Have you ever wondered what kind of math problem solver you are?
Well, you’re about to find out!
Problem solvers typically come in three distinct flavors: Cunning solvers get the problem correct
but their methods are either a bit sketchy or not totally convincing.
They’re sneaky.
Crazy solvers are just the opposite.
They go overboard in their solution with an overzealous approach that is simply overkill.
And then there are the clever solvers.
These guys find the perfect solution which often sheds creative insight into the problem.
We’re going to assess your problem solving skills with the following math competition
problem.
F and g are both cubic polynomials.
You’re given that f(3) = g(5), f(5) = g(7), f(7) = g(9), and f(11) - g(13) = 24.
Based on this information, can you find f(17) - g(19)?
Pause the video to give this problem some thought, and when you’re ready, join me
for the three different solutions to see which category of solver you fall under!
Let’s start with the conniving problem solver.
What’s to prevent him from setting f(3) and g(5) both equal to zero?
And how about f(5) and g(7)?
Yeah, let’s make them both zero.
And hey we’re on a roll now so let’s make f(7) and g(9) also equal to zero.
And, since the difference between f(11) and g(13) must equal 24, let’s assign artificial
values of 25 to f(11) and 1 to g(13).
Now we have all the ingredients to figure out f(x) and g(x).
We’ll attack f(x) first.
Since we forced zeros at 3, 5, and 7, we can write f(x) as a product of these three linear
factors.
And since we assigned f(11) to equal 25, we can solve for the polynomial constant M. Plugging
11 in for x, we get 25 equals M x 8 x 6 x 4.
So M is 25/192.
And f(x) is now fully defined as follows.
Moving on to g.
We’re going to get the formula for g(x) exactly the same way as we did for f.
Now we’re ready to evaluate f(17) - g(19).
For f(x), I’ll plug 17 in for x, then simplifying I’ll get this product.
Here’s a quick trick for multiplying 14 by 12.
Since they can be written as 13 1 times 13 - 1, they’re just the difference of two
squares, so we get 168.
OK, so f(17) is 25x1680 over 192.
We’ll follow the same steps to get g(19) which comes out to 1680 over 192.
So their difference looks like this.
This difference becomes 24x1680 over 192.
And since 24 over 192 is 8, we get 1680 over 8.
Which comes out to 210.
Congratulations to all you conniving problem solvers out there!
You got it correct!
Now let’s move on to the crazy problem solving method.
To crazy’s credit, he’s going to approach this in a more convincing way.
You won’t feel ripped off that it was done by assuming specific values.
Let’s first rewrite the information given by subtracting g from both sides.
Since the g input is always 2 more than the f input, we can create a brand new cubic,
call it h(x), and write it as f(x) – g(x+2).
Since h is also a cubic function, we can write it as ax^3 + bx^2 + cx +d.
So when x equals 3, we get the following equation for h.
When x is 5 we get this, then moving on to x is 7 we get this, and finally when h is
11 we get this.
Notice that these are all linear equations and we can now solve for the coefficients
a, b, c, and d using matrix algebra.
Crazy solver can smell the solution but he first needs to multiply both sides by the
inverse of this 4x4 matrix.
And now he can get the answer to the problem by plugging in 17 for x.
We can get this in one fell swoop by front multiplying both sides of our matrix equation
by the following 1x4 matrix.
And, sparing all the gruesome details, I can assure you, crazy problem solver will eventually
get 210.
Now finally let’s see how the clever solver does it.
He will also define h(x) as f(x) – g(x+2), but instead he’ll write it out as a product
of linear factors.
Just like before, he’ll plug in 3, 5, 7, and 11 for x into h(x) and get the following
four equations.
But now he focuses on the first three equations and notices something interesting.
For the top equation, one of those three constants must equal 3 in order for h(3) to equal zero,
right?
Let’s agree for now that it’s a.
Then for the second equation, one of those constants must equal 5.
Since we already set a=3, let’s make b=5.
And for the third equation, using the same reasoning, we’ll set c=7.
We can easily assign the values like this without loss of generality since there are
exactly 6 different permutations we can select for a, b, and c.
Let’s stick with the values in red and plug them into the fourth equation.
We’ll get h(11) equals K x 8 x 6 x 4 which equals 24.
Solving for K, we get 1/8.
And now h(x) is fully defined and operational.
So let’s plug in 17 to get our answer.
That simplifies to 14 x 12 x 10 over 8.
We know from the conniving approach that the numerator is 1680.
And 1680 over 8 is 210.
Let me know in the comments below which type of problem solver you are!
And as always, thanks for watching and see you soon!