## Wednesday, May 16, 2007

### Helping Students Think 'Outside the Box'...

[Update: I'm adding an additional problem at the end. This question seems to be of interest to some since I've seen it in a Google search for awhile now. Answers and solutions to some of the problems will shortly appear in the comments.]

The issue for me as a math educator has always been:
How do we enable children to think conceptually?

Here are some standardized types of questions each of which can be solved by a variety of methods.

In each of the problems below, there is a conceptual approach that requires skill, knowledge and some insight. We all know as educators we can do something about the first two (provided students are given enough practice and they do it!), but how do we develop insight? I can only tell you how my insight improves: When I tackle harder problems or those requiring me to 'think differently'. I am sure there are those out there who are able to invent these methods on their own, but, as for me, I have to work at it and think about it!

The first 3 questions involve ratios. Since we know how the current generation feels about fractions, these may cause students to feel some frustration!

See if you can find an 'insightful' or conceptual approach. Also, ask yourself how most middle or secondary students would approach these:

1. Background Terminology: For those of the younger generation who may not have heard of the term proper fraction, it means a ratio of positive integers in which the denominator is larger than the numerator. Thus, 4/3 and 3/3 are improper; 2/3 and 1/3 are proper. Also, the phrase 'in lowest terms' means that the greatest common factor of the numerator and denominator is 1. (But everyone knew that, of course!)

If 10/n, 14/n and 15/n are proper fractions in lowest terms, what is the least possible positive integer value of n which is not prime?

2. For how many integer values of n is 11/n between 1/9 and 1/10?
Note: Is this an algebra problem? A guess-test 'plug-in' problem? A calculator problem? Or just a fraction 'exercise'?

3. Fahrenheit temperature is related to Celsius by the equation: F = (9/5)C + 32.
An increase of 36 degree F. is equivalent to an increase of how many degrees Celsius?
Note: Many students struggle with an approach here. Some try it algebraically, most plug in some initial temperature, virtually none I have observed think conceptually about the meaning of ratios.

4. A cube 6 inches on each edge is sliced 'horizontally' to form 2 congruent rectangular solids. If these 2 solids are joined to form a rectangular solid which is not a cube, the surface area of this resulting solid is how many more square inches than the surface area of the original cube?
Note: Some youngsters simply 'see' this with little or no calculation! I guess you could say, they really 'think outside the box!' (sorry 'bout that...)

5. What is the smallest positive integer having exactly seven factors?

Dave Marain said...

1. Ans: 121
Solution: n has to be greater than 15 and have no factor in common with with 2, 3, 5, and 7 (the prime factors of the numerators). Students would naturally think of primes like 17, 19, etc., but these are not allowed. The key is to think differently here and recognize that we can use a product of two primes, not necessarily distinct! The smallest such is 11x11!
Aside: Problems like these are typical of math contest problems such as those found in MathCounts, AMC, etc. Should such problems become part of the regular curriculum to extend all children's thinking? I know some educators do exactly that now...

2. Ans: 10
Algebraic solution--
1/10 < 11/n implies n < 110
Similarly, n > 99.
Many students are not confident in solving inequalities like these. They could avoid them by solving the corresponding equations 1/10 = 11/n and 11/n = 1/9, then choose the integer values between.
Fractional concept approach:
Rewrite 1/9 and 1/10 with a numerator of 11: 1/9 = 11/99, 1/10 = 11/110. The conclusion follows.
This question should be trivial for many. Try it and let me know...

jonathan said...

5. You know, variations on #5, I love them

Denise said...

#1 seems like a good problem for students who have just learned prime factoring, to give them a chance to see how understanding factoring makes the problem easy:
n=(2^a)(3^b)(5^c)(7^d)(11^e)(13^f)...
But a=b=c=d=0, so what is the smallest n we can make?

#2 is a flip-the-fraction problem:
If (a/b)<(c/d),
then (d/c)<(b/a).
Have the students prove this is true---a simple introduction to algebraic proof.

#3...Well, ratios confuse everybody, don't they? Might be an interesting way to introduce notation like delta-F and delta-C.

#5 and related questions are common in math contests, too. Is there a way to solve them other than guess-and-check? With an odd number of factors, it's pretty easy, because you know you're dealing with a square number.

Dave Marain said...

Denise--
For #5, I had an earlier post on a useful factoring principle. For example, 12 = (2^2)(3^1), therefore, 12 has (2+1)(1+1) = 6 factors. Since 7 is prime, the number would have to be of the form p^6, where p is prime. The smallest such number is 2^6 or 64. This same principle applies to the more general problem:
What is the smallest positive integer having exactly p factors, where p is prime?

Denise said...

It would be nice if it were so easy! But I tried that approach on a Math Counts problem that simplified to, "What is the smallest number with exactly 20 factors?"
2^19 = 524,288

Dave Marain said...

Denise,
I'm not sure of your point. I specifically asked for a prime number of factors, namely 7. For a composite number of factors, you're right, the 'power of a prime' method fails.
Dave

Denise said...

Oops! I didn't read your comment carefully enough. (That tends to be my primary downfall in math problems!) So if the number of factors I am looking for is prime, then the answer is a power of 2? That's interesting, but I was hoping for a more general method for any number of factors.

Denise said...

Okay, I think I've got it. Does this work? To find the smallest number N with n factors:
(1) Factorize n. Each prime factor of n will represent the (exponent + 1) of a prime factor of N.
(2) Arrange the prime factors of n from greatest to least, then subtract 1 from each to find e1, e2, e3...
(2) Build N = 2^e1 * 3^e2 * 5^e3...

Dave Marain said...

denise--
your method seems to work if N = a product of DISTINCT primes, however, I'm not sure if it works in all other cases.
Let's say N = 8. If i grasped your method correctly, you would write N = 2 x 2 x 2.
Therefore the smallest with 8 factors would be:
(2^1)(3^1)(5^1) - 30.
However, 8 can also be written as 4x2, leading to (2^3) (3^1) = 24. I found several other examples like this and they're not all powers of 2! Consider N = 24:
24 = 3x2x2x2. By your method (if i understand it), you would get:
(2^2)(3)(5)(7) = 420.
However, writing 24 as 4x3x2 gives:
(2^3)(3^2)(5) = 360.
Am I missing something?

Denise said...

Drats! I obviously didn't try enough examples. So in the general case, is there only "guess and check"?

Dave Marain said...

Denise==
I'm honored to think that you believe I know the full answer! I have not yet solved the general case, nor do I necessarily believe I'm capable of that! I'll think about it some more. I'm sure some mathematician has done this and considers it trivial but who knows! In situations like this, I try to prove it in special cases. For
N = 2^3, 2^5, etc., your formula may not work but I do believe it can be proved in many other cases.
Dave