First, props are due to my friend Natalie Jackucyn of Chicago for introducing me to a variation on this problem a few years ago. I pitched this version of the problem to my Precalculus class this week as a way to review some of their Algebra II topics while enhancing their problem-solving ideas and learning how to use their new CAS handhelds. Here’s the problem:

One of the terms in the expansion of is . If A, B, and n are all integers, what are their values?

For me, the process is critically important. The answers don’t really matter; they just happen. I’ll post my students’ ideas in a few days, but for now I leave it open to any readers to leave their thoughts.

### Like this:

Like Loading...

*Related*

Hm… I guess I can see how that might be a useful question to ask students. I just wrote up some exercises of my own on basic combinatorics, Pascal’s Triangle and Binomial theorem. I’d be interested in the thoughts of someone who teaches those topics, if you could spare a moment.

Darn you for sucking me back into the world of academic challenges! The hardest part of this (so far) was trying to remember how Binomial expansions work. There’s a good chance I’m wrong on this, as I’m a bit rusty, but I believe that the problem boils down to 56A^5x^5B^3y^3=27869184x^5y^3 because it’s the 4th term of a binomial expansion where n=8. So 56A^5B^3=27869184. So A^5B^3=497664. I will admit that I couldn’t think of an easy way to figure out A and B other than trial and error, so I popped up an excel spreadsheet and got A =6 and B=4

>I will admit that I couldn’t think of an easy way to figure out A and B other than trial and error

You might find factoring to be an easier strategy. . goes to the A, and since 11 isn’t divisible by either 5 or 3 (it’s prime) the 2s have to be spread between the coefficients. In fact, the only way to partition 11 into 3s and 5s is . So and .

Pingback: Binomial Expansion Variation | CAS Musings