Update: Answers, solutions have now been posted in the comments.
I thought of these variations on the well-known combinatorial problems involving 3-digit numbers that pop up frequently as I was teaching arithmetic and geometric series yesterday.
These questions are appropriate for grades 6-12 provided students are given definitions and some practice with arithmetic and geometric sequences, topics that are well within the abilities of middle schoolers. A quick intro to these sequences is all that is really needed OR, as I did below, they can be defined in the problem itself. Thus, these questions provide both practice in arithmetic skills and in combinatorial thinking. Of course, all the experienced or budding programmers out there can write simple code to have their graphing calculators count these, but that should only complement and verify their results, not replace the reasoning needed to solve them, unless these are used for a computer science class (even then, programmers should independently verify their code by solving the problems!).
These are not highly challenging and therefore can be used as Problems of the Day, for extra credit, or enrichment. Our readers will hopefully suggest other extensions and further variations (some are suggested below).
1. The digits of 246 form an arithmetic sequence from left to right because 4-2= 6-4. How many positive 3-digit integers satisfy this condition?
2. The digits of 248 form a geometric sequence from left to right because 4/2 = 8/4. How many positive 3-digit integers satisfy this condition?
Now, how could we make these more challenging? 4-digit numbers or will that make one or both easier, i.e., fewer possibilities? What if the digits were allowed to form these sequences in any order? BTW, I apologize for the music pun in the title. I hope you will respond to that with a positive tone!
Thursday, April 5, 2007
Variations on Basic Themes: Digit Problems in the Key of A or G Minor?
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10 comments:
For the arithmetic sequence, could the common difference be zero? Similarly for the geometric sequence?
Allowing a negative common difference or a common ratio < 1 will enable the backwards sequences you talk about.
It may be worthwhile to just say arithmetic or geometric without specifying the order, and see if students are able to come up with all generalizations.
TC
tc--
You should know me better hhan that! I chose not to remind students about a common difference that could be zero or negative! You ruined the surprise!! Similarly for geometric...
As far as disregarding order, I felt this question was sufficient for a warmup or mild enrichment but left it open for embellishment by the instructor. My counting method for the geometric is a bit off-center. I'm guessing you'll come up with it, though! This is not a jd2718 baffler but I'm trying to provide some different ideas for counting problems that are accessible to both middle and high schoolers.
mine is not so much a comment but more of a question. How were you able to generate some of the math symbols that you use in your post (such as the less than or equal to sign)?
What do you expect the kids to do with, let's say the first arithmetic sequence question? Are you looking for some struggle with understanding what's being asked, and then an organized list?
jonathan--
I had many purposes here. First, to apply the definition of arithmetic/geometric sequences. I expect many to forget the possibility that the common difference = 0 or the common ratio = 1, but part of success in problem-solving is to know precise definitions (integer, prime, etc.) which many students do not learn well. The rest is combinatorial reasoning whether it be an organized list or some intuitive approach. I perceived a tone to your comment as if the problem seemed unnecessarily tricky or something else. Is that what you intended?
Answer to #1: 45
#2 to follow later...
Not at all. I just wanted to suggest modifications that were consistent with what you were trying to get them to do.
nnn (9)
abc a = b-1 = c-2 (7) 789...123
abc a = b-2 = c-4 (5) 579...135
abc a = b-3 = c-6 (3) 369...147
abc a = b-4 = c-8 (1) 159
cba a = b-1 = c-2 (8) 987...210
cba a = b-2 = c-4 (6) 975...420
cba a = b-3 = c-6 (4) 963...630
cba a = b-4 = c-8 (2) 951, 840
1+2+3+...+8+9 = 45
That last bit looks like magic. I will think it over. The pattern doesn't drop out directly, so maybe there is a different way to organize the list.
I'd expect many kids to try listing all of them.
#2 looks considerably easier to count, though harder to understand. The answer is prime?
nick--
I don't think I used any special symbols on this post but, in general, if the math symbols are complicated, I type it in Equation Editor in Word, then upload the image, an unsophisticated method but it works. If it's just an exponent, I use simple html code (sup etc). Blogger does not yet enable LaTeX unlike wordpress. This would make life much easier. html does have some basic math symbols. Just Google 'html math symbols' and test it out. Blogger may not allow all of these but try it out.
Update on solutions...
Answer to #2 (correct hopefully): 17
Comment: One counting strategy, which I used, would be to consider that the tens' digit is the geometric mean between the other 2 digits. Thus, in the number 124, 2 is the GM between 1 and 4, or, equivalently, 2^2 = 1x4.
Thus, I began with a tens' digit of 1. Since 1^2 = 1, the only possibility is 111. Moving onto 2, this would lead to 4x1, 1x4 and 2x2 or the numbers 421, 124 and 222. Continue...
As a "programming" exercise you could try to generate a list of all 729 3 digit sequences on a spreadsheet, and test them for common ratio...
Or organize the list by ratio, instead of by middle number.
For the arithmetic sequence, the problem resolves to hunting for three colinear points in a 10x10 lattice. That might be a neat approach, extension.
What would the geometric analog to that lattice approach be? (question for us, not the kids!)
Did we discuss on my site, how many 3 digit numbers consist of strictly increasing digits? That's vaguely related, though the math is quite different.
jonathan--
some interesting insights, particularly the lattice point approach (i need to think about that!)
Here's one way of showingthat the # of arithmetic sequence results is a triangular number 1+2+3+4+5+...+9 (Note that it ends in '9', the largest digit.)
The digits can be expressed as a,a+d and a+2d,in which 'a' is the hundreds digit and d could be pos,neg, or zero. The maximum value of 2d is 8 or d = 4. This leads to 1 ascending number, 159. To build a relationship, note that 9-8=1 and there is one ascending solution. The same value of d leads to '2' descending solutions (951 and 840) and, in general, there will always be one more descending value because of the zero digit.
Similarly, d=3 leads to 9-6 = 3 ascending solutions: 147,258 and 369 and 4 descending: 963,852,741,630.
d=2 leads to 9-4 = 5 ascending:
135,246,357,468,and 579
and 6 descending:975,864,753,642,531 and 420
Continuing this logic produces the result. Note that d=0, produces
9-0 = 9 results!
One could generalize this approach to arithmetic sequences of 3 integers in the range 0 through N (zero is problematic -- see below). For example if the maximum allowable integer is 31, then the maximum value for 2d would be 31-1= 30 so d = 15. There would be 31-30=1 ascending sequence: 1,16,31 and 2 descending sequences:31,16,1, and 30,15,0;...
Decreasing d, we eventually reach
d=1, in which case there will be 31-2 = 29 ascending solutions and 30 descending. The final '31' solutions occur when d = 0 (we're not allowing 0,0,0 for this discussion).
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