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I was asked by an elementary school teacher for a proof that you can ignore all 9's when calculating the digital root of a number.

For instance, when calculating the digital root of 7593329, you start by doing 7+5+9+3+3+2+9 = 38, then 3+8 = 11, and finally 1+1 = 2, so the digital root is 2 - but you can also throw away the 9's and consider the number 75332, where 7+5+3+3+2 = 20, and 2+0 = 2, giving the same digital root.

For proving this, I would use that a number $n$ and its digital root $dr(n)$ are equivalent mod $9$, but I am curious about other (or similar) suggestions for the most straightforward or most pedagogical proof/explanation of this fact.

(Of course there's an exception, when a number consists only of $9$'s)

Formally, the digital root is defined as follows, mimicking http://en.wikipedia.org/wiki/Digital_root

Let $n$ be an integer, and let $S(n)$ be the sum of the digits of $n$. Define recursively $S^k(n) = S(S^{k-1}(n))$. At some point, the sequence $S(n), S^2(n), S^3(n), \ldots$ will be constant, and I define the digital root $dr(n)$ to be this constant.

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  • $\begingroup$ What is your definition of "digital root"? Are you familiar with congruences or modular arithmetic? $\endgroup$ – Bill Dubuque Apr 5 '15 at 20:54
  • $\begingroup$ @BillDubuque I will include it in my post $\endgroup$ – Mankind Apr 5 '15 at 20:55
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    $\begingroup$ One approach would simply be to namedrop "modular arithmetic" and say that it involves advanced math. This would hopefully have the effect of giving the students a sense of wonder and building their interest in mathematics. $\endgroup$ – Jack M Apr 8 '15 at 12:40
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    $\begingroup$ @JackM: It also may have the opposite effect of convincing the students that they won't ever understand it anyway. An elementary proof followed by "but with advanced mathematics, we could have proved it in a single line" is probably much better in building interest. $\endgroup$ – celtschk Apr 9 '15 at 14:28
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    $\begingroup$ this works in other bases aswell -e.g. dropping 3 in base 4 yields the same DR: $1231_4$=13_4=10_4=1_4$. $\endgroup$ – JMP Apr 13 '15 at 9:10
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(Edited to respond to OP comment):

Take a 9 in the original number and put it aside. Add up the remaining digits and iterate, getting a digital root between 1 and 9. Now show that adding the 9 to that root gives a number with the same digital root. More formally:

First, note that $dr(x+9) = dr(x)$. Indeed, suppose first that $x$ consists of all $9$'s; then $dr(x) = 9$, and $x+9 = 19\cdots 98$, which also has a digital root of $9$. If $x$ contains some digit not equal to $9$, say $x = \cdots y9\cdots 9$, then $x+9 = \cdots (y+1)9\cdots 98$, and clearly $dr(x+9) = dr(x)$.

This argument also shows that $dr(x+9\cdot 10^n) = dr(x)$ by considering only the portion of $x$ at or to the left of the $10^n$ position.

So suppose $x = y+9\cdot 10^n$, where $y$ has a zero in the $10^n$ position. Then \begin{equation*} dr(x) = dr(y+9\cdot 10^n) = dr(y). \end{equation*} That is, removing the $9$ had no effect on the digital root.

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  • $\begingroup$ I like this explanation, but it seems to assume that if "dr" is the digital root, then $dr(x+y)=dr(x)+dr(y)$. This is true, of course, but is there an elementary explanation. $\endgroup$ – Mankind Apr 8 '15 at 18:55
  • $\begingroup$ The digital root comes from adding up the digits. Addition is commutative. I'm not sure how much more elementary it can get... $\endgroup$ – rogerl Apr 8 '15 at 19:39
  • $\begingroup$ sorry, I missed something in my comment. What is true is that $dr(x+y)\equiv dr(x)+dr(y) \text{mod} 9$. If one has this result, then the result follows by your argument. $\endgroup$ – Mankind Apr 9 '15 at 8:12
  • $\begingroup$ @HowDoIMath: I think you misunderstood. If you are convinced that $dr(x) = dr(x+9)$, then we can just repeat it as many times as desired to get $dr(x) = dr(x+9) = dr(x+18) = \cdots = dr(x+9\cdot 10^k)$. $\endgroup$ – user21820 Apr 11 '15 at 11:39
  • $\begingroup$ @HowDoIMath It should be dr(x + y) = dr(dr(x) + dr(y)). $\endgroup$ – jkabrg Apr 15 '15 at 11:55
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The only reason I can see immediately is that, in base 10, of course, every number can simply be represented as $$\sum a_{n}10^{n-1}$$ where $a_{n}$ is the nth digit. Since $10^n \equiv 1 \bmod(9)$ for any positive integer n, taking the number modulo 9 is the same exact thing as adding its digits. If one of the digits is 9 though, that will just be congruent to zero, so you can essentially ignore any 9. In the case all digits are 9's, then this whole thing will just be $0 (\bmod9)$ which is also congruent to 9. I don't think there is any kind of intuitive way to explain this. Perhaps because adding any positive integer to 9 is going to yield a new integer in the 10's place, you can prove this.

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Well, to start with, the claim is not completely true: If the number contains of only the digits 0 and 9 (such as 9099090), then clearly ignoring all of them gives you a wrong result (namely 0 instead of 9). So the actual rule is: You can ignore a 9 if there's at least one non-zero non-ignored digit remaining.

OK, so how to prove that? Well, since the digits are summed up, clearly the order of the digits does not matter. Therefore we can safely assume that all 9's are at the end of the number. Also, it is clear that if ignoring them does not change the digital sum, also appending them does not change the digital sum (and vice versa). Finally, it is obvious that to prove that appending arbitrary many 9's doesn't change the digital root, we only have to proof that appending one 9 does not change the digital root. The extra condition is then nicely reduced to the condition that the number we append the 9 to is not 0.

Now what happens if we append a 9 to the number? Well, we increase the digital sum by 9. Since we then calculate the digital root of that digital sum, what we actually have to prove that the digital sum doesn't change if we add 9 to the number we calculate the digital root of.

OK, so we can now consider the following cases:

  • The last digit is a 0 (note that this implies that the number is at least a two-digit, since we excluded that our number is 0). Then this 0 is replaced by a 9, and therefore the next digital sum will again be increased by 9.
  • The last digit is not 0, and the second-to-last digit is not 9 (if the number id single-digit, add a leading 0). Since adding 9 is the same as adding 10 and then subtracting 1, the last digit will be reduced by 1, and the second-to-last digit will be increased by 1. Obviously this doesn't change the sum of the digits.
  • The second-to-last digit is a 9, and the last digit is not a 0. Then the second-to-last digit will be changed to 0, and the rest of the number will change as if the second-to-last digit had been removed (carry!). Clearly removing the second-last digit will again lead to one of the cases listed here (and as soon as we either run out of digits or have reduced the finial digit to 0, we will reach one of the first two cases). Since for each step in that loop but the last one, we subtract 9 once, and in the last one we either add 9, or leave it unchanged, in effect we have changed the digital sum by a multiple of 9.

So we have established that adding 9 to the number changes the digital sum by a multiple of 9. And of course this means that subtracting 9 also changes the digital sum by a multiple of 9, and by iterating, that changing the number by a multiple of 9 also changes the digital sum by a multiple of 9. Also, it is obvious that the digital sum will never be zero unless the number itself was zero.

But the digital root is obtained by doing the digital sum over and over again, so changing the number by a multiple of 9 will change the digital root also by a multiple of 9.

Now the digital root of a non-zero number is always a digit other than 0. Now changing that by a non-zero multiple of 9 will not give a non-zero digit, and therefore the digital sum cannot be changed at all.

And since, as we established in the beginning, ignoring 9's is equivalent to changing the digital sum by a multiple of 9, it follows that ignoring 9's (with the additional condition preventing that we get 0) doesn't change the digital root of the number either.

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One can describe the process of taking the digital root by saying that at each step you take away a multiple of 9 until you reach a number less than 10.

In fact the operation $$ 10^k a_k+...+a_0\mapsto a_k+\dots +a_0 $$ can be broken into $k$ operations $$ -(10^j-1)a_j=-9(10^{j-1}+10^{j-2}+\dots+ 10+1)a_j, $$ each of which is to take away a multiple of 9 (and can be explained to children by saying that you take away $9\cdot 111\dots 1\cdot a_j$).

Every process that take away multiples of 9 of the original number until reaching a number less than 10 must give the same final result (except if you take away all and reach 0 in one step).

But if you first take away all the 9's, you are taking away a multiple of 9.

In fact, deleting a 9 in the j'th place corresponds to the operation $$ -9[(10^{k-1} a_k+...+10^{j}a_{j+1})+10^j a_j] $$ which is to take away a multiple of 9 (which can be explained to children in two steps: first take away $9\cdot 10^j$, which corresponds to replace the 9 by a 0, and then eliminate the zero, which obviously doesn't change the digital root).

Hence, the two operations

  1. Take digital root and
  2. Delete 9's and take then digital root

must give the same result.

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9 is eaten up by the other numbers. E.g. 3+9=12, 1+2=3, you started with 3 and 9 and ended up with 3. So, 3 just ate up the 9. Similarly 9 is eaten by each of the numbers 1,2,3,4,5,6,7,8, and 9 is also eaten up by itself, e.g. start with 9+9, then 18, 1+8=9. (Also the 0's are eaten by everything.) So, you don't really ignore the 9's and 0's but they end up being eaten up, crunchy numbers.

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The essential point is that the operation $n\mapsto S(n)$ does not change the remainder modulo $9$. Write this as $${\rm rem\,}_9\bigl(S(n)\bigr)={\rm rem\,}_9(n)\ .$$ Since $S'(n)$ is defined as $S(n)$, but ignoring the nines in the decimal representation of $n$, the two numbers $S(n)$ and $S'(n)$ differ by a multiple of $9$. Therefore one has $${\rm rem\,}_9\bigl(S'(n)\bigr)={\rm rem\,}_9(n)$$ as well. Since both rules lead to a one digit number $d$ with $${\rm rem\,}_9(d)={\rm rem\,}_9(n)$$ one necessarily has $$dr'(n)=dr(n)\ ,$$ with the sole proviso that in certain very special cases one obtains $dr(n)=9$, but $dr'(n)=0$ (which is more natural than the official $dr(n)=9$ anyway).

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A different way to look at the problem is to notice that if we subtract 9 from a number, the digital sum (and so the digital root) is still the same.

Let's assume the number we are subtracting from has 2 digits, $d_1$ and $d_2$, and is written $[d_1][d_2]$. We establish that we can borrow from $d_1$. Our new number is $[d_1-1][d_2+10]$. Subtracting $9$ gives $[d_1-1][d_2+1]$, which explains why the digital sum and hence the digital root doesn't change. The principle is the same for extended borrowing.

A similar process can be applied to any $9\times 10^k$, and so we can replace all $9$'s in the original number by $0$ without affecting the digital sum.

We must, however, keep at least one non-zero digit, because we need something to borrow from.

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You want a simple answer that anyone can understand immediately I presume. Ok, take a digital sum and add 9 to it. You've added 1 in the tens column and subtracted a 1 from the units so the sum has gone up one and back down. Now point out that the 9 could be anywhere - you can just as easily add 1000 and subtract 100. If you can add 9s you can subtract them just as easily, unless you only have a single 9 left. Think of that case as 09; subtracting 1 from the tens makes no sense, but if you did it you'd get the "number" (-1)(10) which still has a sum of 9

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