In many mental math shortcuts, and much of math, the processes work better with some numbers than with others.

For example, it's easier to work out the square root of 49 in your head than it is to work out the square root of 58. When dividing by 9, 72 is easier to work with than, say, 83.

In this post, you'll learn a simple yet powerful technique that can tame the more challenging numbers. Once you learn this and apply it, you'll find it can make many other math tricks easier!

**BASICS:** The basic idea is simply to break up the more challenging numbers into an addition problem, and deal with each part in a different way. You don't just choose any two numbers, however. The real trick is in creating the addition problem wisely.

The very first step is consider what kind of numbers work easier in the type of problem with which you're dealing. If you're going to be dealing with square numbers, then obviously knowing the perfect squares will be of great help. Perhaps you're dealing with divisibility by 7, in which case the multiples of 7 would be important.

When you have a given number, and you understand the type of numbers you need, find the largest number that is easy to deal with which is less than or equal to the given number, and use that as the first number in the addition. We'll call this first number the “reference number.” The other number is simply the distance from the reference number to the given number, so we'll call that number the “distance.”

To clear things up with a practical example, imagine you're challenged to give the square root of 51. Since we're dealing with square roots, we're looking for a perfect sqaure. What's the largest perfect equal to or less than 51? It's 49, so 49 becomes our reference number. We were given the number 51, which is 2 numbers (the distance) after 49, so we would break 51 down into the addition problem of 49 + 2.

From here, you would deal with 49 in one way, and 2 in another way. If you're actually interested in learning the square root process, I explained it in this blog past last year.

As another example, let's say divisibility by 9 is important in a particular process. In this case, we know we need to focus on multiples of 9. Let's take the same example number from above, 51, and break it down by focusing on divisibility by 9. The largest multiple of 9 less than or equal to 51 is 45, which is 6 away from 51. This means that, for this particular task, you'd break 51 up into 45 (the reference number) + 6 (the distance). As before, you'd deal with the 45 with one approach, and the 6 in another, depending on the process. When multiplying by 10/9, as taught in my recent post, *A Trick For Temperatures*, you'll find that this approach comes in very handy.

Notice that the same number, 51, was broken up in different ways, depending on what features were important. As long as you know what features of a number are important, the actual process of breaking up the number is relatively simple.

**MODULUS:** Modular arithmetic is used in many math tricks, and this approach makes it ridiculously easy. If you need to refresh yourself on the basics of modular arithmetic, read this betterexplained.com post.

Let's say the problem we have to deal with is 51 mod 4. The mod 4 portion of the problem tells us right away that multiples of 4 are important. What's the largest multiple of 4 equal to or less than 51? 48 of course! So, 48 becomes our reference number, and 3 is the distance. We've broken this problem down into (48 + 3) mod 4. What's next?

For modulo calculations, the process is ridiculously simple: Ignore the reference number completely, and use the distance as your answer!

In our example of (48 + 3) mod 4, that means we completely ignore the 48, and that our answer is 3! We went from 51 mod 4 to (48 + 3) mod 4 to simply 3.

How about 51 mod 9? You should already understand why 45 would be the reference number, and 6 is the distance, so this problem becomes (45 + 6) mod 9. Ignoring the reference number, 45, we use the distance, 6, as our answer.

Modulo 9 is very useful in mathematics. In fact, it has a strong connection with what are known as digital roots (YouTube video). Calculating any number mod 9 will give you an answer ranging from 0 to 8, yet digital roots are always in the range of 1 to 9. When you take a number and perform mod 9 on it, and get an answer from 1 to 8, that's also the digital root, but if you get 0 as an answer, the digital root is 9.

For example let's try 72 mod 9. 72 is already a multiple of 9, so the reference number would be 72, and the distance would be 0, turning the problem into (72 + 0) mod 9. Obviously, the answer here is 0. However, the digital root would be 9, as explained in the previous paragraph.

**NUMBER BASES:** This trick becomes essential when working with numbers in different bases. Just last month, I featured an entire post on hexadecimal numbers, which is base 16.

When dealing with 2-digit hexadecimal numbers, it's important to know your multiples of 16 from 0 up to 15 by heart. Combined with your knowledge of this reference number and distance trick, 2-digit hexadecimal number become a lot easier to tame.

What is 189 in hexadecimal? Let's break it down into a sum first. What's the largest multiple of 16 which is less than or equal to 189? 176, so that's our reference number. The distance from 176 to 189 is 13, so 189 becomes 176 + 13.

With hexadecimal, you divide the reference number by 16, and leave the distance as it is. Both of these number are then translated into letters (as taught here), if they represent numbers from 10 to 15. The two number are put side by side, and you have your hexadecimal equivalent!

So, we've broken 189 up into 176 + 13. We take the reference number 176, and divide that by 16 to get 11. Since multiples of 16 are so important, this could actually be done from memory. We have 11 and 13 at this point. 11, in hexadecimal, becomes a B and 13 becomes a D, so now we have BD. That means the decimal number 189 is the same as the hexadecimal number BD!

How about 152? It should break up into 144 + 8. Do you understand why? 144 ÷ 16, as we should recall from knowing our multiples of 16, is 9, so we have 9 and 8. No conversion into letter is required, so decimal 152 equals hexadecimal 98.

This page on hexadecimal colors, which are dealt with as 3 sets of 2-digit hexadecimal numbers, teaches a similar approach.

For working with any base *n*, you should known all the multiples of *n* from 0 to *n*-1. When you get the reference number, knowing all these multiples by memory will help make you division of the reference number by *n* much faster. For example, when dealing with base 8, you should know all the multiples of 8 from 0 to 7 by heart, so dividing the reference number by 8 becomes a matter of memory, and therefore quicker to perform mentally.

**FINAL THOUGHTS:** There are some rare cases where you may need to break up the given number into a subtraction problem, instead of an addition problem. In the year section of my moon phase tutorial, you need to find how close a year is to the nearest multiple of 19, whether that multiple is larger or smaller than the given year.

Using the example of 13 from the tutorial, you might be tempted by the instruction above to break 13 up into 0 (the largest multiple of 19 less than or equal to 13) + 13 (the distance from 0). However, in the case of the moon phase tutorial, you should break it up into 19 - 6, with 19 being the reference number (13 is closer to 19 than it is to 0), and - 6 being the distance.

Again, the key is properly understand any math trick before applying this approach.

With a little thought and creativity, you should be able to adapt this reference number and distance technique to your favorite math tricks, and make them easier in the process. I would love to hear about any uses you have for this approach in the comments below!

## Making Math Tricks Easier

Published on Thursday, July 25, 2013 in math, mental math, site features

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## 1 Response to Making Math Tricks Easier

I agree that with many math "tips" and "tricks" found online, while useful in the examples and context provided, are often difficult to apply to real-world or other problems. Therefore, it is important to realize that you may have to solve these problems in multiple part, such as like you demonstrate in this post. This in itself is a very important math "tip." Thanks for writing this!

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