# Show that $W(10,9x+1)=9$

Definition

Let $$W$$ be the function , defined as $$W(a,b)=r$$

given $$a,b\in \mathbb{Z_+}$$ and $$a>1$$

Take $$m$$ to be the integer s.t. $$a^{m+1} \ge b > a^{m}$$, i.e. $$m = \lceil \log{b}/\log{a} \rceil - 1$$.

Convert number $$a^{m+1} - b$$ in base $$a$$ and add its digits

$$a^{m+1} - b = (r_{l} r_{l-1} ... r_{1} r_{0})_{a}$$

Where $$r=\sum_{i=0}^{l}r_{i}$$.

Show that $$W(10,9x+1)=9$$

Iff $$x=\{(\ \underbrace{ 1\ 1\cdots\ 1\ 1}_{\text{n terms}}\ \ 0 \ \ \underbrace{ \alpha_t\ \alpha_{t-1} \cdots \alpha_1 \ \alpha_0}_{\text{u terms, u=t+1}}) \mid\ n,u\ge 0\ and \ 9 \ge \alpha_j\ge \alpha_{j-1} \ge 1 \ for \ t\ge j \ge 1 \}$$

Example

x= \begin{align} 5 \\ 432 \\ 1108552 \\ 111110777322 \\110111 \\ 11103221 \\ 11110 \\ \vdots \end{align}

Note: $$x$$ have at most only one '0' digit means $$111...111$$ not alowd

Python programming for calculate $$W$$ function

n1=5
n2=77
rem_array = []
while n2 != 1:
mod = n2%n1
if mod != 0:
rem = n1-mod
n2 = n2 + rem
rem_array.append(round(rem))
n2=n2/n1
else:
n2 = n2/n1
rem_array.append(0)
print(rem_array[::-1])
print(sum(rem_array))


• Are you sure? Can you explain how to calculate $W(10,10) = 9$ or $W(10, 100 ) = 9$, or $W(10, 109) = 9$? Dec 1 '19 at 20:33
• @CalvinLin in $W(10,10) \implies x=1,W(10,100) \implies x=11,W(10,110) \implies x=12$ which not satisfied. Dec 1 '19 at 20:56
• Ah ic. I missed the "iff" condition as it wasn't indicated in the title. Dec 2 '19 at 4:00
• Another question about $W$ from this user, mathoverflow.net/questions/347796/… Dec 7 '19 at 5:28

First, some definitions of sets that are crucial to this problem.

1. Let $$S(k)$$ be the set of $$k$$ digit numbers with digit sum of 9.
2. Let $$X(k)$$ be the set as given in the problem, namely$$\{ x=(\ \underbrace{ 1\ 1\cdots\ 1\ 1}_{\text{n terms}}\ \ 0 \ \ \underbrace{ \alpha_t\ \alpha_{t-1} \cdots \alpha_1 \ \alpha_0}_{\text{k terms, k=t+1}}) \mid\ n,k\ge 0\ and \ 9 \ge \alpha_j\ge \alpha_{j-1} \ge 1 \ for \ t\ge j \ge 1 \}$$
3. Let $$D(k)$$ be the set of $$k$$ digit numbers whose digits are non-increasing, namely $$\{x=(\ \underbrace{ \alpha_t\ \alpha_{t-1} \cdots \alpha_1 \ \alpha_0}_{\text{u terms, u=t+1}}) \mid\ u\ge 0\ and \ 9 \ge \alpha_j\ge \alpha_{j-1} \ge 1 \ for \ t\ge j \ge 1 \}$$.
These are the tail end of the set $$X$$.

Lemma: For $$k\geq 2$$, given $$s_k \in S(k)$$, $$10^k - s_k - 1 = 9 d_{k-1}$$ iff $$d_{k-1} \in D(k-1)$$.

Proof: Given $$d_{k-1} \in D(k-1)$$

$$9 d_{k-1} = (10-1) d_{k-1} = \underbrace{ (\alpha_t -1 )\ (9 -\alpha_{t-1}+\alpha_{t-1}) \cdots (9 - \alpha_1+\alpha_0) \ (10 - \alpha_0})$$.
Observe that each place value is nonnegative, so this is indeed the base 10 representation (possibly ignoring leading 0's).

$$10^k -1 - 9d_{k-1} = \underbrace{ (10-\alpha_t )\ (\alpha_{t}-\alpha_{t-1}) \cdots ( \alpha_1-\alpha_0) \ (\alpha_0} - 1)$$ The sum of the digits is $$10-\alpha_t +\alpha_{t}-\alpha_{t-1} + \ldots + \alpha_0 -1 = 9$$.

For the converse, just reverse these steps.

Corollary: Given $$s_k \in S(k)$$, $$10^{k+n} - s_k - 1 = 9 x_{k-1}$$ iff $$x_{k-1} \in X(k-1)$$.

Proof: $$\frac{ 10^{k+n} - 10^k } { 9} = 10^k \frac{ {\underbrace {9\ 9 \ 9 }_\text{n terms}}} {9} = \underbrace {1\ 1 \ 1 }_\text{n terms} \times 10^k$$ as desired.

Corollary $$W(10, 9x+1) = 9$$ iff $$x \in X(k)$$ for some $$k$$.

Proof: This is a restatement of the previous corollary.

• Please elaborate set $S(k)$ with example Dec 2 '19 at 11:13
• @Pruthviraj What part of "$k$ digit numbers with digit sum of 9" needs elaborating on? What are you confused by? Can you state some examples of what you are thinking is in / not in this set? Dec 2 '19 at 16:11
• I apologize, i just got a bit confused.now I'm clear. I will give you responce in few hours. Dec 2 '19 at 16:45
• In lemma proof, why $1\le \alpha_0 <9$, why not $1\le \alpha_0 \le 9$ Dec 2 '19 at 18:17
• Edited/Removed. That wasn't required. Dec 2 '19 at 23:38