Surprising Formulas

Published on Thursday, September 25, 2008 in , ,

FormulasMany people think of formulas as one of the most boring parts of math, when they really should be considered one of the most interesting parts. When you get right down to it, a formula is the mathematical expression of a pattern. Plug in the particular numbers you need, and the formula will show you the results you need.

Often, it isn't immediately apparent why a particular formula works, with the result that plugging numbers into the formula can seem like magic. If there's a formula for something that doesn't seem to have a coherent pattern, such as the Day of the Week For Any Date feat, the feeling of wonder may be increased even more.

Learning why a particular formula works, since it requires learning about and understanding a pattern, can also be eye-opening. For example, we all now that the area of a circle can be calculated with the formula pi*r2, but why does that work? In this BetterExplained article, you'll be walked through the process of discovering the formula. You might just accidentally introduce yourself to calculus along the way, too.

Here's a question for you that needs a formula for an answer. At what temperature are Fahrenheit and Celsius the same number? To get Fahrenheit, you're usually taught to the Celsius temperature, multiply it by 9/5, and then add 32. To go the other way, you're taught to subtract 32, then multiply by 5/9.

Given this approach, the answer to the question isn't readily apparent, and would require some trial and error to work out. However, there's actually a much simpler approach that makes it immediately clear!

Try this much simpler formula on for size: (F + 40) = 1.8 × (C + 40). One look at this formula, and it's not hard to understand that the only temperature at which Fahrenheit and Celsius are the same number is -40 (-40 is the only number that turns both sides into 0).

Taking a closer look, though, and you can make the formula even easier. First, note that regardless of whether you're starting from Celsius or Fahrenheit, the first step will always be adding 40, and the last step will always be subtracting 40. To multiply by 1.8, simply multiply by 2 and subtract 10% of the result. To divide by 1.8, divide by 2 and add 10% (the actual percentage would be 11.11111%). If you think of Fahrenheit as the “bigger” number of the two, it's easy to remember that you need to multiply to go from Celsius to Fahrenheit, and divide when you're going from Fahrenheit to Celsius.

For example, what is the equivalent of 23 degrees Celsius? Add 40, to get 63. Double that, giving 126. Subtract 10% of that number, which is 126-12.6, which gives 113.4. Don't forget to subtract 40! This last step gives 73.4 degrees Fahrenheit. That's much easier, isn't it? Even if you round the 10% estimate (13 instead of 12.6), you'd still get 73 degrees!

We're going to move on to a very unusual formula now. Assuming every page in a given book is numbered starting with 1, how many digits total are required to number that book?

With a book of up to 9 pages, it would just be the number itself. A book with 10 numbered pages, however, requires 11 digits to number. There's the 9 digits from 1 through 9, plus the 2 digits in the number 10. If you have a book with n pages, and n is a number with k digits, then the formula for the number of digits required to number that book is kn-[((10k-1)/9) - k].

That seems complicated, doesn't it? It's much simpler than it appears. Let's find out how many digits would be required to number a 20-page book, and I'll explain it along the way.

First, we multiply 2 by 20, because we have 20 pages, and 20 is a 2-digit number. So far, we have 40. Next, we subtract 11. Why 11? For this step, you need a number consisting of all 1s, and has the same number of digits (k), as the original number in question. If we started with a 3-digit number in the first step, we'd subtract 111 here. If we started with a 4-digit number of pages, we'd subtract 1,111, and so on. So, 40-11 is 29. Finally, we add the number of digits the number of pages (2, as 20 is a 2-digit number), giving us 31. This means that a 20-page book would require 31 digits to number from 1 to 20!

As you practice, this gets easier to understand, too. How many digits would a 256-page book require? 256 × 3 = 768. 768 - 111 (remember why?) = 657. 657 + 3 (256 is a 3-digit numb = 660, so a 256-page book requires a total of 660 digits to number! If you need practice multiplying 3 digit numbers by 3 and 4 digit numbers by 4 in your head, I highly recommend Arthur Benjamin's book, Secrets of Mental Math: The Mathemagician's Guide to Lightning Calculation and Amazing Math Tricks.

One of the classic uses for formulas and patterns is what are known as “divisibility tests”. These are tests to determine whether one number divided by another number will result in a whole number. Most numbers from 1-9 have easy divisibility tests. 1? Every number is divisibly by 1! 2? If the rightmost digit is 0, 2, 4, 6, or 8, it's divisible by 2. 3? Add up the digits of the number, and if their total is divisible by 3, then the number is divisible by 3. 4? Are the rightmost 2 digits divisible by 4? Then the whole number is wholly divisible by 4! 5? If the last digit is 5 or 0, you can evenly divide a number by 5. 6? It's a little tricky, but if it's an even number AND passes the divisibility test for 3, you can divide it by 6. 8? If the last 3 digits are divisible by 8, then the number is evenly divisible by 8. 9? Add up the digits of the number. If they total a multiple of 9, then you can divide the original number evenly by 9.

Notice that I skipped right over 7. That's because 7 doesn't offer divisibility tests as easy as any of the others. Let's say we want to find out whether 398,594,966 is evenly divisible by 7. How can we do this?

This first test is really bizarre. Starting from the rightmost digit, and moving to the left, you multiply the first digit by 1, the next digit by 3, the next digit by 2, the next digit by -1, the next digit by -3, the next digit by -2, the next digit by 1, and so on, continuing on in the pattern of 1, 3, 2, -1, -3, -2, and so on. If the sum of these numbers is 7, then the whole number will be divisible by 7. Applying this to 398,594,966, we get:

(1×6)+(3×6)+(2×9)+(-1×4)+(-3×9)+(-2×5)+(1×8)+(3×9)+(2×3)=6+18+18-4-27-10+8+27+6=42. Since we know 42 is divisible by 7, so is 398,594,966.

It doesn't quite “click”, does it? There are more methods that are almost as weird for 7, such as the X+10Y+9Z method and the 3X+Y method. Few of the divisibility tests for 7 can be turned into something you can do in your head, but there are a few that are interesting enough that they're fun and easy enough to work out on paper.

The first of these divisibility tests involves splitting the rightmost digit off from the rest of the number, doubling it, and then subtracting that result from the rest of the number. With 398,594,966 (better written as 398594966 for this example), we break off the rightmost 6, double it to get 12, and then subtract that 12 from 39859496, giving 39859484. Now, it still isn't apparent as whether 39859484 is a multiple of 7, but we can apply the test again to this new number! Performing this over and over until we get a number that's easily recognizable as a multiple of 7, we get:

39859484 --> 3985948 - (2×4) = 3985948 - 8 = 3985940
3985940 --> 398594 - (2×0) = 398594 -0 = 398594
398594 --> 39859 - (2×4) = 39859 - 8 = 39851
39851 --> 3985 - (2×1) = 3985 - 2 = 3983
3983 --> 398 - (2×3) = 398 - 6 = 392
392 --> 39 - (2×2) = 39 - 4 = 35

35 is a known multiple of 7, so we know that 398,594,966, and all the other numbers in that left column, are multiples of 7! As an aside, instead of doubling the number and subtracting the result, you can multiply the number by 5 and add the result, but the doubling and subtracting method is usually preferred, as it continually decreases the number.

That last method takes a while, though, as it only eliminates 1 digit from the number at each step. Wouldn't it be nice to eliminate several digits in one step, while still testing for even divisibility of 7? Yes, that is also possible!

You'll love this approach. First, break the number in question into groups of 3. Thanks to commas, this is usually very easy. Starting from the rightmost trio, we subtract the next group, add the group after that, and continue on adding and subtracting groups of 3, until you've used up the whole number.

In our example, 398,594,966 becomes 966 (the rightmost trio) - 594 + 398, which totals 770. It's easy to see that 770 is a multiple of 7, so it passes the test in one step!

Let's try that again. What about 258,894,048? That becomes 48 - 894 + 258, which is -588. For this test, you can ignore a minus sign, so we just focus on 588. Just at a glance, it's not easy to see if 588 is a multiple of 7, so what do we do? This is where the previous test comes in handy! With the previous test, 588 becomes 58 - (2×8) = 58 - 16 = 42. In just two steps, instead of the 9 steps that would otherwise be required, we've determined the number is a multiple of 7!

Finally, let's try this approach on 832,089,053. That becomes 53 - 89 + 832, or 796. 796 becomes 79 - (2×6) = 79 - 12 = 67, which is most definitely NOT a multiple of 7 (It's between 63 and 70). Now we know that 832,089,053 isn't a multiple of 7 with only two steps.

Of course, the quickest divisibility test for 7 these days is simply to enter the number in a calculator, and divide it by 7. If you get a whole number, it's evenly divisible by 7! However, if you're in a situation where you can't use a calculator, these tests can come in handy!

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