Henry Ernest Dudeney was an English mathematician who was a prolific creator of logic puzzles and mathematical games. His "Perplexities" column was featured in the Strand Magazine from 1910 until his death in 1930. The Strand was a very popular publication during this time and was probably best know for The Sherlock Holmes short stories written by Arthur Conan Doyle. Below are four problems that were written by Dudeney and printed in the Strand. Each of them are problems that I think would be approachable by students and related to Nova Scotia mathematics outcomes. I've shown both Dudeney's solution as well as my attempts to solve them.
Inside a rectangular room, measuring 30 feet in length and 12 feet in width and height, a spider is at a point on the middle of one of the end walls, 1 foot from the ceiling, as at A; and a fly is on the opposite wall, 1 foot from the floor in the centre, as shown at B. What is the shortest distance that the spider must crawl in order to reach the fly, which remains stationary? Of course the spider never drops or uses its web, but crawls fairly. My Solution: I knew that the straight path up the side of the room and across the ceiling was unlikely to be the shortest path (of 42 ft). I created a net diagram, labeled points A and B and drew a right triangle to show the spiders path. I used Pythagorean Theorem to solve for the length of the path and got 40.7185 ft. I thought I was quite clever until I read Dudeney's solution and realized that there are several different ways to draw the net diagram and that I hadn't found the shortest path. I had stumbled upon a variation Dudeney's No. 3 solution. Students would probably benefit from modeling the net of the prism with Polydron pieces to see if different students come up with different net diagrams. Dudeney's Solution: Imagine the room to be a cardboard box. Then the box may be cut in various different ways, so that the cardboard may be laid flat on the table. I show four of these ways, and indicate in every case the relative positions of the spider and the fly, and the straightened course which the spider must take without going off the cardboard. These are the four most favourable cases, and it will be found that the shortest route is in No. 4, for it is only 40 feet in length (add the square of 32 to the square of 24 and extract the square root). It will be seen that the spider actually passes along five of the six sides of the room! Having marked the route, fold the box up (removing the side the spider does not use), and the appearance of the shortest course is rather surprising. If the spider had taken what most persons will consider obviously the shortest route (that shown in No. 1), he would have gone 42 feet! Route No. 2 is 43.174 feet in length, and Route No. 3 is 40.718 feet. Juggling with Digits  The Strand Magazine, Volume 77 (1929) page 312 Arrange the ten digits in three arithmetical sums, employing three of the four operations of addition, subtraction, multiplication, and division, and using no signs except the ordinary ones implying those operations. Here is an example to make it quite clear: 3 + 4 = 7; 9  8 = 1; 30 + 6 = 5. But this is not correct, because 2 is omitted, and 3 is repeated. My Solution: This puzzle reminded me of an Open Middle problem. Since you need three equations, only one of the numbers involved can be a two digit number. The rest must be single digits. Also, since you can't use any digit multiple times, the digit 1 can't be used to multiple or divide a number. Similarly, the digit 0 can't be used to add or subtract or multiple a number. This means that the 0 must be used as the ones digit of a twodigit number. For example, it could be used as 6 + 4 = 10 or as 2 x 5 = 10. After playing around with the numbers a bit I ended up with 4 x 5 = 20, 9  3 = 6 and 1 + 7 = 8. I was also came up with 3 + 7 = 10, 8  2 = 6 and 4 + 5 = 9 but then realized that I hadn't used three of the four operations. I haven't found any additional solutions, but i'm not positive that I've exhausted all the possibilities. Dudeney's Solution: 7 + 1 = 8; 9  6 = 3; 4 X 5 = 20 The Russian Motorcyclists  The Strand Magazine, Volume 53 (1917), pages 9597 Two Army motorcyclists, on the road at Adjbkmlprzll, wish to go to Brczrtwxy, which, for the sake of brevity, are marked in the accompanying map as A and B. Now, Pipipoff said: "I shall go to D, which is six miles, and then take the straight road to B, another fifteen miles." But Sliponsky thought he would try the upper road by way of C. Curiously enough, they found on reference to their cyclometers that the distance either way was exactly the same. This being so, they ought to have been able easily to answer the General's simple question, "How far is it from A to C?" It can be done in the head in a few moments, if you only know how. Can the reader state correctly the distance? My Solution: I labeled the distance from A to C as x and the distance from B to C as z. I then created a system of equations to solve algebraically. Since the distance both ways is the same, I knew that x + z = 21. Since this is a right triangle I also used Pythagorean Theorem to create the equation (x + 6)^2 + (15)^2 = z^2. Using the first equation I solved for z to find z = 21  x. I then substituted this equation into the second equation (x+6)^2 + 225 = (21x)^2 and then solved for x. The x^2 terms cancel out so you are not left with a quadratic to solve. Solving for x results in a distance of 3 1/3 miles. Dudeney's question implies that there is a simple way to solve this in your head so he is not talking about my method of solution! Dudeney's Solution: The two distances given were 15 miles and 6 miles. Now, all you need do is to divide 15 by 6 and add 2, which gives us 4 1/2. Now divide 15 by 4 1/2, and the result (3 1/3 miles) is the required distance between the two points. This pretty little rule applies to all such cases where the road forms a rightangled triangle. A simple solution by algebra will show why that constant 2 is added. We can prove the answer in this way. The three sides of the triangle are 15 miles, 9 1/3 miles (6 plus 3 1/3 miles) and 17 2/3 miles (to make it 21 miles each way). Multiply by 3 to get rid of the fractions, and we have 45, 28, and 53. Now, if the square of 45 (2,025) added to the square of 28 (784) equal the square of 53 (2,809) then it is correctand it will be found that they do so. I have not idea why Dudeney's "pretty little rule" works. If you have any idea what he is talking about, please let me know. I'm very curious and I'm hoping I haven't missed something obvious. I found a formula that relates the perimeter of a right triangle (P) to its two legs (a and b). The formula states that a = P * (P2b)/(2*(Pb)). Using this fomula a = 42 *(423)/(2*(4215)) = 21 * 12 / 27 = 28/3 = 9 1/3. This does not appear to be the relationship that Dudeney is using however. Does it have something to do with a generating function for Pythagorean Triples (such as m^2 – n^2 , 2mn , m^2 + n^2)?
My Solution: This question also reminds me of an Open Middle problem. To minimize the solution, we'd like to have the most single digit numbers possible, the rest will be two digit numbers. We will need at least 3 twodigit numbers since the digits 4, 6 and 8 must be in the tens place. If they were in the digits place they would be even and hence not prime. The digits 1 and 9 are not prime and therefore need to be the unit digit with 4, 6 or 8 in the tens place. The remaining four digits {2, 3, 5, 7} could be added as individual units but one of them must join the 1 and 9 as the unit digits in one of the 3 twodigit numbers. The digits 2 and 5 would make any of these two digit numbers composite, so they will be a prime on their own. Next, I looked at all the twodigit prime numbers that start with 4, 6 and 8 and found {41, 43, 47, 61, 67, 83, 89}. Since 89 is only number with a 9 in it, 89 must be one of the numbers in our answer. The remaining digits can be placed in three possible ways, each giving a total sum of 207 since we're just swapping the units digits in different numbers: 2+3+5+47+61+89, 2+3+5+41+67+89, 2+5+7+43+61+89. Dudeney's Solution: The 4, 6, and 8 must come in the tens place, as no prime number can end with one of these, and 2 and 5 can only appear in the units place if alone. When those facts are noted the rest is easy, as here shown: 47+61+89+2+3+5=207. I think that it is important, as mathematics educators, that we occasionally work on solving challenging problems. There was recently an article posted about how Math Teachers' Circles give educators a chance to remember why they love math and puts them in the learners' seat where they can better empathize with their students. When there are no organized Math Teachers' Circle in your area, sometimes trying out an occasional interesting logic puzzle or mathematics problem can have similar benefits. EL
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