+128/143 min., the latter being the time when the minute hand is nearest of all to the point IX--in fact, it is only 15/143 of a minute distant.

Readers may find it instructive to make a table of all the sixty-six pairs of times when the hands of a clock change places. An easy way is as follows: Make a column for the first times and a second column for the second times of the pairs. By making a = 0 and b = 1 in the above expressions we find the first case, and enter hr. 5+5/143 min. at the head of the first column, and 1 hr. 0+60/143 min. at the head of the second column. Now, by successively adding 5+5/143 min. in the first, and 1 hr. 0+60/143 min. in the second column, we get all the eleven pairs in which the first time is a certain number of minutes after nought, or mid-day. Then there is a "jump" in the times, but you can find the next pair by making a = 1 and b = 2, and then by successively adding these two times as before you will get all the ten pairs after 1 o'clock. Then there is another "jump," and you will be able to get by addition all the nine pairs after 2 o'clock. And so on to the end. I will leave readers to investigate for themselves the nature and cause of the "jumps." In this way we get under the successive hours, 11 + 10 + 9 + 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1 = 66 pairs of times, which result agrees with the formula in the first paragraph of this article.

Some time ago the princ.i.p.al of a Civil Service Training College, who conducts a "Civil Service Column" in one of the periodicals, had the query addressed to him, "How soon after XII o'clock will a clock with both hands of the same length be ambiguous?" His first answer was, "Some time past one o'clock," but he varied the answer from issue to issue. At length some of his readers convinced him that the answer is, "At 5+5/143 min. past XII;" and this he finally gave as correct, together with the reason for it that at that time _the time indicated is the same whichever hand you may a.s.sume as hour hand!_ 62.--THE CLUB CLOCK.

The positions of the hands shown in the ill.u.s.tration could only indicate that the clock stopped at 44 min. 51+1143/1427 sec. after eleven o'clock. The second hand would next be "exactly midway between the other two hands" at 45 min. 52+496/1427 sec. after eleven o'clock. If we had been dealing with the points on the circle to which the three hands are directed, the answer would be 45 min. 22+106/1427 sec. after eleven; but the question applied to the hands, and the second hand would not be between the others at that time, but outside them.

63.--THE STOP-WATCH.

The time indicated on the watch was 5+5/11 min. past 9, when the second hand would be at 27+3/11 sec. The next time the hands would be similar distances apart would be 54+6/11 min. past 2, when the second hand would be at 32+8/11 sec. But you need only hold the watch (or our previous ill.u.s.tration of it) in front of a mirror, when you will see the second time reflected in it! Of course, when reflected, you will read XI as I, X as II, and so on.

64.--THE THREE CLOCKS.

As a mere arithmetical problem this question presents no difficulty. In order that the hands shall all point to twelve o'clock at the same time, it is necessary that B shall gain at least twelve hours and that C shall lose twelve hours. As B gains a minute in a day of twenty-four hours, and C loses a minute in precisely the same time, it is evident that one will have gained 720 minutes (just twelve hours) in 720 days, and the other will have lost 720 minutes in 720 days. Clock A keeping perfect time, all three clocks must indicate twelve o'clock simultaneously at noon on the 720th day from April 1, 1898. What day of the month will that be?

I published this little puzzle in 1898 to see how many people were aware of the fact that 1900 would not be a leap year. It was surprising how many were then ignorant on the point. Every year that can be divided by four without a remainder is biss.e.xtile or leap year, with the exception that one leap year is cut off in the century. 1800 was not a leap year, nor was 1900. On the other hand, however, to make the calendar more nearly agree with the sun's course, every fourth hundred year is still considered biss.e.xtile. Consequently, 2000, 2400, 2800, 3200, etc., will all be leap years. May my readers live to see them. We therefore find that 720 days from noon of April 1, 1898, brings us to noon of March 22, 1900.

65.--THE RAILWAY STATION CLOCK.

The time must have been 43+7/11 min. past two o'clock.

66.--THE VILLAGE SIMPLETON.

The day of the week on which the conversation took place was Sunday. For when the day after to-morrow (Tuesday) is "yesterday," "to-day" will be Wednesday; and when the day before yesterday (Friday) was "to-morrow," "to-day" was Thursday. There are two days between Thursday and Sunday, and between Sunday and Wednesday.

67.--AVERAGE SPEED.

The average speed is twelve miles an hour, not twelve and a half, as most people will hastily declare. Take any distance you like, say sixty miles. This would have taken six hours going and four hours returning. The double journey of 120 miles would thus take ten hours, and the average speed is clearly twelve miles an hour.

68.--THE TWO TRAINS.

One train was running just twice as fast as the other.

69.--THE THREE VILLAGES.

Calling the three villages by their initial letters, it is clear that the three roads form a triangle, A, B, C, with a perpendicular, measuring twelve miles, dropped from C to the base A, B. This divides our triangle into two right-angled triangles with a twelve-mile side in common. It is then found that the distance from A to C is 15 miles, from C to B 20 miles, and from A to B 25 (that is 9 and 16) miles. These figures are easily proved, for the square of 12 added to the square of 9 equals the square of 15, and the square of 12 added to the square of 16 equals the square of 20.

70.--DRAWING HER PENSION.

The distance must be 6 miles.

71.--SIR EDWYN DE TUDOR.

The distance must have been sixty miles. If Sir Edwyn left at noon and rode 15 miles an hour, he would arrive at four o'clock--an hour too soon. If he rode 10 miles an hour, he would arrive at six o'clock--an hour too late. But if he went at 12 miles an hour, he would reach the castle of the wicked baron exactly at five o'clock--the time appointed.

72.--THE HYDROPLANE QUESTION.

The machine must have gone at the rate of seven-twenty-fourths of a mile per minute and the wind travelled five-twenty-fourths of a mile per minute. Thus, going, the wind would help, and the machine would do twelve-twenty-fourths, or half a mile a minute, and returning only two-twenty-fourths, or one-twelfth of a mile per minute, the wind being against it. The machine without any wind could therefore do the ten miles in thirty-four and two-sevenths minutes, since it could do seven miles in twenty-four minutes.

73.--DONKEY RIDING.

The complete mile was run in nine minutes. From the facts stated we cannot determine the time taken over the first and second quarter-miles separately, but together they, of course, took four and a half minutes. The last two quarters were run in two and a quarter minutes each.

74.--THE BASKET OF POTATOES.

Multiply together the number of potatoes, the number less one, and twice the number less one, then divide by 3. Thus 50, 49, and 99 multiplied together make 242,550, which, divided by 3, gives us 80,850 yards as the correct answer. The boy would thus have to travel 45 miles and fifteen-sixteenths--a nice little recreation after a day's work.

75.--THE Pa.s.sENGER'S FARE.

Mr. Tompkins should have paid fifteen s.h.i.+llings as his correct share of the motor-car fare. He only shared half the distance travelled for 3, and therefore should pay half of thirty s.h.i.+llings, or fifteen s.h.i.+llings.

76.--THE BARREL OF BEER.

Here the digital roots of the six numbers are 6, 4, 1, 2, 7, 9, which together sum to 29, whose digital root is 2. As the contents of the barrels sold must be a number divisible by 3, if one buyer purchased twice as much as the other, we must find a barrel with root 2, 5, or 8 to set on one side. There is only one barrel, that containing 20 gallons, that fulfils these conditions. So the man must have kept these 20 gallons of beer for his own use and sold one man 33 gallons (the 18-gallon and 15-gallon barrels) and sold the other man 66 gallons (the 16, 19, and 31 gallon barrels).

77.--DIGITS AND SQUARES.

The top row must be one of the four following numbers: 192, 219, 273, 327. The first was the example given.

78.--ODD AND EVEN DIGITS.

As we have to exclude complex and improper fractions and recurring decimals, the simplest solution is this: 79 + 5+1/3 and 84 + 2/6, both equal 84+1/3. Without any use of fractions it is obviously impossible.

79.--THE LOCKERS PUZZLE.

The smallest possible total is 356 = 107 + 249, and the largest sum possible is 981 = 235 + 746, or 657+324. The middle sum may be either 720 = 134 + 586, or 702 = 134 + 568, or 407 = 138 + 269. The total in this case must be made up of three of the figures 0, 2, 4, 7, but no sum other than the three given can possibly be obtained. We have therefore no choice in the case of the first locker, an alternative in the case of the third, and any one of three arrangements in the case of the middle locker. Here is one solution:-- 107 134 235 249 586 746 --- --- --- 356 720 981 Of course, in each case figures in the first two lines may be exchanged vertically without altering the total, and as a result there are just 3,072 different ways in which the figures might be actually placed on the locker doors. I must content myself with showing one little principle involved in this puzzle. The sum of the digits in the total is always governed by the digit omitted. 9/9 - 7/10 - 5/11 -3/12 - 1/13 - 8/14 - 6/15 - 4/16 - 2/17 - 0/18. Whichever digit shown here in the upper line we omit, the sum of the digits in the total will be found beneath it. Thus in the case of locker A we omitted 8, and the figures in the total sum up to 14. If, therefore, we wanted to get 356, we may know at once to a certainty that it can only be obtained (if at all) by dropping the 8.

80.--THE THREE GROUPS.

There are nine solutions to this puzzle, as follows, and no more:-- 12 483 = 5,796 27 198 = 5,346 42 138 = 5,796 39 186 = 7,254 18 297 = 5,346 48 159 = 7,632 28 157 = 4,396 4 1,738 = 6,952 4 1,963 = 7,852 The seventh answer is the one that is most likely to be overlooked by solvers of the puzzle.

81.--THE NINE COUNTERS.

In this case a certain amount of mere "trial" is unavoidable. But there are two kinds of "trials"--those that are purely haphazard, and those that are methodical. The true puzzle lover is never satisfied with mere haphazard trials. The reader will find that by just reversing the figures in 23 and 46 (making the multipliers 32 and 64) both products will be 5,056. This is an improvement, but it is not the correct answer. We can get as large a product as 5,568 if we multiply 174 by 32 and 96 by 58, but this solution is not to be found without the exercise of some judgment and patience.

82.--THE TEN COUNTERS.

As I pointed out, it is quite easy so to arrange the counters that they shall form a pair of simple multiplication sums, each of which will give the same product--in fact, this can be done by anybody in five minutes with a little patience. But it is quite another matter to find that pair which gives the largest product and that which gives the smallest product.

Now, in order to get the smallest product, it is necessary to select as multipliers the two smallest possible numbers. If, therefore, we place 1 and 2 as multipliers, all we have to do is to arrange the remaining eight counters in such a way that they shall form two numbers, one of which is just double the other; and in doing this we must, of course, try to make the smaller number as low as possible. Of course the lowest number we could get would be 3,045; but this will not work, neither will 3,405, 3,45O, etc., and it may be ascertained that 3,485 is the lowest possible. One of the required answers is 3,485 2 = 6,970, and 6,970 1 = 6,970.

The other part of the puzzle (finding the pair with the highest product) is, however, the real knotty point, for it is not at all easy to discover whether we should let the multiplier consist of one or of two figures, though it is clear that we must keep, so far as we can, the largest figures to the left in both multiplier and multiplicand. It will be seen that by the following arrangement so high a number as 58,560 may be obtained. Thus, 915 64 = 58,560, and 732 80 = 58,560.

83.--DIGITAL MULTIPLICATION.

The solution that gives the smallest possible sum of digits in the common product is 23 174 = 58 69 = 4,002, and the solution that gives the largest possible sum of digits, 9654 =18327=5,886. In the first case the digits sum to 6 and in the second case to 27. There is no way of obtaining the solution but by actual trial.

84.--THE PIERROT'S PUZZLE.

There are just six different solutions to this puzzle, as follows:-- 8 multiplied by 473 equals 3784 9 " 351 " 3159 15 " 93 " 1395 21 " 87 " 1287 27 " 81 " 2187 35 " 41 " 1435 It will be seen that in every case the two multipliers contain exactly the same figures as the product.

85.--THE CAB NUMBERS.

The highest product is, I think, obtained by multiplying 8,745,231 by 96--namely, 839,542,176.

Dealing here with the problem generally, I have shown in the last puzzle that with three digits there are only two possible solutions, and with four digits only six different solutions.

These cases have all been given. With five digits there are just twenty-two solutions, as follows:-- 3 4128 = 12384 3 4281 = 12843 3 7125 = 21375 3 7251 = 21753 2541 6 = 15246 651 24 = 15624 678 42 = 28476 246 51 = 12546 57 834 = 47538 75 231 = 17325 624 78 = 48672 435 87 = 37845 ------ 9 7461 = 67149 72 936 = 67392 ------ 2 8714 = 17428 2 8741 = 17482 65 281 = 18265 65 983 = 63985 ------ 4973 8 = 39784 6521 8 = 52168 14 926 = 12964 86 251 = 21586 Now, if we took every possible combination and tested it by multiplication, we should need to make no fewer than 30,240 trials, or, if we at once rejected the number 1 as a multiplier, 28,560 trials--a task that I think most people would be inclined to s.h.i.+rk. But let us consider whether there be no shorter way of getting at the results required. I have already explained that if you add together the digits of any number and then, as often as necessary, add the digits of the result, you must ultimately get a number composed of one figure. This last number I call the "digital root." It is necessary in every solution of our problem that the root of the sum of the digital roots of our multipliers shall be the same as the root of their product. There are only four ways in which this can happen: when the digital roots of the multipliers are 3 and 6, or 9 and 9, or 2 and 2, or 5 and 8. I have divided the twenty-two answers above into these four cla.s.ses. It is thus evident that the digital root of any product in the first two cla.s.ses must be 9, and in the second two cla.s.ses 4.

Owing to the fact that no number of five figures can have a digital sum less than 15 or more than 35, we find that the figures of our product must sum to either 18 or 27 to produce the root 9, and to either 22 or 31 to produce the root 4. There are 3 ways of selecting five different figures that add up to 18, there are 11 ways of selecting five figures that add up to 27, there are 9 ways of selecting five figures that add up to 22, and 5 ways of selecting five figures that add up to 31. There are, therefore, 28 different groups, and no more, from any one of which a product may be formed.

We next write out in a column these 28 sets of five figures, and proceed to tabulate the possible factors, or multipliers, into which they may be split. Roughly speaking, there would now appear to be about 2,000 possible cases to be tried, instead of the 30,240 mentioned above; but the process of elimination now begins, and if the reader has a quick eye and a clear head he can rapidly dispose of the large bulk of these cases, and there will be comparatively few test multiplications necessary. It would take far too much s.p.a.ce to explain my own method in detail, but I will take the first set of figures in my table and show how easily it is done by the aid of little tricks and dodges that should occur to everybody as he goes along.

My first product group of five figures is 84,321. Here, as we have seen, the root of each factor must be 3 or a multiple of 3. As there is no 6 or 9, the only single multiplier is 3. Now, the remaining four figures can be arranged in 24 different ways, but there is no need to make 24 multiplications. We see at a glance that, in order to get a five-figure product, either the 8 or the 4 must be the first figure to the left. But unless the 2 is preceded on the right by the 8, it will produce when multiplied either a 6 or a 7, which must not occur. We are, therefore, reduced at once to the two cases, 3 4,128 and 3 x 4,281, both of which give correct solutions. Suppose next that we are trying the two-figure factor, 21. Here we see that if the number to be multiplied is under 500 the product will either have only four figures or begin with 10. Therefore we have only to examine the cases 21 843 and 21 834. But we know that the first figure will be repeated, and that the second figure will be twice the first figure added to the second. Consequently, as twice 3 added to 4 produces a nought in our product, the first case is at once rejected. It only remains to try the remaining case by multiplication, when we find it does not give a correct answer. If we are next trying the factor 12, we see at the start that neither the 8 nor the 3 can be in the units place, because they would produce a 6, and so on. A sharp eye and an alert judgment will enable us thus to run through our table in a much shorter time than would be expected. The process took me a little more than three hours.

I have not attempted to enumerate the solutions in the cases of six, seven, eight, and nine digits, but I have recorded nearly fifty examples with nine digits alone.

86.--QUEER MULTIPLICATION.

If we multiply 32547891 by 6, we get the product, 195287346. In both cases all the nine digits are used once and once only.

87.--THE NUMBER CHECKS PUZZLE.

Divide the ten checks into the following three groups: 7 1 5--4 6--3 2 8 9 0, and the first multiplied by the second produces the third.

88.--DIGITAL DIVISION.

It is convenient to consider the digits as arranged to form fractions of the respective values, one-half, one-third, one-fourth, one-fifth, one-sixth, one-seventh, one-eighth, and one-ninth. I will first give the eight answers, as follows:-- 6729/13458 = 1/2 5823/17469 = 1/3 3942/15768 = 1/4 2697/13485 = 1/5 2943/17658 = 1/6 2394/16758 = 1/7 3187/25496 = 1/8 6381/57429 = 1/9 The sum of the numerator digits and the denominator digits will, of course, always be 45, and the "digital root" is 9. Now, if we separate the nine digits into any two groups, the sum of the two digital roots will always be 9. In fact, the two digital roots must be either 9--9, 8--1, 7--2, 6--3, or 5--4. In the first case the actual sum is 18, but then the digital root of this number is itself 9. The solutions in the cases of one-third, one-fourth, one-sixth, one-seventh, and one-ninth must be of the form 9--9; that is to say, the digital roots of both numerator and denominator will be 9. In the cases of one-half and one-fifth, however, the digital roots are 6--3, but of course the higher root may occur either in the numerator or in the denominator; thus 2697/13485, 2769/13845, 2973/14865, 3729/18645, where, in the first two arrangements, the roots of the numerator and denominator are respectively 6--3, and in the last two 3--6. The most curious case of all is, perhaps, one-eighth, for here the digital roots may be of any one of the five forms given above.

The denominators of the fractions being regarded as the numerators multiplied by 2, 3, 4, 5, 6, 7, 8, and 9 respectively, we must pay attention to the "carryings over." In order to get five figures in the product there will, of course, always be a carry-over after multiplying the last figure to the left, and in every case higher than 4 we must carry over at least three times. Consequently in cases from one-fifth to one-ninth we cannot produce different solutions by a mere change of position of pairs of figures, as, for example, we may with 5832/17496 and 5823/17469, where the 2/6 and 3/9 change places. It is true that the same figures may often be differently arranged, as shown in the two pairs of values for one-fifth that I have given in the last paragraph, but here it will be found there is a general readjustment of figures and not a simple changing of the positions of pairs. There are other little points that would occur to every solver--such as that the figure 5 cannot ever appear to the extreme right of the numerator, as this would result in our getting either a nought or a second 5 in the denominator. Similarly 1 cannot ever appear in the same position, nor 6 in the fraction one-sixth, nor an even figure in the fraction one-fifth, and so on. The preliminary consideration of such points as I have touched upon will not only prevent our wasting a lot of time in trying to produce impossible forms, but will lead us more or less directly to the desired solutions.

89.--ADDING THE DIGITS.

The smallest possible sum of money is 1, 8s. 9d., the digits of which add to 25.