Treebeard's Homepage : Stumpers # Treebeard's Stumper Answer27 October 2000

Halloween Number Trick (& Treat)

Here's some tricky number magic for Halloween. Take the year of your birth, or any number with at least two different digits. Then scramble the digits of your number any way you like to make a different number. Subtract the smaller from the larger. Finally add up all the digits of your answer and look at the picture below. Start at the skull on top as number one, and keep counting clockwise around the circle until you get to your number. I'm sure you'll land on the pumpkin! How does this Halloween trick work, and how is it a useful treat? (Halloween clipart from Roxy's Renditions Halloween Holiday Graphic Collection.)

The sum of the digits of a whole number must have the same remainder divided by nine as the number itself. The digits in the sum can be added again until only a single digit remains, which is the remainder. For example, 1987 = 1+9+8+7 = 25 = 2+5 = 7, and 1987 / 9 = 220 r 7. Rearranging a number's digits doesn't change this sum, so the difference must be divisible by nine with no remainder, which takes us around the nine-step circle to the pumpkin every time. This technique of casting out nines is a useful way to catch math mistakes.

Notes:

This is a tough stumper to explain in a small space, because this simple trick invokes some serious math. We first learn division with whole quotients and remainders. But remainders become less important in Middle and High School after we learn decimals. Later in our math education, we learn about congruence modulo k. That's just a fancy way of talking about the remainder after a division. In many real-world problems, the remainder is more important than the actual quotient, but calculators still don't usually have a [Remainder] or [Mod] key, and you have to go through hoops (how?) to find the remainder after a division.

If a circle has 9 steps (like the stumper), and I walk 100 steps around it, then where do I end up? 100 / 9 = 11.111..., but that's not the answer! Every nine steps takes me back to my start. After 100 steps, I do 11 complete circles with a remainder of one step more, so I end up one step beyond my start. Another way of saying this is that 100 modulo 9 is congruent to 1 or 100 mod 9 = 1. That sounds fancy, but we're really just saying that 100 / 9 = 11 r 1. It's just good old quotient and remainder, but it's the remainder that's important.

The sum of the digits of a whole number (in base 10) must have the same remainder divided by nine as the number itself. An example shows why and suggests how to develop formal proof.

Consider any number, say the year of my birth, 1947. Write this digit-by-digit with powers of 10 to show place value:

```  1947 =    1000    +   900     +    40    +    7
= (1 x 1000) + (9 x 100) + (4 x 10) + (7 x 1)
```

Subtract 1 from each power of 10 with the distributive law, and expand and rearrange the terms:

```       = [1 x (999+1)] + [9 x (99+1)] + [4 x (9+1)] + [7 x (0+1)]
= [(1 x 999)+1] + [(9 x 99)+9] + [(4 x 9)+4] + [(7 x 0)+7]
= (1 x 999) + (9 x 99) + (4 x 9) + (7 x 0) + [1 + 9 + 4 + 7]
```

All the terms in parentheses are now divisible by nine, which leaves the original digits in the brackets as the remainder. So our number now has the form:

```       = (9 x something) + [1 + 9 + 4 + 7]
```

So the original number 1947 equals some multiple of nine plus the sum of the original digits. [1+9+4+7] = 21. We can now do the same process with this number: 2+1 = 3, and 1947 / 9 = 216 with a remainder of 3. So the remainder of the original number divided by nine is the same as the sum of the digits (or nine in case the remainder is zero). This ultimate digit-sum is known as the digital root of the original number.

The sum of the digits of any whole number in base 10 will always have the same remainder divided by nine as the number itself. You can extend this to show that the sum of the digits of any whole number in base k must have the same remainder divided by k-1 as the number itself. Base 2 numbers are an interesting case, since the digital root of a binary number is 1 if there are an odd number of 1 digits and 0 if there are an even number. Computers can use this fact to make quick parity checks of calculations and transmitted data.

Now back to the stumper. Any number abcd... is equal to:

```       (9 x something1) + [a + b + c + d + ...]
```
Rearranging the digits gives another number with all the same digits. But order doesn't matter with addition, so the sum is the same:
```       (9 x something2) + [a + b + c + d + ...]
```
The sums of the digits in brackets are the same, so they will cancel out when we subtract, leaving:
```       (9 x something3)
```

The result must be a multiple of nine, which will take us around the nine-step circle to the pumpkin every time.

That's my Halloween trick. The treat is that this ancient method of casting out nines is a useful way to catch arithmetic mistakes. Add the digits of your numbers (repeatedly) to find the digital root of each number. Whether you're adding, subtracting, multiplying, or dividing your numbers, the digital root of your answer should be the same as the same operation on the digital roots of your original numbers.

```         123      1+2+3 = 6
456      4+5+6 = 15 = 1+5 = 6
+ 789      7+8+9 = 24 = 2+4 = 6   digital root = 6+6+6 = 18 = 9
-----
1368      1+3+6+8 = 18           digital root = 1+8 = 9, looks good!
```
This doesn't guarantee a correct answer, but it is a good check! It doesn't work very well with division and decimals.

Digital roots provide an easy way to solve some classic math stumpers. Can you use the digits 1 - 9 to make three three-digits numbers that add as close as possible to 1500?

```         abc
def
+ ghi
-----
1500,     as close as possible
```
The sum must have the same digital root as the addends, but the digital root of 1500 = 1+5 = 6, and the sum of a+b+c+...+i = 1+2+3+...+9 = 54 and 54 = 5+4 = 9. So an exact answer is not possible. 1503 and 1494 both have digital roots of nine, and 1503 is closest to 1500. Some digit twiddling gives the answer:
```         519
748
+ 236
-----
1503
```

Graybear got this stumper easily, and he adds a neat trick that I've used in class as a calculator exercise:

Do schools still teach the mathematical concept of "casting out nines"? If so, the solution to this weeks stumper is fairly obvious. No matter what number you originally choose, or how you rearrange it, the digits of the two numbers will add to the same amount. The number you get by subtracting one original number from the other must be evenly divisible by nine, and so will the sum of its digits.

A related problem is as follows - choose a three digit number where no two digits are the same. Reverse the three digits to form a new three digit number and subtract the smaller from the larger to get a new three digit number. If this number is less than 100, put a zero in front to give it three digits. Now, take this number, reverse the digits, and ADD it. The result is ALWAYS 1089! If these instructions are confusing, look at these examples:

```847 - 748 = 099; 099 + 990 = 1089
684 - 486 = 198; 198 + 891 = 1089
931 - 139 = 792; 792 + 297 = 1089
```

If your kids haven't heard of casting out nines, by all means tell them! It is a great way to catch math mistakes and works for addition, subtraction, multiplication and division as long as only integers are used.

I don't think I ever learned this technique as a kid, at least not in school, but I did learn how to use a slide rule and find square roots by hand. Methods change, but the math behind them stays the same. I used to prefer subtraction to addition because it was easier to check the answer. But this Halloween treat provides a way!

Here are some other procedures that will always result in a number divisible by nine. Start with any number, and...

• Multiply by nine, or a multiple of nine.
• Add the digits and subtract the sum from the original number.
• Add the digits and multiply by eight, and add the result to the original number.
• Add two scrambled versions to the original number, then square the result.
Many "Mathemagic" tricks can be based on these manipulations!

Here are some links for further research:

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