Very sorry for the disappearing and reappearing blog the past 24 hours or so. Blogger has had some difficulties, but I’ve used it for the better part of a decade without incident, so this one won’t send me running for the hills. I’m assured that the super long post I wrote and posted yesterday about obsessive vocabulary studying will be back shortly. Or I will seriously freak out.

Prize for answering this weekend’s challenge question: I’ll use an image of your choice in a future post. Image can’t be copyrighted, profane, or have anything to do with the Philadelphia Phillies. I reserve the right not to post anything I think sucks.

Your weekend challenge:

If

pandqare positive integers, what isp+q?

Put your answers in the comments; I’ll post the solution Monday. Good luck!

UPDATE: Congrats to JD for getting it the long way, and then the short way. Solution below the cut.

It’s funny: I write these questions in a vacuum and sometimes I don’t realize exactly how hard they’ll be until I discuss them with a few people. This one, as these challenge questions go, was especially difficult. I knew it was devious to use 243 in the exponent and the denominator, since the exponent literally could have been anything, but I just couldn’t help myself. So I’m sorry if this one drove you nuts.

There are two insights required to solve this:

**15**. This is easier to see when you’re dealing with variables (ex: (^{243}= 3^{243}5^{243}*xy*)^{2}=*x*^{2}*y*^{2}), but you can always factor numbers that are raised to exponents, if that’ll help you towards a solution. In this case we’re trying to get to 3^{p}5^{q}; that’s a clue that you’re going to want to factor the 15 out.**243 = 3**. Yeah. That’s gonna be important.^{5}

^{p}5

^{q}.

Now you can cancel the 3^{5} out of the denominator by subtracting 5 from the exponent in 3^{243} in the numerator. (For a review of this and other exponent rules, click here.)

And there you have it. If *p* and *q* are positive integers, then they have to be 238 and 243, respectively. So *p* + *q* = 481.

## Comments (5)

p + q = 3

Normally I don’t try to solve these until Monday morning but this has been bothering me because my initial try (drastically reducing the 243 and trying to find a pattern) was way off base.

Three notes:

1) 243 = 3^5 ∴ 1/243 = 3^-5

2) 15^243 = (3^243) * (5^243)

3) (3^p)/(3^x) = 3^(p-x)

So, with that in mind, I would divide both side of you equation by 15^243…

1/243 = [(3^p) * (5^q)] / [(3^243) * (5^243)] which reduces to…

1/243 = 3^(p-243) * 5^(q-243)

So, from note #1, we know the right side must work out to 3^-5 in order for the equation to hold. The only way for this to happen is if…

(p-243) = -5 and (q-243) = 0

because then we will have…

1/243 = (3^-5) * (5^0)

so…

p=238 and q=243 ∴

p+q = 481

???

Ok, so that was much more work than necessary since 15^243/243 reduces to…

3^238 * 5^243

which gives you the answer directly instead of going through all the needless steps I went through originally. Funny how the brain works, I did not even see that until a day later…

Yeah, that’s more like it! 🙂 Shoot me an email if you’ve got any great images that should be included in future posts.

481 it is! 15^243 = (3^243)(5^243) And 243 = 3^5, so the problem becomes:

(3^243) (5^243) divided by 3^5. (3^243) divided by (3^5) = 3^238.

So (3^p)(5^q) = (3^238)(5^243). So p + q = 238 + 243, 0r 481.