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 p and q are positive integers, what is p + 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:

  1. 15243 = 32435243. This is easier to see when you’re dealing with variables (ex: (xy)2 = x2y2), 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 3p5q; that’s a clue that you’re going to want to factor the 15 out.
  2. 243 = 35. Yeah. That’s gonna be important.
Given those things, let’s attack the problem (remember, every step you take should bring you closer, somehow, to 3p5q.

Now you can cancel the 35 out of the denominator by subtracting 5 from the exponent in 3243 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)

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)


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…

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.

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