tag:blogger.com,1999:blog-8231784566931768362.post3042129870137560710..comments2023-09-09T08:21:55.454-04:00Comments on MathNotations: Percent Increase vs. Decrease Redux: A Rich Investigation for Middle Schoolers and Algebra StudentsDave Marainhttp://www.blogger.com/profile/13321770881353644307noreply@blogger.comBlogger13125tag:blogger.com,1999:blog-8231784566931768362.post-35975573488497130852009-08-21T08:44:58.556-04:002009-08-21T08:44:58.556-04:00Sorry, Matt, that I didn't acknowledge you ear...Sorry, Matt, that I didn't acknowledge you earlier...<br /><br />Nice observations and I would definitely like to see more of your thinking on this. Could you email me details at dmarain "at" gmail dot com?Dave Marainhttps://www.blogger.com/profile/13321770881353644307noreply@blogger.comtag:blogger.com,1999:blog-8231784566931768362.post-30284397351414589082009-08-17T18:24:44.244-04:002009-08-17T18:24:44.244-04:00Oops, forgot to mention that the harmonic mean is ...Oops, forgot to mention that the harmonic mean is the square on one of the legs.Matt Badleynoreply@blogger.comtag:blogger.com,1999:blog-8231784566931768362.post-22261642766213687862009-08-17T17:49:49.841-04:002009-08-17T17:49:49.841-04:00Wow, I know I'm really late to this party, but...Wow, I know I'm really late to this party, but I just now was directed here. I did a little investigation on this and found that each pair of unequal integers are related to Pythagorean triples. The 45 & 5 and the 10 & 40 are from a 3-4-5. The arithmetic mean is always the square on the hypotenuse. <br />I was hoping to explore this more in depth, if anyone is interested...Matt Badleynoreply@blogger.comtag:blogger.com,1999:blog-8231784566931768362.post-33619875227798835002008-08-18T04:56:00.000-04:002008-08-18T04:56:00.000-04:00What if we add the condition that, find A and B su...What if we add the condition that, find A and B such that abs(A - B ) is minimum ?! Then clearly we can do better than A = 45, B = 5...Jayanthttps://www.blogger.com/profile/09502265636397042024noreply@blogger.comtag:blogger.com,1999:blog-8231784566931768362.post-65972561174568271642007-11-16T10:25:00.000-05:002007-11-16T10:25:00.000-05:00HI Dave,The point of my last comment was that ther...HI Dave,<BR/><BR/>The point of my last comment was that there was bound to be a flaw in the proof because the statement is wrong. <BR/><BR/>Some other doubles include (13,325), (34, 544), (104,234).<BR/><BR/>While trying to generalize these, I arrived at a couple of other conjectures: <BR/>For any integer m<BR/>(1) 7 does not divide m^2+1<BR/>(2) 11 does not divide m^2+1<BR/><BR/>TCUnknownhttps://www.blogger.com/profile/06449079338919787252noreply@blogger.comtag:blogger.com,1999:blog-8231784566931768362.post-165286837179233922007-11-16T09:49:00.000-05:002007-11-16T09:49:00.000-05:00tc--Is the multiple of 5 idea a conjecture based o...tc--<BR/>Is the multiple of 5 idea a conjecture based on observation or do you have an algebraic argument? This is impressive if it turns out to be a necessary condition. Certainly, we can generate an infinitude of solutions from A=5, B=45 by taking integer multiples, i.e., A=5m, B=45m will also satisfy all 3 conditions. Proving necessity in general seems more formidable.Dave Marainhttps://www.blogger.com/profile/13321770881353644307noreply@blogger.comtag:blogger.com,1999:blog-8231784566931768362.post-87371290906115664412007-11-16T08:45:00.000-05:002007-11-16T08:45:00.000-05:00If someone can actually prove that in order for th...If someone can actually prove that in order for the arithmetic, geometric and harmonic means of two integers to be integers, the two numbers have to be multiples of five, I would really like to see that proof (since I can then have the challenge of finding the flaw in the proof:-))<BR/><BR/>TCUnknownhttps://www.blogger.com/profile/06449079338919787252noreply@blogger.comtag:blogger.com,1999:blog-8231784566931768362.post-28268366593162059102007-11-16T08:10:00.000-05:002007-11-16T08:10:00.000-05:00Hi Dave,I realized very soon that I had not expli...Hi Dave,<BR/><BR/>I realized very soon that I had not explicitly ruled out the trivial case, but trusted that our readers can be discerning enough to eschew this. <BR/><BR/>Yes, 5 & 45 is the bonus answer. In my typical rambling fashion, I first found 10 & 40, and then found 5 & 45. <BR/><BR/>You realize very soon that both numbers may have to be multiples of 5. <BR/><BR/>The extra challenge is to find two such integers (with all three integer means) such that both are not multiples of 5. Alternatively, prove that they both have to be multiples of 5. <BR/><BR/>TCUnknownhttps://www.blogger.com/profile/06449079338919787252noreply@blogger.comtag:blogger.com,1999:blog-8231784566931768362.post-53484797915389675122007-11-16T07:58:00.000-05:002007-11-16T07:58:00.000-05:00tc--I think your extra problem should be entitled,...tc--<BR/>I think your extra problem should be entitled, "By All Means!" (sorry, I couldn't resist that). <BR/>This is a nice problem and, in its more general form, perhaps could stand alone as a "tc challenge!"<BR/><BR/>For those who might have forgotten the definition of the harmonic mean of A and B, the formula is 2AB/(A+B). I realize that these days we go right to Wikipedia or MathWorld to get this background!<BR/>In general, the HM is defined as the reciprocal of the Arithmetic Mean of the reciprocals of A and B (or, in general, for n numbers).<BR/><BR/>My analysis is fairly unsophisticated and I suspect, tc, you have a more penetrating analysis. Well, first of all, I assume you intended the problem to read <B>"unequal integers"</B>, otherwise we have trivial solutions like A=1, B=1. I will refer to this as the <B>trivial</B> case below. The rest of the discussion below generally assumes A and B are not equal.<BR/><BR/>My initial analysis was fairly basic. A and B had to have the same parity from the Arithmetic Mean condition. I decided to work from the Geometric Mean condition and restrict attention to perfect squares, i.e., AB = C^2. I concluded that C^2 cannot be the square of an odd prime (which produces only the trivial solution A=B) or a power of 2, so I was able to get to 225 = 15^2 fairly quickly. This led to the factors A=5, B=45, which satisfy all 3 conditions. You can tell me if this satisfies your bonus in the non-trivial case. <BR/><BR/>My instinct is that a general analysis of conditions insuring that the harmonic mean is an integer is non-trivial.Dave Marainhttps://www.blogger.com/profile/13321770881353644307noreply@blogger.comtag:blogger.com,1999:blog-8231784566931768362.post-12658404851116754342007-11-15T15:53:00.000-05:002007-11-15T15:53:00.000-05:00An offshoot problem: Find two positive integers wh...An offshoot problem: Find two positive integers whose arithmetic, geometric and harmonic means are all integers. For bonus points, find the two integers for which the sum of the three integer means is lowest.Unknownhttps://www.blogger.com/profile/06449079338919787252noreply@blogger.comtag:blogger.com,1999:blog-8231784566931768362.post-63420671706426092212007-11-15T15:08:00.000-05:002007-11-15T15:08:00.000-05:00OK, One thought seems to be : If we had scored X t...OK, One thought seems to be : If we had scored X times higher, and they had scored X times lower, we would be equal.' For the right value of X, we get the geometric mean.<BR/><BR/>TCUnknownhttps://www.blogger.com/profile/06449079338919787252noreply@blogger.comtag:blogger.com,1999:blog-8231784566931768362.post-3754767803146830652007-11-15T14:50:00.000-05:002007-11-15T14:50:00.000-05:00Beautiful generalization, tc! How many hs students...Beautiful generalization, tc! How many hs students never get to see the harmonic mean! I'll have to play with the geometric mean also. I'm sure there is an interesting interpretation we can make based on 60:x = x:90.Dave Marainhttps://www.blogger.com/profile/13321770881353644307noreply@blogger.comtag:blogger.com,1999:blog-8231784566931768362.post-55001276024536346362007-11-15T08:49:00.000-05:002007-11-15T08:49:00.000-05:00Interesting!Let us consider the number of points s...Interesting!<BR/><BR/>Let us consider the number of points scored in each case.<BR/><BR/>(1) If the student thought: If we had scored X more points, and the opponents had scored X fewer points, we would have been equal, For the right value of X, the points scored by each team would have been the Arithmetic mean of A&B.<BR/><BR/>(2) If the student thought (as in this problem), that if we had scored X% more points, and the opponents had scored X% fewer points, then we would have been equal. For the right value of X, the points scored by each team would be the Harmonic mean of A&B.<BR/><BR/>I am wondering - 'What is the similar thought to get the geometric mean?' I can write it as an equation, but that does not seem to lend itself to a cogent thought. <BR/><BR/><BR/>TC.Unknownhttps://www.blogger.com/profile/06449079338919787252noreply@blogger.com