Saturday, July 07, 2012

Problem of the Day: Factorials

I was going through old papers and torn magazine pages stored in folders that I hadn't looked at in years, finding stuff to recycle. I happened to glanced at a calendar of daily problems, most of which were either too advanced for the classes I was teaching at the time, or just a little too involved. However, one simple problem jumped out at me, and I decided to file that one away for next year. It's definitely a question that needs the solver to explain how he arrived at the answer, and would give me insight into their thinking.

I made a second problem based on the original. Here they both are:

1. Find the largest prime factor of (87!)(88!).

2. Find the largest prime factor of 87! + 88!

Why I love the problems: first, students needed to know something about factors, prime numbers and factorials. Second, seeing a number like 88!, that student whose first instinct is to reach for the calculator will have to put it down and find a new approach.

Answers below. Stop reading here if you didn't figure them out yet.

The answer to problem 1 is fairly straightforward. The factors of 87!*88! are
1 * 2 * 3 * ... * 87 * 1 * 2 * 3 * ... * 87 * 88.

There is no need to make factor trees to find the prime factorization (an approach students might take). The largest prime factor would be the largest prime number not greater than 88. That would be 83.

(If a student guessed 87, show them that 8+7 = 15, which is divisible by 3, so 87 is divisible by 3. Or just have them divide 87 / 3 and see that they'll get 29.)

The answer to problem 2 requires a little work. We want a factor of the sum, but factors are for products, not sums. No problem. Let's make a multiplication problem out of it.

87! + 88!
= 87! + 87!(88)
= 87!(1 + 88)
= 87!(89)

The largest prime factor of 87! is still 83. But 89 is a prime number, so it's the largest prime factor, which is another reason why this is such a neat little problem: 89 isn't in the initial problem, but every whole number less than 89 is!

I think that this is a great journal-type question to assess their understanding of concepts and their ability to communicate a solution. And it will fall in nicely with whatever they're calling "differentiating of instruction" this semester. (Yes, they mentioned a new term at the last department meeting, but I neglected to write it down or even really care.)

P.S. The first problem is mine. The second is the original.

1 comment:

GregT said...

That IS a neat couple problems, with elegant solutions... must remember to, at the least, toss them at my higher achieving students in data management (the only high school course where they actually have factorials).

Possible extension could even be to have them generate their own problem of this type, with a different operation.