tag:blogger.com,1999:blog-32146088185845272362024-03-13T10:23:52.000-07:00Peter Ash's Thoughts on Math and EducationMusings on doing and teaching mathematics, book reviews, math problems. Information about my math education business, Cambridge Math Learning.Peter Ashhttp://www.blogger.com/profile/07604759637404727566noreply@blogger.comBlogger149125tag:blogger.com,1999:blog-3214608818584527236.post-77067982846418146282023-06-16T09:37:00.000-07:002023-06-16T09:37:47.613-07:00E36 Area of a Right Triangle<p> <span style="font-family: arial;">A friend, Teddy O'Connell, sent me this geometry problem he found on the Web, which is not too hard and has a pretty neat solution.</span></p><p><span style="font-family: arial;">Let ABC be a right triangle, with a right angle at C. Suppose that a circle has been inscribed in the triangle (the incircle) which is tangent to ABC at one point on each side including the point D on the hypotenuse AB, such that |AD| = 3 and |DB| = 5. Find (ABC), the area of triangle ABC. </span></p><p></p><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgCuqkMVSxBHIEcSF3sf-JrKh5z6M4EOcJFqvfGt8ISmZlkhmFDr9DSmz4sKxDbnnbaf8S_K_fEyUJh20JkhYRaouDzqw8kV0kchW4Hh_5k7NLSBCy94Tu45Hb2oOL8JYUBh6llsw3TW5iPNGbXTt33uKkJ4APLBzYRtZw-gCoeHhaOTIu2XMu_V7bZWw/s3526/incircle%20for%20blog.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="2038" data-original-width="3526" height="185" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgCuqkMVSxBHIEcSF3sf-JrKh5z6M4EOcJFqvfGt8ISmZlkhmFDr9DSmz4sKxDbnnbaf8S_K_fEyUJh20JkhYRaouDzqw8kV0kchW4Hh_5k7NLSBCy94Tu45Hb2oOL8JYUBh6llsw3TW5iPNGbXTt33uKkJ4APLBzYRtZw-gCoeHhaOTIu2XMu_V7bZWw/s320/incircle%20for%20blog.png" width="320" /></a></div><br /><span style="font-family: arial;">(drawing not to scale)</span><p></p><p><span style="font-family: arial;">I will follow this posting with two comments. The first will contain a couple of hints, and the second (to be posted later) is my solution and a related question that I don't have an answer to.</span></p><p><span style="font-family: arial;"><br /></span></p>Peter Ashhttp://www.blogger.com/profile/07604759637404727566noreply@blogger.com2tag:blogger.com,1999:blog-3214608818584527236.post-30723088433005881332021-03-13T08:43:00.000-08:002021-03-13T08:43:50.793-08:00A 19. Tiling an equilateral triangle<p><span style="font-family: arial;">Consider an equilateral triangle. We wish to tile it with n congruent triangular tiles. We will call such a triangulation a CT-n tiling. Clearly there is a CT-n tiling if n = 2 (two 30-60-90 right triangles), n = 3 (three 30-120-30 isosceles triangles) or n = 4 (four equilateral triangles).</span></p><p><span style="font-family: arial;">(1) Show that there is a CT-n for the following values of n: 1,2,3,4,6,8,9,12,16,18.</span></p><p><span style="font-family: arial;">(2) Find 4 infinite sequences of n such that there is a CT-n for all values of n in the sequence. (The sequences may not overlap.)</span></p><p><span style="font-family: arial;">Unsolved (by me) problems</span></p><p><span style="font-family: arial;">(3) Find (with proof) at least one value of n for which there is no CT-n.</span></p><p><span style="font-family: arial;">(4) Determine exactly which values of n yield a CT-n.</span></p>Peter Ashhttp://www.blogger.com/profile/07604759637404727566noreply@blogger.com0tag:blogger.com,1999:blog-3214608818584527236.post-78550693517910053772021-03-13T06:48:00.006-08:002021-03-13T08:27:18.169-08:00E 36. How many solutions?<p><span style="font-size: 14pt;"><span style="font-family: arial;">(2020 Putnam
Exam, problem A1)</span></span></p><p class="MsoNormal"><span style="font-size: 14pt; line-height: 107%;"><span style="font-family: arial;">
How many positive integers N satisfy all of the following 3 conditions:<o:p></o:p></span></span></p>
<p class="MsoNormal"><span style="font-size: 14pt; line-height: 107%;"><span style="font-family: arial;">(i) N is
divisible by 2020<o:p></o:p></span></span></p>
<p class="MsoNormal"><span style="font-size: 14pt; line-height: 107%;"><span style="font-family: arial;">(ii) N has
at most 2020 decimal digits<o:p></o:p></span></span></p>
<p class="MsoNormal"><span style="font-size: 14pt; line-height: 107%;"><span style="font-family: arial;">(iii) The
decimal digits of N are a string of consecutive ones followed by a string of
consecutive zeros.</span><o:p></o:p></span></p><p class="MsoNormal"><span style="font-size: 14pt; line-height: 107%;"><span style="font-family: arial;">Note: This only requires pre-college level mathematics, but like most Putnam problems it is not easy.</span></span></p>Peter Ashhttp://www.blogger.com/profile/07604759637404727566noreply@blogger.com0tag:blogger.com,1999:blog-3214608818584527236.post-42323852812666399092021-03-13T06:43:00.002-08:002021-03-13T06:43:34.828-08:00A 18. Setting up breakout rooms<p><span style="font-family: Calibri, sans-serif; font-size: 14pt;">A Zoom
session has N = 2</span><sup style="font-family: Calibri, sans-serif;">k </sup><span style="font-family: Calibri, sans-serif; font-size: 14pt;">people. There are n breakout sessions, each into two
equal-sized breakout rooms (i.e., N/2</span><span style="font-family: Calibri, sans-serif; font-size: 14pt;"> people in each) The
facilitator wishes that every person in the session meets with every other person
in a breakout room at least once. Show this can be done if n = k + 1.</span></p><p><span style="font-family: Calibri, sans-serif; font-size: 14pt;">Note: This is the simplest version. If you want more of a challenge, you could try the case where N is not a power of 2. Obviously if N is odd, the breakout rooms cannot be exactly equal in size, but there is always an efficient solution where the numbers of people in each breakout room are always within 2 of one another.</span></p>Peter Ashhttp://www.blogger.com/profile/07604759637404727566noreply@blogger.com0tag:blogger.com,1999:blog-3214608818584527236.post-76102592344482430232020-12-27T16:00:00.002-08:002020-12-28T11:11:45.176-08:00A 17. Square inscribed in a triangle<p><span style="background-color: white; color: #282829; font-family: "Times New Roman", serif;"><span style="font-size: large;">Given a triangle ABC with acute angles B and C, how do you construct a square PQRS with PQ in BC and vertices S and R in AB and AC, respectively?</span></span></p><p><span style="background-color: white; color: #282829; font-family: "Times New Roman", serif;"><span style="font-size: large;">I found this problem on Quora. It took me a while to solve it, but the construction turned out to be pretty simple and elegant.</span></span></p>Peter Ashhttp://www.blogger.com/profile/07604759637404727566noreply@blogger.com0tag:blogger.com,1999:blog-3214608818584527236.post-20688882717754549022020-09-20T15:45:00.000-07:002020-09-20T15:45:57.948-07:00A16. Treasure search<p>A treasure hunter is on an island which is in the shape of the unit disc, and they are located on the boundary at (1,0). The treasure hunter has a treasure-detector, which will light up if the detector is within 1/2 unit of any buried treasure. The treasure hunter can move along any path (say, continuous and piecewise infinitely differentiable) that starts at (1,0) and stays within the closed unit disk. The treasure hunter claims that they can discover whether or not there is treasure buried on the island (or in other words, that the union of all disks of radius 1/2 centered on points along the path covers the unit disk) by traveling along a path of length π. Prove or disprove the claim.</p><p>This is my revision of a problem sent to me by Apratim Roy.</p>Peter Ashhttp://www.blogger.com/profile/07604759637404727566noreply@blogger.com0tag:blogger.com,1999:blog-3214608818584527236.post-83204887605909862542020-09-20T15:17:00.004-07:002020-09-20T15:17:25.592-07:00A15. A Problem by Henry Dudeney<p>The houses on a long street have addresses 1, 2, 3, ... n. (Dudeney was British. Unlike the custom in most of the US, odd and even house numbers in Britain occur on the same side of the street. Assume all houses are on the same side of the street.) Call a house a half-way house if the sum of the numbers of the houses before it are equal to the sum of the numbers of the houses after it. Depending on n, there may or may not be a half-way house. For example, if n = 8, then there is a half-way house, namely 6, because 1 + 2 +3 + 4 + 5 = 7 + 8. You can check there is no half-way house if 1 < n < 8.</p><p>Find the value of n if you know that 50 < n < 500 and there is a half-way house.</p><p>You could do this easily by a brute-force search with a computer, so to make it interesting, no computer/calculator use is allowed.</p><p>I saw this problem presented online by the Mathologer. It connects with many fascinating parts of number theory, and there is an interesting connection with Ramanujan.</p>Peter Ashhttp://www.blogger.com/profile/07604759637404727566noreply@blogger.com0tag:blogger.com,1999:blog-3214608818584527236.post-72505883163603354272020-08-03T09:23:00.002-07:002020-08-03T09:23:12.775-07:00E35 - Three Triangles in a Regular HexagonHere is another problem from the Catriona Shearer collection. ABCDEF is a regular hexagon. Lines have been drawn to create three triangles, whose interiors have been shaded. What fraction of the hexagon is shaded?<div><br /></div><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEggUPKCccgGrgq3YB-_wsJIKBFgZrEIFxyQI0eVjip2ugsrB4Aak9Tvx4WIIedtoZ11R-IhYX3CHpyONE6jNV8Fb448DSBR3TK6alXkt_5NAYQeu0m38bN6FT9GeYW6V7vgvFzUO9fm-lIW/s1032/hex_3triangles.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="954" data-original-width="1032" height="296" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEggUPKCccgGrgq3YB-_wsJIKBFgZrEIFxyQI0eVjip2ugsrB4Aak9Tvx4WIIedtoZ11R-IhYX3CHpyONE6jNV8Fb448DSBR3TK6alXkt_5NAYQeu0m38bN6FT9GeYW6V7vgvFzUO9fm-lIW/w320-h296/hex_3triangles.png" width="320" /></a></div><div class="separator" style="clear: both; text-align: center;"><br /></div><div class="separator" style="clear: both; text-align: center;"><br /></div><div>The points labeled <span style="font-family: calibri, sans-serif; font-size: 11pt;">M</span><sub style="font-family: calibri, sans-serif;">i </sub>are the midpoints of their sides, and the unlabeled point marks the center of the hexagon.</div>Peter Ashhttp://www.blogger.com/profile/07604759637404727566noreply@blogger.com0tag:blogger.com,1999:blog-3214608818584527236.post-79849092485747284132020-08-03T09:01:00.000-07:002020-08-03T09:01:23.972-07:00E34. Tangent circles in a rectangleThis problem was called to my attention to Apratim Roy. It comes from a collection, on Twitter, of many dozens of interesting geometry problems, at the level of high school geometry, by Catriona Shearer, a math teacher in the UK, which she has posted on Twitter, @Cshearer41.<div><br /></div><div><p class="MsoNormal">In the diagram below. you are given circles O and P, externally
tangent at B. Circle O is tangent to two sides of rectangle DEFG, and circle P
is tangent to three sides of the rectangle, as shown. Find the measure of angle
ABC.</p></div><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhGUDf07lB1OXnHmW07XsYyxre_bYg6EER7cba3BQTo6cHcVNWq_G3rUN2wgmamvqn0XIRrt895NoTDTb4voxZIwKvGnyFmCgdztaWDwdvpiBW_Gfa-EI-A6rKcZKdNSRT9KCjWThz5ZAFv/s1204/Shearer1.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="960" data-original-width="1204" height="255" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhGUDf07lB1OXnHmW07XsYyxre_bYg6EER7cba3BQTo6cHcVNWq_G3rUN2wgmamvqn0XIRrt895NoTDTb4voxZIwKvGnyFmCgdztaWDwdvpiBW_Gfa-EI-A6rKcZKdNSRT9KCjWThz5ZAFv/w320-h255/Shearer1.png" width="320" /></a></div><div class="separator" style="clear: both; text-align: center;"><span style="text-align: left;">It is quite surprising that you don’t need to be given any measurements to answer the problem. (But no fair using that fact in your proof; that is, you can’t just prove a special case such as two circles of equal size.) I also like this problem because it is hard, but not too hard, and because there are several different nice ways to solve it.</span></div><div class="separator" style="clear: both; text-align: center;"><o:p style="text-align: left;"></o:p></div><div><span style="text-align: left;"><br /></span></div>Peter Ashhttp://www.blogger.com/profile/07604759637404727566noreply@blogger.com0tag:blogger.com,1999:blog-3214608818584527236.post-28638730462492138382020-07-22T08:52:00.000-07:002020-07-22T09:06:00.724-07:00A 14. Triangles in a regular hexagonI came across a neat problem in Quora. When I found it, there were no solutions, and with crucial help from my student Apratim Roy, I was able to find a very elegant solution. Unfortunately, when I went back to Quora I was unable to find the original problem, so I cannot give credit to the proposer. If anyone wants, send me an email and I will send you my solution, or provide a hint.<br />
<br />
Here is the problem: Let ABCDEF be a regular hexagon, and P a point inside the hexagon. Suppose Area(PAB) = 3, Area(PCD) = 5, and Area(PEF) = 8. Find Area(PBC).Peter Ashhttp://www.blogger.com/profile/07604759637404727566noreply@blogger.com0tag:blogger.com,1999:blog-3214608818584527236.post-81094311106115113282016-11-11T06:15:00.000-08:002016-11-11T06:20:16.851-08:00E33. Tiling a checkerboard<span style="font-family: "arial" , "helvetica" , sans-serif;">This problem was shown to me by my student, Apratim Roy. Though it involves only elementary concepts, I found it rather difficult, and thought the solution was very surprising and elegant.</span><br />
<span style="font-family: "arial" , "helvetica" , sans-serif;"><br /></span>
<span style="font-family: "arial" , "helvetica" , sans-serif;">You are given a standard 8 x 8 checkerboard, with one square removed, and 21 3 x 1 tiles. In other words, there are exactly enough tiles to cover the modified board. Your task is to find a way to do this, without cutting any tile.</span><br />
<span style="font-family: "arial" , "helvetica" , sans-serif;"><br /></span>
<span style="font-family: "arial" , "helvetica" , sans-serif;">(a) Find out what square must be removed for the task to be possible. (4 possible answers).</span><br />
<span style="font-family: "arial" , "helvetica" , sans-serif;">(b) Describe the tiling. (Many possible answers).</span><br />
<span style="font-family: "arial" , "helvetica" , sans-serif;"><br /></span>
<span style="font-family: "arial" , "helvetica" , sans-serif;">You might guess that the removed square needs to be one of the four corners of the checkerboard, but you would be wrong.</span>Peter Ashhttp://www.blogger.com/profile/07604759637404727566noreply@blogger.com0tag:blogger.com,1999:blog-3214608818584527236.post-55243395957105100222016-05-12T19:33:00.000-07:002016-05-12T19:33:38.010-07:00A13. Points of tangency of an ellipse and a circleLet E be the ellipse with equation x<sup>2</sup>/4 + y<sup>2</sup> = 1 and C(r) be the circle with center (1,0) and radius r. For which values of r do the curves E and C(r) have point(s) of tangency?<br />
<br />
This is fairly routine, but still a bit challenging to find all solutions.Peter Ashhttp://www.blogger.com/profile/07604759637404727566noreply@blogger.com1tag:blogger.com,1999:blog-3214608818584527236.post-84517000043583432992016-04-13T07:14:00.001-07:002016-05-12T19:34:33.376-07:00A12. Fifth powers final digit - generalized.If numbers are expressed in base b, for which b is it true that n<sup>5</sup> and n end in the same digit for all positive integers n?<br />
This is an obvious generalization of Problem E32.Peter Ashhttp://www.blogger.com/profile/07604759637404727566noreply@blogger.com0tag:blogger.com,1999:blog-3214608818584527236.post-90184756604730075452016-04-13T07:08:00.000-07:002016-05-12T19:34:11.465-07:00E32. Fifth powers final digitThe following rather neat problem occurs in <i>Challenging Problems in Algebra</i> by Alfred S. Posamentier and Charles T. Salkind. I think it is suitable for a bright high school student or, with some hints, even for average high school students.<br />
<br />
Prove that n and n<sup>5</sup> always end in the same digit (in ordinary base-10 representation).Peter Ashhttp://www.blogger.com/profile/07604759637404727566noreply@blogger.com1tag:blogger.com,1999:blog-3214608818584527236.post-42264816607635642022015-12-25T08:45:00.001-08:002015-12-25T08:45:46.682-08:00Hippasus & The Discovery of Irrational NumbersIn November I gave a speculative talk to the New England Section of the Mathematical Association of America on the discovery of irrational numbers by the Pythagorean Hippasus through an examination of the mystic pentagram, the sacred symbol of the Pythagoreans. I have expanded it a bit, adding a method of recursively computing the golden ratio, phi. If you want to see how these topics are related, please check my paper on Scribd: <a href="https://www.scribd.com/doc/294006886/Irrational-Numbers-the-Mystic-Pentagram-and-Eigenvectors">https://www.scribd.com/doc/294006886/Irrational-Numbers-the-Mystic-Pentagram-and-Eigenvectors</a>. Peter Ashhttp://www.blogger.com/profile/07604759637404727566noreply@blogger.com0tag:blogger.com,1999:blog-3214608818584527236.post-8926804951192411072015-09-28T06:39:00.000-07:002015-09-28T06:40:03.838-07:00E32. ChimesThe following problem is given in Jacobs' <i>Geometry</i>:
If it takes a clock 3 seconds to ring 3:00 (3 chimes), how long does it take the same clock to ring 6:00 (6 chimes)? The answer is not 6 seconds.
The answer depends on making some assumptions, which I think are reasonable ones.Peter Ashhttp://www.blogger.com/profile/07604759637404727566noreply@blogger.com0tag:blogger.com,1999:blog-3214608818584527236.post-63776643375118192422015-05-16T08:11:00.001-07:002015-05-16T08:11:36.393-07:00Embarrasing mistakeI just realized that my last post was far off the mark. There is a super-obvious example that shows that f(n) >= n/2, and so in fact f(n) = n/2. The example is the subset {n/2 + 1, n/2 + 2, ... , n} which contains n/2 elements and clearly has the non-divisible property. The problem has very little to do with prime numbers. The previously-published result was sharp.Peter Ashhttp://www.blogger.com/profile/07604759637404727566noreply@blogger.com0tag:blogger.com,1999:blog-3214608818584527236.post-17337902223148015362015-05-11T06:33:00.000-07:002015-05-11T06:33:57.534-07:00A simple (?) number theory conjectureCall a set S of positive integers non-divisible if, whenever a and b belong to S, it is not true that (a|b or b|a). For example, the set of prime numbers is non-divisible. Let S(n) be the set of the first n positive integers, and let f(n) be the cardinality of the largest non-divisible subset of S(n). Then clearly f(n) >= pi(n), where pi(n) denotes the number of prime numbers <= n. Furthermore, I know a pretty proof (published, but not by me) that shows f(n) <= n/2 (n even). That is, in any subset of n/2 + 1 positive integers all <= n, at least one integer divides another integer in the set. This proof involves the pigeonhole principle. So, pi(n) <= f(n) <= n/2. (n even)
My conjecture is that, in fact, f(n) = pi(n) for all n > 1. Does anyone have a counterexample or a proof? It seems like it should be very simple.Peter Ashhttp://www.blogger.com/profile/07604759637404727566noreply@blogger.com0tag:blogger.com,1999:blog-3214608818584527236.post-2055625816035279382015-03-24T08:08:00.001-07:002020-09-20T15:55:35.296-07:00E31. A packing polynomialFunny how one comes across math problems. I was reading my Reed College alumni magazine and came across an article about Maddie Grant, class of '15, whose undergraduate thesis is apparently a substantial generalization of a 1923 result by Fueter and Polya. Feuter and Polya evidently discovered a polynomial function that maps the non-negative integers one-to-one and onto the pairs of non-negative integers. That is, they found a simple formula to express the inverse Cantor's "diagonal" mapping of pairs of non-negative integers to non-negative integers. Their function ‐ a so-called packing polynomial ‐ is given by<br />
<br />
f(x,y) = (1/2)[(x + y)<sup>2</sup> + x + 3y],<br />
where x and y are non-negative integers. <br />
<br />
The proof that this function is one-to-one and onto the non-negative integers is actually pretty elementary if you look at it the right way. I will print a hint as a comment in a few days.Peter Ashhttp://www.blogger.com/profile/07604759637404727566noreply@blogger.com1tag:blogger.com,1999:blog-3214608818584527236.post-87408692402594244782015-01-18T10:47:00.000-08:002015-01-18T10:47:03.502-08:00A11. Splitting a triangle and a tetrahedronJános Kurdics published this interesting problem in The Math Connection Linkedin group:<br />
Given a triangle, find the shortest line segment that divides the triangle into two regions of equal area. A solution that does not involve calculus is preferred.<br />
The solution that I know is very nice, and gives quite a workout in elementary trigonometry.<br />
<br />
János says that he is working on three-dimensional analog (plane that divides a tetrahedron into two regions of equal volume and gives smallest possible cross-sectional area with the tetrahedron) but has only solved it for a regular tetrahedron.Peter Ashhttp://www.blogger.com/profile/07604759637404727566noreply@blogger.com0tag:blogger.com,1999:blog-3214608818584527236.post-6751063628461364622015-01-17T09:37:00.000-08:002015-01-17T09:37:42.485-08:00A10. Triangle with sides in arithmetic sequence.Find a number n such that there is a triangle with sides n, n + 1, and n + 2 in which the largest angle is twice the smallest angle. How many such numbers n are there? Note that n does not have to be an integer.<br />
<br />
This problem came up in a high school textbook during a tutoring session, except there were several hints given that made the problem much easier. See if you can do it without hints.Peter Ashhttp://www.blogger.com/profile/07604759637404727566noreply@blogger.com0tag:blogger.com,1999:blog-3214608818584527236.post-17077603135529183282015-01-13T09:21:00.000-08:002015-01-13T09:21:39.072-08:00E 30. A Cryptarithm<b>Introduction:</b> Here is a problem that I think would be good for any child who knows how to use the standard algorithm to do multi-digit multiplication. It is a cryptarithm from <i>The Moscow Book of Puzzles</i>, an amusing book by Boris Kordemsky, available in a very inexpensive Dover edition. For those of you who don't know (as I didn't) a cryptarithm is like a cryptogram, except the unknowns are digits, not letters. One you may have seen is (S E N D) + (M O R E) = (M O N E Y), where each letter stands for a digit, with different letters standing for different digits, and the expressions in parentheses standing for multi-digit numbers.<br />
<br />
<b>Problem:</b> In the following multiplication, each * stands for a prime number digit (2, 3, 5, 7). Replace each * by a prime digit so that the multiplication is correct.<br />
<br />
* * *<br />
<u><span style="font-size: small;"> <span style="font-family: Arial,Helvetica,sans-serif;">x</span></span> * *</u><br />
* * * *<br />
<u>* * * * </u><br />
* * * * *<br />
Strangely enough, you have all the information you need to find the unique solution.<br />
<br />Peter Ashhttp://www.blogger.com/profile/07604759637404727566noreply@blogger.com0tag:blogger.com,1999:blog-3214608818584527236.post-67316394152691337132015-01-11T09:24:00.000-08:002015-01-13T09:25:38.402-08:00Multiplication table problemSometimes mathematics problems that are interesting to the professional mathematician can arise out of the simplest questions. We all learned a multiplication table in school, which lists all 100 products of integers from 1 * 1 to 10 * 10. I've been told that in England, students are required to memorize up to 12 * 12. In the multiplication table for 1 to 10, there are 100 products, but it is only necessary to learn 55, because of the commutative property. In other words, we only need to learn the products
a * b, where a ≤ b, and the number of such products is 1 + 2 + ... + 10 = T(10) = (11)(10)/2, where T(n) is the n-th triangular number, (n + 1) * n / 2. Therefore, the number of different products is at most 55. Actually it is less. For example, 6 appears twice in the triangular table, as 1 * 6 and as 2 * 3. The number of different products in a 10 by 10 multiplication table is 42. (Shades of Douglas Adams!)<br />
<br />
A mathematician would naturally be interested in knowing how many different products there are in an n by n multiplication table, in other words, the number of different products of the form ab, where a and b are positive integers less than or equal to n. Call this sequence a(n). Then a(n) ≤ T(n), so the lim sup of a(n)/n^2 as n goes to infinity is less than or equal to 1/2. In fact, Erdös gave a very nice proof that the limit is 0, and he and others obtained more accurate asymptotic formulas for a(n).<br />
<br />
See the Online Encyclopedia of Integer Sequences, where a(n) is given as sequence A027424. Links provide much information about the sequence. Also, the question of how many different numbers appear in a multiplication table could be given to students at almost any level.Peter Ashhttp://www.blogger.com/profile/07604759637404727566noreply@blogger.com0tag:blogger.com,1999:blog-3214608818584527236.post-43220372831311199852015-01-08T18:46:00.000-08:002015-01-08T18:46:45.245-08:00E29. Divisible by SixteenHere's an easy one:<br />
<br />
You have sixteen balls, numbered consecutively 1 through 16, and 4 boxes. You are asked to put 4 of the balls into each box, so that the product of the numbers in each box is divisible by 16. Give a solution, or prove that this cannot be done.Peter Ashhttp://www.blogger.com/profile/07604759637404727566noreply@blogger.com0tag:blogger.com,1999:blog-3214608818584527236.post-82876152852368156392014-12-31T06:29:00.000-08:002014-12-31T20:42:57.484-08:00E28. A balancing cubeA sculpture is in the form of a cube, one meter on a side, balancing unstably on one vertex. Let A be the bottom vertex (on the ground), B one of the three adjacent vertices, and C the vertex at the other end of the space diagonal of the cube from A. Assume the ground is a horizontal plane, and AC is perpendicular to to the ground. Find the distance from B to the ground, with answer in exact form.Peter Ashhttp://www.blogger.com/profile/07604759637404727566noreply@blogger.com0