# Product rule for exponents

A very annoying parrot says "O" on the 1st day, "OK" on the 2nd day, "OKKO" on the 3rd day, and "OKKOKOOK" on the 4th day. What is the pattern, and if the parrot squawks one letter per second, will the parrot have enough time to finish his squawking on the 16th day?

Notice that on the third day, the parrot squawks OKKO|KOOK. The second half is the first half, with Os and Ks interchanged. So the pattern seems to be: start by squawking the same thing as the previous day, then squawk the same thing, but with Os and Ks interchanged. Now let's figure out whether the parrot has enough time to finish squawking on the 16th day. On the first day, the parrot squawks for 1 second, and every day after, the duration doubles. We know that 1 = 2^0, so on the nth day, the parrot will squawk for 2^(n - 1) seconds. Thus, on the 16th day, in particular, the parrot will squawk for 2^15 seconds. There are 24 hours in a day, 60 minutes per hour, and 60 seconds per hour. Thus, there are 24 * 60 * 60 seconds in a day. Factoring this, we get 2^7 * 3^3 * 5^2. We're now comparing 2^15 and 2^7 * 3^3 * 5^2, to see if they are equal, and if not, which is bigger. Right away, we can divide both by 2^7. Now we're comparing 2^8 with 3^3 * 5^2. The nearest power of 2 less than 3^3 = 27 is 2^4 = 16. The nearest power of 2 less than 5^2 = 25 is also 2^4 = 16. But 2^8 = 2^4 * 2^4. Thus, it's obvious that 2^4 * 2^4 will be less than 3^3 * 5^2. Therefore, the bird will finish his squawking on the 16th day.

This problem was slightly modified from a problem in "The Inquisitive Problem Solver." I didn't like their solution, so I came up with my own.