Earlier this year, I posted about calculating powers of *e* in your head, as well as powers of Pi.

This time around, I thought I'd pass on a method for calculating powers of a much more humble number: 2. It sounds difficult, but it's much easier than you may think!

**BASICS:** For 2^{0} up to 2^{10}, you'll memorize precise answers. For answers to 2^{11} and higher integer powers, you'll be estimating the numbers in a simple way that comes very close.

First, you must memorize the powers of 2 from 0 to 10 by heart. Here they are, along with some simply ways to memorize each of them:

```
Problem Answer Notes
2
```^{0} = 1 Anything to the 0th power is 1
2^{1} = 2 Anything to the 1st power is itself
2^{2} = 4 2^{2} = 2 × 2 = 2 + 2
2^{3} = 8 3 looks like the right half of an 8
2^{4} = 16 2^{4} = 4^{2}
2^{5} = 32 5 = 3 + 2
2^{6} = 64 2^{6} begins with a 6
2^{7} = 128 2^{6} × 2^{1}
2^{8} = 256 Important in computers
2^{9} = 512 2^{8} × 2^{1}
2^{10} = 1024 2^{10} begins with a 10

Take a close look at 2^{10}, which is 1024. It's very close to 1,000, so we're going to take advantage of the fact that 2

^{10}≈ 10

^{3}!

When multiplying 2

^{x}× 2

^{y}, remember that you simply add the exponents together. For example, 2

^{3}(8) × 2

^{7}(128) = 2

^{7 + 3}= 2

^{10}(1024). Similarly, you can break up a single power of 2 into two powers which add up to the original power, such as 2

^{9}(512) = 2

^{6 + 3}= 2

^{6}(64) × 2

^{3}(8).

**TECHNIQUE:**We'll start with 2

^{15}as an example.

Start by breaking up the given power of 2 into the largest multiple of 10 which is equal to or less than the given power, and multiply it by whatever amount is leftover, which will be a number from 0 to 9.

Using this step, 2

^{15}becomes 2

^{5 + 10}, which becomes the problem 2

^{5}× 2

^{10}.

For an powers from 0 to 9, you should know by heart, so you can convert these almost instantly. In the example problem we've been doing, we know that 2

^{5}is 32, so the problem is now 32 × 2

^{10}.

Now we deal with the multiple of 10. For every multiple of 10 involved, you can replace 2

^{10}with 10

^{3}. With our problem which is now 32 × 2

^{10}, there's only a single multiple of 10 in the power, so we can replace that with 10

^{3}. This turns our current problem into 32 × 10

^{3}.

At this point, it's best to represent the number in scientific notation. In this feat, that simply refers to moving the decimal point to the left, so that the left number is between 0 and 10, and then adding 1 to the power of 10 for each space you moved the decimal. Converting to scientific notation, 32 × 10

^{3}becomes 3.2 × 10

^{4}.

That's all there is to getting our approximation!

How close did we come? 2

^{15}= 32,768, while 3.2 × 10

^{4}= 32,000. I'd say that's pretty good for a mental estimate!

**EXAMPLES:**Over 6 years ago, I related the story of Dr. Solomon Golomb. While in college, he took a freshman biology class. The teacher was explaining that human DNA has 24 chromosomes (as was believed at the time), so the number of possible cells was 2

^{24}. The instructor jokingly added that everyone in the class knew what number that was. Dr. Golomb immediate gave the exact right answer.

Can you estimate Dr. Golomb's answer? Let's work through the above process with 2

^{24}.

First, we break the problem up, so 2

^{24}= 2

^{4 + 20}= 2

^{4}× 2

^{20}.

Next, replace the smaller side with an exact amount. In this step, 2

^{4}× 2

^{20}becomes 16 × 2

^{20}.

Replace 2

^{10x}with 10

^{3x}, which turns 16 × 2

^{20}into 16 × 10

^{6}.

Finally, adjust into scientific notation, so 16 × 10

^{6}becomes 1.6 × 10

^{7}.

If you know your scientific notation, that means your estimated answer is 16 million. Dr. Golomb, as it happened, had memorized the 1st through 10th powers of all the integers from 1 to 10, and new that 2

^{24}was the same as 8

^{8}, so he was able to give the exact answer off the top of his head: 16,777,216. 16 million is a pretty good estimate, isn't it?

Below is the classic

*Legend of the Chessboard*, which emphasizes the powers of 2. In the video, the first square has one (2

^{0}) grain of wheat placed on it, the second square has 2 (2

^{1}) grains of wheat on it, with each square doubling the previous number of grains.

The 64th square, then, would have 2

^{63}grains of wheat on it. About how many is that? I'm going to run through the process a little quicker this time.

Step 1: 2

^{63}= 2

^{3 + 60}= 2

^{3}× 2

^{60}

Step 2: 2

^{3}× 2

^{60}= 8 × 2

^{60}

Step 3: 8 × 2

^{60}≈ 8 × 10

^{18}

While 2

^{63}is 9,223,372,036,854,775,808, our estimate of 8,000,000,000,000,000,000 works.

**TIPS:**If you're really worried about the error, there is a way to improve your estimate. Percentage-wise, the difference between 1,000 (10

^{3}) and 1,024 (2

^{10}) is only 2.4%. So, for every multiple of 10 to which you take the power of 2 (or every power of 3 to which you take 10), you can multiply that by 2.4% to get a percentage difference. You can then multiply that percentage difference by your estimate to improve it.

Just above, we converted 2

^{63}into 8 × 10

^{18}. Since we started with six 10s, our percentage difference would be 6 × 2.4%, or 14.4%. In other words, our estimate of 8 × 10

^{18}could be made closer by adding 14.4% to 8.

Assuming your comfortable with doing percentages like this in your head, 8 increased by 14.4% is 8 + 1.152 = 9.152, so our improved estimate would be 9.152 × 10

^{18}. Considering the actual answer is roughly 9.223 × 10

^{18}, that's quite close!

Practice this, and you'll have an impressive skill with which to impress family, friends, and computer geeks!

## 2 Response to Calculate Powers of 2 In Your Head!

Thank you! These tricks really help!

I thought this would be about the art of memorizing/calculating powers of two.

I feel anyone who needs to approximate 2^10 as 1000 for the purpose of computing 2^15 shouldn't be bothering learning tricks for mental calculation.

I have all the powers of two memorized up to 2^64. And then the power of two powers of two memorized up to 2^2^10 or 2^1024 (unambiguously if you understand exponentiation is right associative or else trivially reducible)

Post a Comment