strager's blag

Yesterday, our math teacher announced there was to be held a county-sponsored math competition the following day. Long story short, I had a day to prepare for the competition. Sadly, I didn't know what to prepare for until the following day (that is, the day of the test). I was given a sample test and worked those math problems.

One of the problems was of the following sort:

Give the first 4 digits of 2009²⁰⁰⁹.

This is a crazy problem, and I was sure there was some stupid math trick which wasn't really math to accomplish this.

I found a mathy way, though. I'm so proud.

Begin by writing the equation:

x = 2009²⁰⁰⁹

Take the common logarithm (log₁₀) of both sides:

log x = log 2009²⁰⁰⁹

Turn the power into multiplication:

log x = 2009 * log 2009

Evaluate the right-hand expression:

log x = 6635.68669 (approx.)

The key to the problem is the next step. We split the right-hand expression into two, along the decimal point:

log x = 6635 + 0.68669 (approx.)

Then raise both sides as exponents with the base 10:

x = 10⁶⁶³⁵ * 10^0.68669 (approx.)

Okay, we're almost done. Let's revisit the original problem:

Give the first 4 digits of 2009²⁰⁰⁹.

How do we find the first four digits? Simple: because multiplying by 10ⁿ for any integer n simply moves the decimal place, not affecting any of the digits, we can disregard it. This reduces the problem to simply:

x = 10^0.68669 (approx.)

Then use a calculator to finish:

x = 4.86060131 (approx.)

We now have our four digits: 4, 8, 6, 0. (Note we aren't rounding.) Checking using Wolfram|Alpha, this answer is correct.

Simply change the numbers to suit any problem. You can even solve for the first digits using a different base by changing the base of the logarithm (using the change-of-base forumula if your calculator isn't godly).