Casting Out Nines

What happens when the sum of the digits of your number is not a mul-tiple of 9? For example, consider the number 3457. Following the above process, we can write 3457 (whose digits sum to 19) as 3(⁹⁹⁹) +⁴(⁹⁹) + 5(⁹) +⁷+12, so 3457 is 7+¹² = 19 bigger than a multiple of 9. And

since 19=¹⁸+1, this indicates that 3457 is just 1 bigger than a multiple of 9. We can reach the same conclusion by adding the digits of 19, then adding the digits of 10, which I represent as

3457→¹⁹→¹⁰→¹

The process of adding the digits of your number and repeating that process until you are reduced to a one-digit number is called casting out nines, since at each step of the process, you are subtracting a multiple of 9. The one-digit number obtained at the end of the process is called the digital root of the original number. For example, the digital root of 3457 is 1. The number 3456 has digital root 9. We can succinctly summarize our previous conclusions as follows. For any positive number n:

Ifn has a digital root of 9, then n is a multiple of 9.

Otherwise, the digital root is the remainder obtained whenn is divided by 9.

Or expressed more algebraically, if n has digital root r, then n=^9x+^r

for some integer x. The process of casting out nines can be a fun way to check your answers to addition, subtraction, and multiplication prob-lems. For example, if an addition problem is solved correctly, then the digital root of the answer must agree with the sum of the digital roots.

Here’s an example. Perform the addition problem

!!

! 5

6 11 32 24 91787 + 42864 134651

51

! ! 2

! 2 20

#

Notice that the numbers being added have digital roots of 5 and 6 and their sum, 11, has digital root 2. It is no coincidence that the digital root of the answer, 134,651, also has a digital root of 2. The reason this process works in general is based on the algebra

(^9x+^r1) + (^9y+^r2) =⁹(^x+^y) + (^r1+^r2)

If the numbers didn’t match, then you have definitely made a mis-take somewhere. Important: if the numbers do match, then your answer is not necessarily right. But this process will catch most random errors about 90 percent of the time. Note that it will not catch errors where you’ve accidentally swapped two of the correct digits, since the swap-ping of two correct digits does not alter the digit sum. But if a single digit is in error, it will detect that mistake, unless the error turned a 0 into a 9 or a 9 into a 0. This process can be applied even when we are adding long columns of numbers. For example, suppose that you purchased a number of items with the prices below:

! 5

Adding the digits of our answer, we see that our total has digital root 5, and the sum of the digital roots is 32, which is consistent with our answer, since 32 has digital root 5. Casting out nines works for subtraction as well. For instance, subtract the numbers from our earlier addition problem:

The answer to the subtraction problem 48,923 has digital root 8.

When we subtract the digital roots of the original numbers we see that 5−⁶ = −1. But this is consistent with our answer, since−¹+⁹ = ^8, and adding (or subtracting) multiples of 9 to our answer doesn’t change the digital root. By the same logic, a difference of 0 is consistent with a digital root of 9.

Let’s take advantage of what we’ve learned to create another magic trick (like the one given in the book’s introduction). Follow these in-structions; you may use a calculator if you wish.

1. Think of any two-digit or three-digit number.

2. Add your digits together.

3. Subtract that from your original number.

4. Add the digits of your new number.

5. If your total is even, then multiply it by 5.

6. If your total is odd, then multiply it by 10.

7. Now subtract 15.

Are you thinking of 75?

For example, if you started with the number 47, then you began by adding 4+⁷ = 11, followed by 47−¹¹ = ^{36. Next 3}+⁶ = ^{9, which} is an odd number. Multiplying it by 10 gives you 90, and 90−¹⁵=^75.

On the other hand, if you started with a 3-digit number like 831, then 8+³+¹ =^{12; 831}−¹²= ^{819; 8}+¹+⁹= 18, an even number. Then 18×5=90, and subtracting 15 gives us 75, as before.

The reason this trick works is that if the original number has a digit sum of T, then the number must be T greater than a multiple of 9. When we subtract T from the original number, we are guaranteed to have a multiple of 9 below 999, so its total will be either 9 or 18. (For example, when we started with 47, it had a digit sum of 11. We subtracted 11 to get to 36, with digit sum 9.) After the next step, we will be forced to have 90 (as 9×^{10 or 18}×5) followed by 75, as in the examples above.

Casting out nines works for multiplication too. Let’s see what hap-pens as we multiply the previous numbers:

32 24

5 6 30

!!

!!

! 3

! 3

91787

× 42864 3,934,357,968

#

57 12 !

The reason that casting out nines works for multiplication is based on FOIL from Chapter 2. For instance, in our last example, the digi-tal roots on the right tell us that the numbers being multiplied are of the form 9x+^{5 and 9y}+6, for some integers x and y. And when we multiply these together, we get

(^9x+⁵)(^9y+⁶) = ^81xy+^54x+^45y+³⁰

= ⁹(^9xy+^6x+^5y) +³⁰

= (a multiple of 9)+ (²⁷+³)

= (a multiple of 9)+³

Although casting out nines is not traditionally used to check divi-sion problems, I can’t resist showing you an utterly magical method for dividing numbers by 9. This method is sometimes referred to as the Vedic method. Consider the problem

12302÷⁹ Write the problem as

9

)

^{1 2 3 0 2}

Now bring the first digit above the line and write the letter R (as in remainder) above the last digit, like so.

9)⁾1 2 3 0 R 9

)

^{1 2 3 0 2}

1

Next we add numbers diagonally like in the circled positions below.

The circled numbers 1 and 2 add to 3, so we put a 3 as the next number in the quotient.

9))1 3 3 0 R 9

)

1 2 3 0 2

3

Then 3+³=^6.

9))1 3 6 0 R 9

)

^{1 2 3 0 263}

Then 6+⁰=^6.

9))1 3 6 0 R 9

)

1 2 3 0 2

6 3

Finally, we add 6+2=8 for our remainder.

9))1 3 6 6 R 9

)

1 2 3 0 2

8 0

R

And there’s the answer: 12,302÷⁹ = 1366 with a remainder of 8.

That seemed almost too easy! Let’s do another problem with fewer de-tails.

31415÷⁹ Here’s the answer!

9 3 4 8 9 R 9)3 1 4 1 5

14

Starting with the 3 on top, we compute 3+¹ = ^{4, then 4}+⁴ = ^8, then 8+¹=^{9, then 9}+⁵=14. So the answer is 3489 with a remainder of 14. But since 14=9+5, we add 1 to the quotient to get an answer of 3490 with a remainder of 5.

Here’s a simple question with an attractive answer. I’ll leave it to you to verify (on paper or in your head) that

111,111÷⁹=^{12,345 R 6}

We saw that when the remainder is 9 or larger, we simply added 1 to our quotient, and subtracted 9 from the remainder. The same sort of thing happens when we have a sum that exceeds 9 in the middle of the division problem. We indicate the carry, then subtract 9 from the total and continue (or should I say carry on?) as before. For example, with the problem 4821÷9, we start off like this:

9 4 3 5 R 9

)

^{4 8 2 1}

1

4

Here we start with the 4, but since 4+⁸ = 12, we place a 1 above the 4 (to indicate a carry), then subtract 9 from 12 to write 3 in the next spot. This is followed by 3+² = ^{5, then 5}+¹ = 6 to get an answer of 535 with a remainder of 6, as illustrated on the next page.

9 4 3 5 R 6 9

)

^{4 8 2 1}

1

Here is one more problem with lots of carries. Try 98,765÷9.

9 9 8 6 3 R 8 9

)

^{9 8 7 6 5}

1 1 1

Starting with 9 on top, we add 9+⁸ = 17, indicate the carry and subtract 9, so the second digit of the quotient becomes 8. Next, 8+⁷= 15; indicate the carry and write 15−⁹=^{6. Then 6}+⁶=12; indicate the carry and write 12−⁹=3. Finally, your remainder is 3+⁵=^{8. Taking} all the carries into account, our answer is 10,973 with a remainder of 8.

Aside

If you think dividing by 9 is cool, check out dividing by 91. Ask for any two-digit number and you can instantly divide that number by 91 to as many decimal places as desired. No pencils, no paper, no kidding! For example,

53÷ 91=0.582417 . . .

More specifically, the answer is 0.582417, where the bar above the digits 582417 means that those numbers repeat indefinitely. Where do these numbers come from? It’s as easy as multiplying the original two-digit number by 11. Using the method we learned from Chapter 1, we calcu-late 53×11=583. Subtracting 1 from that number gives us the first half of our answer, namely 0.582. The second half is the first half subtracted from 999, which is 999−582 =417. Therefore, our answer is 0.582417, as promised.

Let’s do one more example. Try 78÷91. Here, 78×11=858, so the answer begins with 857. Then 999−857 = 142, so 78÷91= 0.857142.

We have actually seen that number in Chapter 1, because 78/91 simpli-fies to 6/7.

This method works because 91×11 = 1001. Thus, in the first ex-ample, ⁵³₉₁ = ^53×11_91×11 = ₁₀₀₁⁵⁸³. And since 1/1001 = 0.000999, we get the repeating part of our answer from 583×999=583,000−583=582,417.

Since 91 = 13×7, this gives us a nice way to divide numbers by 13 by unsimplifying it to have a denominator of 91. For instance, 1/13= 7/91, and since 7×11=077, we get

1/13=7/91=^0.076923

Likewise, 2/13=14/91=0.153846, since 14×11=154.

In document [Arthur_Benjamin]_The_Magic_of_Math_Solving_for_x & Figuring out why .pdf (Page 64-71)