I teach students to solve math problems by appreciating the big picture, or by noticing the problem’s inner symmetries, or through a deep understanding of the problem. In the long run, one thing leads to another: such training structures their minds so that they are better at understanding mathematics and, as a consequence, they perform well at math competitions.
That is why, when AMC is still far away, I do not give my students a lot of AMC problems; rather, I pick problems that contain useful ideas. When I do give AMC problems, I remove the multiple choices, so they understand the problems completely, instead of looking for shortcuts. For example, this problem from AHSME 1999 is a useful problem with or without choices.
What is the largest number of acute angles that a hexagon can have?
As AMC approaches, we start discussing how to solve problems given multiple choices. Training students for AMC is noticeably different from teaching mathematics. For example, some problems are very specific to AMC. They might not even exist without choices. Consider this problem from the 2001 AMC12:
A polynomial of degree four with leading coefficient 1 and integer coefficients has two zeros, both of which are integers. Which of the following can also be a zero of the polynomial?
Here we really need the choices in order to pick one complex number, such that its real part is an integer or a half integer and, in addition, the product of the number with its conjugate produces an integer.
Sometimes the choices distract from solving the problem. For example, in the following problem from the 2005 AMC12, having choices might tempt students to try to eliminate them one by one:
The sum of four two-digit numbers is 221. None of the eight digits is 0 and no two of them are same. Which of the following is not included among the eight digits?
Without the choices, students might start considering divisibility by 9 right away.
On some occasions, the choices given for the problems at AMC make the problem more interesting. Here is an example from the 2000 AMC10:
Two different prime numbers between 4 and 18 are chosen. When their sum is subtracted from their product, which of the following numbers could be obtained?
The choices are 21, 60, 119, 180 and 231. We can immediately see that the answer must be odd. Because the span of the three remaining choices is so wide, we suspect that we can eliminate the smallest and the largest. Trying for 5 and 7 — the two smallest primes in the range — we can eliminate 21. Similarly, checking the two largest primes in the range, we can eliminate 231. This leaves us with the answer: 119. If the choices were different, we might have lost the interplay between the solution and the list of choices. Then, solving the problem would have been slower and more boring. There are ten pairs of prime numbers to check. And we would need on average to check five of them until we stumbled on the correct choice.
In other cases having multiple choices makes the problem more boring and less educational. Here is another problem from the same competition.
Two non-zero real numbers, a and b, satisfy ab = a – b. Find a possible value of a/b + b/a – ab.
Solving this problem without choices can teach students some clever tricks that people use when playing with expressions. Indeed, when we collect a/b + b/a into one fraction (a2 + b2)/ab, we might remember that a2 + b2 is very close to (a – b)2, and see from here that a/b + b/a – 2 is (a – b)2/ab, which, given the initial condition, equals ab. Thus, we can get the answer: 2.
On the other hand, if you look at the multiple choices first: -2, -1/2, 1/3, 1/2, 2, you might correctly assume that the answer is a number. Thus, the fastest way to solve it is to find an example. If a = 1, then b must be 1/2, and the answer must be 2. This solution doesn’t teach us anything new or interesting.
My next example from the 2002 AMC10 is similar to the previous one. The difference is that the solution with multiple choices is even more boring, while the solution without these choices is more interesting and beautiful.
Let a, b, and c be real numbers such that a – 7b + 8c = 4 and 8a + 4b – c = 7. Then a2 – b2 + c2 is: 0, 1, 4, 7, or 8.
Given the choices, we see that the answer is a number. Hence we need to find any solution for the system or equations: a – 7b + 8c = 4 and 8a + 4b – c = 7. For example, if we let c = 0, we have two linear equations and two variables a and b that can be solved by a straightforward computation. Then we plug the solution into a2 – b2 + c2.
Without knowing that the result is a number, we need to look at the symmetries of our two initial equations. We might discover a new rule:
If we have two expressions ax + by and bx – ay, where a and b are switched between variables and there is a change in sign, it is a good idea to square each of them and sum them up, because the result is very simple: (a2 + b2)(x2 + y2).
Hence, in our initial problem we need to move the term with b in our two linear equations to the right; then square them and sum up the results. This way we may get a very simple expression. And indeed, this trick leads to a solution and this solution provides insight into working with algebraic expressions.
This is the perfect problem to linger over, assuming you’re not in the middle of a timed competition. It might make you wonder for which parameters this problem works. You might discover a new theorem that allows you to create a very similar problem from any number that can be represented as a sum of two non-trivial squares in two different ways.
To prepare my students for AMC I need to teach them tricks that are not useful at USAMO, or in mathematics in general for that matter. Many tricks distract from new ideas or from understanding the problem. All they give us is speed.
This bothers me, but to pacify myself, I keep in mind that most of my students will not become mathematicians and it might be useful in their lives to be able to make split-second decisions among a small number of choices.
However, it seems like Americans have the opposite problem: we make quick decisions without thinking. I’m concerned that training for multiple choice tests and AMC competitions aggravates this problem.Share: