Keep those skills fresh

A student and I were slaying a tricky geometry problem from an old AMC.  She hadn’t taken geometry yet, but she had excellent insights, and we were making good progress.  We were close to the end and we needed to simplify \sqrt{12}.  Since 12 is a multiple of a square (4), and since the square root of a product is the product of the square roots, we can rewrite \sqrt{12}  as  2\sqrt{3}.
But she couldn’t do it.  I thought maybe she hadn’t learn how to simplify square roots, but she admitted she had learned them, but hadn’t solved a square root problem for a long time and was out of practice.   Fair enough.
As a student progresses through a math curriculum over the years, they are gradually adding tools to their problem solving tool box.  Simplifying square roots is another tool.  The textbook teaches you how to use it, you practice using it on a bunch of squares, then you throw into your tool box for some unspecified time when you will need it later.   Then they go on to learn another tool.  And another tool.
The trouble is, like my student, you can forget how to use your tools if you don’t practice.
That’s another reason why practicing math contests can be so helpful in keeping those skills fresh.  Math contests typically have all kinds of problems: geometry, probability, number theory, and yes solving square roots.  Sometimes, like in that geometry problem, you need to deploy several of your tools to arrive at a solution.

Factorials –> Factoring

I was working through this year’s MathCounts State Level exam, and found 2 problems involving factorials.  Your reflex when seeing a factorial problem is to see how you can factor the expression.  For example we are asked to find the value of this expression:

\frac{5!+6!}{4!+3!}

Notice that the 2 terms in the numerator are both products of 5! and the 2 terms in the denominator are both products of 3!.  So we can factor these both out:

\frac{5!+6!}{4!+3!}=\frac{5!(1+6)}{3!(4+1)}

If you notice that 5! and 3! both have a factor of 3! then the fraction is much easier to evaluate.

Another problem on the same exam asks us to solve for n:

(n+1)! - n! = 4320

Again you can factor n!  from the 2 terms on the left hand side:

(n+1)! - n! = n!(n+1 - 1)= 4320

n!\cdot n = 4320

To finish out this problem we can find the prime factorization of 4320 and match that up to the product of consecutive integers.

4320 = 2^5 \cdot 3^3 \cdot 5 = 2 \cdot 3 \cdot 2^2 \cdot 5 \cdot (2 \cdot 3) \cdot (2 \cdot 3) = 6! \cdot 6

n = 6

 

Try it and see if it works

I was tutoring a student who is using the AoPS Prealgebra textbook, and we were learning how to solve linear equations with one variable (p. 215).  This is the sort of problem that looks like this:
21n + 28 = 10n - 40.
My student understood that in order to solve the equation we first need to combine like terms and isolate the variable, which he successfully did:
11n = -68.
My student also understood that we need our variable n to have a coefficient of 1 so that we get something that looks like
n = some number.
But how to turn 11n = into n = ?
It turned out he had a lot of interesting ideas.  “How about we subtract 10n from both sides?” he suggested.
Those of you well-versed in algebraic manipulations can foretell that this will lead you farther away from a solution.  But rather than cut him off and tell him this is wrong, I went along with it.
“Okay, let’s give it a try and see what happens.  Subtracting 10n from both sides…”
11n - 10n = -68 - 10n
So: n = -68 - 10n .
Well, that didn’t work.  It turned out my student had many, many creative approaches to isolating n and solving this problem, none of which got us closer to actually solving the problem.  But rather than cutting him off and prematurely telling him he was wrong, I went along with the playful exploration.  This approach of “try it and see if it works” experimentation is something we want to cultivate in our students, not just in math, but in many academic fields.   Eventually he remembered that we could divide both sides by 11 to solve the problem.   While this sort of discovery approach can be inefficient and tedious, it’s fun to pop into a rabbit hole occasionally.
My hope is that in the future, if he sees another problem like this, he won’t panic, but will instead try to reason it out, just as he did when he learned it for the first time.

Mental Math Shortcuts

Today my MathCounts team practiced their first Countdown Round.  This is the spelling bee style competition in which students compete head-to-head on stage to answer the question before their opponent.  Identifying elegant shortcuts is critical to getting the answer asap.  For example:

ink

(Source: 2016 MathCounts Chapter Countdown Round)

The question asks for the difference between the average time and 10 minutes.  While one could find the average of the 3 times and then subtract from 10 minutes, a more clever approach (using smaller numbers) is to average the 3 differences from the start.  Since 10:13 > 10:00t_1 = -13.  The other time differences are 9 and 22.  The sum is -13+9+22=18 and the average is 6.  Smaller numbers means quicker, more accurate calculations.

Answer the Question that was Asked

When I coach my MathCounts teams, I make sure to remind them to answer the question that is being asked.  That is, after all the calculations have been performed and you have a value for x, reread the question, and make sure x is what the problem is asking for.  Only when you are sure you have answered the question should you write it down (or bubble it in).
For example, today I’m reviewing materials for my team, and I came across this problem:
 “The three-digit integer 5A4 is a multiple of six.  What is the sum of all the possible values for the digit represented by A?”
After some calculations, I found 4 possible values for A: 0, 3, 6, and 9.  My first thought was: “The answer is 4!  There are four possible values of A.”  But then I reread the question, and it asks for the sum of all possible values.  The correct answer is
0+3+6+9 = 18. 
Sometimes you can avoid this error by setting  your variable equal to the value the problem is asking for, so that when you do solve for x, you do not need to perform additional calculations to generate the correct answer.
This is a well-known error, and Richard Rusczyk has written about it in the context of developing good problem solving habits.  Some people may describe this as a “trick question” but I think it’s just a matter of taking care to reread the question and answer the question that is being asked, not the question that you wish had been asked.

Mixed up functions

I recently came across an AIME problem that compared what I call “mixed up functions:”

How many real numbers x satisfy the equation:

\frac{1}{5}(\log_2 x)  =sin(5\pi x)

By mixed up, I mean we have the periodic sine curve equated with the monotonically increasing log function.  Often with logs we can replace them with an exponential function, in this case with base 2, but that looks messy.

The approach I often see with these mis-matched functions is to draw a graph and see where the 2 functions intersect.  But first, I’m going to multiply both sides by 5 because who likes fractions?

\log_2 x  =5 sin(5\pi x)

Ah that’s better.  Notice that the maximum and minimum values of y=5 sin(5\pi x) occur when x=\frac{1}{10} and x=\frac{-1}{10}, and the maximum and minimum values are 5 and -5.  And of course this is a periodic function that is defined for all real numbers.

y=\log_2 x on the other hand is defined only over positive reals, and since it is always increasing from left to right, at some point the value of the function will exceed 5.  Specifically, this will occur for x>32.  Likewise, y=\log_2 x will be less than -5 for x<\frac{1}{32}.

And it’s apparent that in the interval \frac{1}{32}<x<32 the log function is going to intersect the sine function as it goes up and down on its periodic path.  How many times does it intersect?  AIME writers like to catch students on “off by one” errors, so take care to examine what is happening close to x=\frac{1}{32} and x=32 before entering your answer!