Multi-Level Approaches to the Solution of Problems

How a student answers a question is more important than the answer given by the student. For example, the student may have randomly guessed, the student may have used a rote and unimaginative method for solution, or the student may have used a very creative method. It seems that one should judge the student by the “way” he or she answers the question and not just by the answer to the question.

Example:

Question: Without using a calculator, which is greater:

355 x 356 x 354 or 354 x 357?

Case 1: Rote Memory Approach (a completely mechanical approach not realizing the fact that there may be a faster method that takes into account patterns or connections of the numbers in the question): The student multiplies 355 x 356 gets 126,380, and then multiplies 354 x 357 and gets 126 x 378.

Case 2: Observer's Rot Approach (an approach that makes use of a mathematical strategy that can be memorized and tried for various problems): The student does the following:

Divide both quantities by 354:

He or she then gets 355 x 356/354 compared with 354 x 357/354.

He or she then divides these quantities by 356 and then gets 355/354 compared with 357/356.

Now he or she realizes that 355/354 = 1 and 1/354; 357/356 = 1 and 1/356.

He or she then reasons that since the left side 1 and 1/354 is greater than the right side, 1 and 1/356, the left side of the original quantities, 355 x 356 is greater than the right side of the original quantities, 354 x 357.

Case 3: The Pattern Seeker's Mind (most mathematically creative method-- an approach in which the student looks for a pattern or sequence in the numbers and then is astute enough to represent the pattern or sequence in more general algebraic language to see the pattern or sequence more clearly):

Look for a pattern. Represent 355 x 356 and 354 x 357 by symbols.

Let x = 354.

Then 355 = x + 1, 356 = x + 2, 357 = x + 3.

So 355 x 356 = (x + 1) (x + 2) and 354 x 357 = x(x + 3).

Multiplying the factors we get

355 x 356 = (x times x) + 3x + 2 and 354 x 357 = (x times x) + 3x + 2 minus

(x times x) minus 3x, which is just 2.

So 355 x 356 is greater than 354 x 357 by 2.

Note: You could have also represented 355 by x. Then 356 = x + 1;354 = x – 1;357 = x + 2. We could then get 355 x 356 = (x) (x + 1) and 354 x 357 = (x – 1) (x + 2). Then we would use the method above to compare the quantities.
 * OR –

You could have written 354 as a and 357 as b. Then 355 = a + 1 and 356 = b – 1. So 355 x 356 = (a + 1) (b – 1) and 354 x 357. Let's see what (355 x 356) – (354 x 357) is. This is the same as (a + 1) (b – 1) – ab, which is (ab + b – a – 1) – ab, which is in turn b – a – 1. Since b – a – 1 = 357 – 354 – 1 = 2, the quantity 355 x 356 – 354 x 357 = 2, so 355 x 356 is greater than 354 x 357 by 2. Case 4: The Astute Observer's Approach (the simplest approach-- an approach that attempts to figure out a connection between the numbers and uses that connection to figure out the solution):

355 x 356 = (354 + 1) x 356 =(354 x 356) + 356 and

354 x 357 = 354 x (356 + 1) = (354 x 356) + 354

One can see that the difference is just 2.

Case 5: The Observer's Common Relation Approach (this is the approach that people use when they want to connect two items to a third to see how the two items are related):

355 x 356 is greater than 354 x 356 by 356.

354 x 357 is greater than 354 x 356 by 354.

So this means that 355 x 356 is greater than 354 x 357.

Case 6: Scientific, Creative, and Observational Generalization Method (a highly creative method and the most scientific method, as it spots a critical and curious aspect of the sums being equal and provides for a generalization to other problems of that nature):

Represent 354 = a, 357 = b, 355 = c, and 356 = d

We have now that (1) a + b = c + d We want to prove ab > dc
 * b – a| > |d – c|

Proof:

Square inequality (2): (b – a)2 > (d – c)2

Therefore: (3) b2 – 2ab + a2 > d2 – 2dc + c2

Multiply (3) by (– 1) and this reverses the inequality sign: or Now square (1): (a + b) = (c + d) and we get: Add inequality (4) to inequality (5) and we get
 * (b2 – 2ab + a2) < – (d2 – 2dc – c2)
 * 1) – b2 + 2ab – a2 < – d2 + 2dc –c2
 * 1) a2 + 2ab + b2= c2+ 2dc + d2

4ab < 4dc

Divide by 4 and we get:

ab > dc

The generalization is that for any positive numbers a, b, c, d when |b – a| > |d – c| and a + b = c + d, then ab < dc.

This also generalizes in a geometrical setting where for two rectangles whose perimeters are the same (2a + 2b = 2c + 2d), the rectangle whose absolute difference in sides |d – c| is least has the greatest area.