Day: October 14, 2014

Same problem 2 years apart

I saw an old note in our Number Theory book about a problem I used for a movie with my older son.  Seeing that note made me want to try the same problem today with my younger son and then go and look at the difference two years had made.

The problem is:  Let N be a positive integer with exactly 11 divisors.  How many divisors does N^3 have?

From October 17, 2012:

From October 14, 2014:

Similarities:

(1) Both kids recognize that only perfect squares can have an odd number of divisors.
(2) Both kids recognize the formula for calculating the number of divisors.  I found it interesting that my younger son seemed to understand the idea behind this formula at the beginning of the video, but the large power at the end of his video gave him a little trouble.

(3) Both kids recognize that 11 is prime and hence is only divisible by 11 and1, and seem to understand how to use that idea to solve the problem.

(4) I do more work in the first video compared to the second, so I’ve learned to slow down a little which is nice.
(5) Both kids recognize some basic powers about powers, though my older son seems more comfortable with those concepts.

(6) and Ha – we had a fire in the fireplace on October 17th 2012 and on October 12th 2012.  Some things don’t change 🙂

A request from Nathan Kraft tied together a bunch math ideas from my day today

I had a nice time teaching the kids this morning.  We were covering some exciting material – stuff that is  fun all by itself while also providing great examples of mathematical thinking.  Unfortunately I goofed up one of the examples with my older son, but who knows, maybe the lesson was made more memorable because of the goof up 🙂

Anyway, these fun ideas were kicking around in my mind all day and  then a fun question that Nathan Kraft put on twitter tonight just happened to tie them all together.  Love it when that happens. Here is Nathan’s question:

The interesting thing about this question is that you don’t have to actually find the value of the shaded area to find the answer.  In fact, trying to find that area makes the problem much harder.  (I’m not going to give away the answer, but a hint would be that this sure looks an awful lot like a Venn diagram problem, doesn’t it?)    It is pretty amazing that you can find the answer to this problem without knowing the area of the shaded regions – but that shows how powerful some ideas in math can be.    By coincidence, I touched on this point in three different ways with my kids this morning.

My younger son and I started a new chapter in our Number Theory book this morning.  The first topic in the chapter is counting divisors.   A typical problem would be how many different divisors does the number 100 have?  Prior to this chapter he would have solved that problem by making a list and counting each element.  This new chapter shows a different approach – one that allows us to say how many divisors there are without actually knowing what all of them are.  It is pretty amazing that you can count the divisors without knowing what they are!!  Quite a beautiful illustration of the power of mathematics:

Next, my older son had a little difficulty with this problem from an old AMC 8:

http://www.artofproblemsolving.com/Wiki/index.php/1992_AJHSME_Problems/Problem_21%20

The difficulty he was having was that he didn’t see how to solve this problem without checking the percent increase on all 5 pairs of bar charts.   But, a little bit of mathematical thinking helps you see that you don’t need to do every computation.   There are 3 examples were the difference between the two sales charts is 2 boxes and there are 2 examples where the difference between the two sales charts is 3 boxes.    Since you are looking for the largest percent increase, you only need to compare the largest percent increase from each of these two sets.    The amazing thing here is that you can know which chart has the largest percent increase in each of those two sets without actually doing any computation since the percent increase will be largest when the starting point is the smallest.  So, just eyeballing the charts now we can see that February’s percent increase (an increase of about 66%) is the one that we need to compare to March’s (an increase of 50%) and we see that February had the largest percent increase.  Again, we can see that February has the largest percent increase without calculating what the percent increase was in January, April, or May.    An amazing idea when you are seeing it for the first time – or at least that’s the impression I got this morning!

Finally an old V.I. Arnold problem that I saw on Tanya Kovanova’s blog a while ago.  Unfortunately her blog has been damaged by a spam attack so the problem isn’t up right now.  Hopefully she’s able to save the blog.

Now, this problem shows basically the exact opposite of what I’ve been talking about up to now.  In the prior three examples we found (or asserted) solutions without doing lots of calculation.  Here we’ll see the danger of computing without thinking.  The problem itself is easy to state:  Suppose you have a right triangle with a hypotenuse of 10 and that the altitude to that hypotenuse has a length of 6.  What is the area of this triangle?

See here for the fun discussion I had about this problem with my older son today:

It is funny sometimes how little seemingly unrelated ideas can suddenly tie a bunch of other ideas together.  Thanks to Nathan Kraft for providing that little spark tonight.