## Math magic with the number 9

9 is a most interesting number. I'm sure that's largely because 9 is 1 less than 10 and most of us have 10 fingers (or digits) and we do arithmetic in a base 10 system. I've seen an amazing number of math tricks that take advantage of something called "digital roots", which is closely related to the idea of "casting out nines." I want to introduce you to these two concepts and share some fun Math tricks you can do with this "9 stuff."

The digital root of a number, and this only makes sense for whole numbers, is what you get when you add up all of its digits. So, the digital root of 112 is 1+1+2, or 4. The digital root for 1234 is 1+2+3+4, which is 10. Now, when you're computing digital roots you only want a single digit so in the case of 1234, you add up its digits to get 10 then add 1+0 to get 1. So, 1 is the digital root of 1234.

A few more examples:

314159 --> 3+1+4+1+5+9 --> 23 --> 2+3 --> 5

2718281828 --> 2+7+1+8+2+8+1+8+2+8 --> 47 --> 4+7 --> 11 --> 1+1 --> 2

1+2+3+4+5+6+7+8+9 --> 45 --> 4+5 --> 9

The digital root, it turns out, is the remainder when you divide a number by 9. If you divide 43 by 9 you get 4 remainder 7 and you could have quickly determined that the remainder was going to be 7 by adding up the digits of the number 43. Note that, in this world of digital root 9 and 0 are equivalent. So, the digital root of 18 is 1+8=9 but when you divide 18 by 9 you don't get 9 as a remainder you get 0.

"Casting out nines" is a shortcut you can take to computing the digital root of a number. You can ignore 9's when you are adding up the digits. So, if I want to know the digital root of 129 I can compute 1+2+9=12, then do 1+2=3 and the digital root is 3. Or, I can ignore the 9 altogether and compute the digital root as 1+2=3. Another example: 1996 --> 1+9+9+6 --> 25 --> 2+5 --> 7. If I ignore, or "cast out" the two 9's, I get 1+6=7 and I've computed the digital root much more quickly.

Another thing I can do with the idea of casting out 9's is to cast out pairs of digits that add up to 9. So, to compute the digital root of 18273645 I can do it the long way: 1+8+2+7+3+6+4+5 = 36 and 3+6 = 9. Or, I can notice that 1 and 8 add up to 9, as do 2 and 7, 3 and 6, and 4 and 5. In casting out all those pairs of numbers that add up to 9 I get 0 left and, as mentioned before, 0 and 9 are equivalent in this digital root business.

A final thought on casting out 9's: You can cast out any number of digits that add up to 9, not just one or two, and the digits being cast out don't need to be next to one another. So, to compute the digital root of 213487 I notice that the digits 2, 3, and 4 add up to 9 as do the digits 1 and 8. I cast all of them out and I'm left with 7, which is the digital root of this number.

Ok, I've introduced digital roots and casting out 9's. What fun things can we do with these ideas?

1. Here's a very simple trick. Ask someone to write down a number and to tell you the digits as they're writing it down. Make it a 3-digit number so it's not too hard for either of you to work with. Ask them to divide it by 9 and to not tell you the answer. You compute the digital root, convert 9 to 0 if you get 9 as the root, and tell them the answer. If, for example, they tell you that the number is 521 you quickly add up the digits and know that the remainder will be 8, much quicker than they can do the division.

2. Here's another simple trick. Write down the number 0 on a piece of paper and put that piece of paper in your pocket. Have someone write down a 4-digit number then have them mix up the digits any way they want and write down the permuted digits as a second number. Have the person subtract the smaller number from the larger number and divide the result by 9. Pull out the piece of paper and show them that you magically predicted that the remainder would be 0.

Here's an example of this trick. Let's say the person picks the number 1672 and that they mix up the digits to get 2167. Then they subtract the smaller number from the larger number, 1672 from 2167. So, 2167-1672 = 495. Lo and behold, the digital root of 495 is 9 so the remainder is 0.

There's some algebra involved in explaining how this and other digital root tricks work. I'm not going to cover the general algebra in this article but for this particular trick one thing to notice is that when you take two numbers that have the same remainder and you divide each by 9 that the difference of those two numbers is divisible by 9. So, if two numbers have a remainder of 7 when divided by 9 and one of the numbers is 9x+7 and the other number is 9y+7 then the difference of the two numbers is (9x+7) - (9y+7) = 9x-9y = 9(x-y), which is divisible by 9 thus the difference leaves no remainder when divided by 9.

3. Here's another trick. This one is a variation on the last one. Have someone pick a 4-digit number, mix up the digits to form a new number, subtract the larger number from the smaller one, and tell you all the digits of the difference, in any order, omitting any digit they want as long as it's not 0. So, if the person picked 2345 as their number and 5243 as the permuted version of the number then the difference is 2898. Then they tell you that 8, 9, and 8 are digits of the difference. You then tell them that 2 is the missing digit.

The way you know that 2 is the missing digit is that you know that the digital root of the difference of the two numbers they subtracted will be 0 or 9. Adding the digits they told you, 8+9+8 gives you a digital root of 7. You need to add 2, which is the missing digit, to 7 to get a digital root of 9 for all of the digits of the difference.

Let's do one more example. Let's say the person picks 3456 as their number and 6543 as the mixed up version. The difference is 3087. We don't allow the person to pick 0 as the missing digit because then we wouldn't be able to know if 0 or 9 was the missing digit. Let's say the person picks 3 and tells you the other digits, 0, 8, and 7. You add up the the known digits, get 15, which reduces to 6 and then you know that 3 is the digit you need to get the total to 9.

4. Here's a variation on the last trick. Tell a person to pick any 3 digit number and to multiply it by 1089. Then have them tell you all the digits of the number, in any order, except for one. Again, tell them not to pick 0. Let's work out an example. The person picks 473. 473x1089=515097. The person tells you that the digits are 0, 5, 1, 7, and 9. You add the digits as the person tells them to you and you get 22, which reduces to 4. So, 5 must be the missing digit. This trick is related to the last trick in that 1089 is a multiple of 9 and any number multiplied by a multiple of 9 will always be a multiple of 9. So, the digital root of the product will be 9 or 0. So, you have enough information to determine the missing digit if you know what all the other digits are.

5. Here's a final trick. This one is classic and impressive but it takes a bit more work for the person doing the arithmetic than the other tricks. Have the person think of a 3-digit number where the digits are in decreasing order, like 542, 651, or 871. Have the person reverse the digits of the number and subtract it from the first. So, if the person picked 542 then they calculate 542-245 = 297. Then, have the person take the difference they got, 297, in this case, and reverse its digits. In this case they get 792. Finally, have them add the number to its reversal. They add 297 to 792. The answer will always be 1089.

Let's do one more example: The person picks 651. They reverse the digits to get 156. 651-156=495. They reverse the digits in 495 to get 594. 495+594=1089.

I hope you enjoy these Math tricks and that you share them with your friends.

Alex KayFebruary 4th, 2008 - 15:59

Enjoyed the post as always Sol!

I’ll go throw out some 9’s…

Haha. Interesting concepts and fun reading. Thanks 🙂

SolFebruary 5th, 2008 - 09:13

Alex,

Nice to see you here. Don’t throw those 9’s out too far!

Eric NovakFebruary 13th, 2008 - 11:58

Thanks for your submission to the Homeschool Carnival. You can find you post at http://ericnovak.com/?p=145

Eric Novak

SaphireeyesFebruary 16th, 2008 - 02:20

I love this site! I just bought two books from Amazon, but I can’t see how that’s supporting your work. Unless that’s another math trick you’ve got up your sleeve 🙂

Seriously, though, do we have to click through from your page, or visit a site you hold, on Amazon in order for you to earn the credit? Your work herein is awesome and I’d like to give credit where credit is due.

SolFebruary 16th, 2008 - 11:50

@Eric: Thank you for publishing my post.

@Saphireeyes: What supports my work is when people click on the square Amazon.com box underneath the “show your appreciation” heading on the right sidebar. Then, whatever people buy at Amazon after clicking on that link gets me a small commission. Every little bit helps. Also, if I review a book and people click on the link and buy the book, I get a small commission. But, don’t worry, I’ll never give a positive review for a product I don’t believe in. Thanks for asking.

LuisaNovember 15th, 2009 - 11:43

I’ll go do the 9 tricks to my little cousin. They’ll think i’m a genius!!:)

AnonymousAugust 3rd, 2011 - 04:43

wow that was really understandable, i really like dit

george stewartFebruary 5th, 2012 - 15:23

wow! amazing dude.9-9=9 wahaa woadda wubbish 2+2 =2