My 9th grade daughter just took a Maryland Math League test. This is a “fun” competition where the problems always seem much harder than they are. She got 2 out of six right. I got 4 out of six in about two minutes, but two questions have me stumped.
Write the only interger n > 32 that satisfies: (n-23)!(23)! = (n-32)!(32)!. I am guessing the answer is 33 since the answers are always simple and 23 + 9 =32 and any other number would make things very complicated. Assuming I am right, what is the rule regarding factorials that make this so.
If r(sub 1), r(sub 2), r(sub 3), r(sub 4) are the roots of x^4 - 4x^2 +2 = 0 what is the value of (1 + rsub1)(1+rsub2)(1+ rsub3)(1+rsub4)?
From what you’re told, (x - r[sub]1[/sub])(x - r[sub]2[/sub])(x - r[sub]3[/sub])(x - r[sub]4[/sub]) = x[sup]4[/sup] - 4x[sup]2[/sup] + 2. Substitute x = -1, take out the factor of (-1)[sup]4[/sup] = 1 that appears, and you have (1 + r[sub]1[/sub])(1 + r[sub]2[/sub])(1 + r[sub]3[/sub])(1 + r[sub]4[/sub]) = 1 - 4 + 2 = -1.
Both of these use facts that kids in 9th grade have seen. Like most such tests, you have to see the “trick,” then they’re easy.
The 2nd one’s answer seems to be -1, but I don’t know how you do it without a calculator. I mean, it’s pretty easy to substitute y=x^2 and then use the quadratic equation to get all the answers (x = +/- sqrt(2 +/- sqrt(2))) but I don’t know how to do the rest.
solkoe, these are “easy” in the sense that they don’t require any math concepts beyond elementary algebra. They rely (as math-league problems tend to) on some insight rather than just on the cookbook solution methods that are usually taught. (Of course, then students practice on similar questions to enlarge their cookbooks so that they can solve the problems more quickly, a never-ending arms race of problem designers vs. problem solvers.)
Cunctator has already posted a solution to the first one. Here you might notice that the definition of “!” is not really that important. If you look at the problem and see that it’s asking for n such that
f(n-23)f(23) = f(n-32)f(32)
then you might see that you can always find a solution by setting n=23+32; this pairs up the arguments of f in a way that always works because multiplication is commutative. (For arbitrary f, it’s not guaranteed that this is the only solution; you can prove uniqueness for !, though, e.g. using properties of binomial coefficients.)
For the second problem, it is also useful to consider a generalization. What if you consider more general values of “1”? That is, what is
(x+r[sub]1[/sub])(x+r[sub]2[/sub])(x+r[sub]3[/sub])(x+r[sub]4[/sub]) ?
It’s actually not classwork, but the Maryland Math League, one of many in the United States (www.mathleague.com). One takes them optionally, and one person in each year gets a certificate if they are the high scorer. I think they are fun, my daughter not so much. The problems look impossible until you think about them, then most are very simple.
I was in the New Jersey Math League back in high school, and these are ctypical of the questions we’d get. You’d get one hour to solve ten questions like that. That’s only six minutes a problem. They’re eminently doable, once you get a handle on them. “Getting a handle” means not taking the brute force solution.
The first one can be done without evaluating the factorial;s explicitly, but by looking for a solution that reduces each side of the equation to a simple result. The second is solved by setting y = x^2 and solving the resulting quadratic equation to get y, then taking the two roots of that. Having the roots, you just plug them into the (1 + r1)(1+r2(1+r3)(1+r4) equation and solve – most of the terms cancel or rationalize themselves pretty quickly.
If you were a high school math geek, then this kind of thing was fun. We used to go to another high school after classes once a month for this stuff. We had a chemistry league, too, for solving chemistry problems.
I got a special patch for Math League, which I proudly wore on my Band Jacket.
Probably explains why I didn’t get laid until I was 74.
The second is easier than CalMeacham makes it. I can’t be arsed to dick around with r(sub)1 and so forth so I’ll use a, b, c and d as the roots of the equation.
Then (x-a)(x-b)(x-c)(x-d) = the original equation.
Reverse the sign of all the terms and the sign of the solution doesn’t change, so (a-x)(b-x)(c-x)(d-x) still = the original equation.
Then let x=-1 and this becomes (a+1)(b+1)(c+1)(d+1).
In other words this is identical to the original equation when x=-1. That gives you 1 - 4 + 2 = -1.