Tomorrow's problems focus on sequences and are of varying levels of difficulty. Although #4 may be more appropriate for Algebra 1/2 students, middle schoolers should be able to handle the others. Again, read the comments later in the evening for the answers, comments and solutions. There were some profound ideas expressed about today's questions particularly that innocent-looking quadrilateral problem with the 60 degree angles!
Tuesday, January 30, 2007
Problems 1-31-07
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9 comments:
For problem (1), are the students expected to be familiar with Arithmetic series? If so, do you expect them to actually write out the general term of the series and then find the differences as asked, or infer that the differences between the 100th and 96th term are the same as the differences between two other terms? Anyway, it does not seem to me that the difficulty levels of the two parts are different.
The logic in (2) can be tricky for some. Good problem.
I personally do not care for (3) mainly because only an exhaustive approach works. You can be smart in culling the choices, and I guess the skill as such is important for the kids to develop. Also, the great mathematicians (like Gauss) did not shy away from grunge work.
On (4), I assume one of the multiple choice values will be the value of N rather than what you have asked.
(4) got me thinking about algebraic identities. To see if the students actually get the algebraic identities like (a+b)^2, (a+b)(a-b) etc., I have used problems like 102*102, and 63*77 when tutoring. Some kids still persist in doing the direct multiplication rather than recognizing that the identity applies to numbers too.
Totally_clueless
'Official Answers' for 1-31-07:
1) (a) 14 (b) 348
2) 69
3) 23
4) 2450
Comments:
totally--
very insightful thoughts as always...
My replies:
(1) I'm absolutely suggesting we need to teach a recursive approach here. Some would call this strategy 'Make the problem simpler' but it really is recognition of the relationship between any term and the term whose subscript is 4 greater; thus a-sub-5 minus a-sub-1 will always equal 14 for this sequence. Middle school students can recognize this with patterning and practice. However, one needs to develop a general formula to find the 100th or 1000th term. Students in their development need to understand the arithmetic sequence formula: initial value + (number of 'spaces')(common difference) or some such idea. They should also relate this to b + mx, where m is the common difference, even though (x-1) makes more sense here.
One of my purposes in #3 was to focus on terminology and reading. However, the main clue was that the 'sum of the digits was prime'. Students need to identify the most helpful clue rather than just read vertically. This clue forced the number to be in the 20's since the number could not have 2 odd digits and '2' could not be the units' digit. Yes, many children will use the method of 'exhaustion, but it is the educator's role to ask questions and guide them or explain alternate methods when reviewing.
You're right on #4. If this were multiple-choice, I'd surely have the value for N, namely 49. Now here's where it gets scary. I was thinking precisely the same way about algebraic identities! In fact, here's what you can ask:
Calculate 49^2 + 49 mentally. There are several ways but, beyond factoring it into 49(49+1), let's go further: (49)(50) = (50-1)(50) = 50^2 - 50, which is much easier to do in your head!
Looks like we're having a 2-way dialogue for now!
Problem (3) teaches an important lesson: think before you compute.
a. Each of the digits is prime, so the digits are 2, 3, 5, or 7.
b. The sum of the two digits is prime, so one of them had better be 2. Then, the other is 3 or 5.
c. The one's digit had better be odd. The remaining possibilities are 23 and 25.
d. 25 isn't prime. 23 is left.
e. 2²+3²=13. The last condition is satisfied.
Anyone who starts by enumerating the primes between 11 and 99 will waste a lot of time; this is punishment enough.
This doesn't directly relate to anything, but I found the equations at this comic fascinating. It probably might not be appropriate for the classroom, but I find the relationship pretty interesting.
eric--
thanks for the comments on the prime puzzle; actually, i tell my students to beign by listing all of the prime digits: 2,3,5,7.
Anyone out there care to guess what % of high school students would make at least one error here! That's the weakness in knowledge we all see every day. I threw in the extra condition on squares just because...
darmok--
those equations and graphs actually work but i do need to be discreet since some of my students may be aware of my blog; as one student put it yesterday, "He's probably going to write about us and this lesson in his MySpace!"
There used to be simpler versions of (1) on the battery of "Iowa" tests that they gave me each year in elementary school. (early/mid 70s). Just now, and all those years ago, I divided this into two separate parts (taking turns). Concentrating on the even terms, this is not so hard.
I do (2) in 2 steps (remove votes from first candidate, recalculate, add votes to 2nd candidate, recalculate). Adults have trouble with this.
I like #3, but you probably knew that. Since the sum is prime, we know one of the digits. There are only two candidates for the second digit. Test.
I disagree with TC here: I think it is worth kids learning how to do exhaustive searches. Related, in my senior elective, I teach them how to make exhaustive lists, but they need help (at first)
Anecdotally, I have heard of a survey wherein a sample of people was asked to pick a random prime number less than 100, and a significant number picked 91.
I also find that I've been had by problem 3. I neglected the second condition and made things more complicated.
Now you know the reason for my nom-de-keyboard :-)
tc--
never ever apologize! Considering how many errors i make...
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