Proof. Let $k_0(b,m,r)\in \mathbb {N}$ be such that for all integers $k\geq k_0$ ,

(2.1) $$ \begin{align} k\geq(b-1)\bigg\lceil\log_b\bigg(\frac{s_b(m)}{d}\cdot k+\frac{s_b(r)}{d}\bigg)\bigg\rceil+(b-2)\bigg((b-1)\bigg\lceil\log_b\bigg(\frac{s_b(m)}{d}\cdot k+\frac{s_b(r)}{d}\bigg)\bigg\rceil-1\bigg). \end{align} $$

Note that $k_0$ is well defined since $b,m,r$ are constants and the right-hand side of (2.1) is of order $O(\log k)$ . Using Dirichlet’s theorem on primes in arithmetic progressions, let $k\in \mathbb {N}$ be such that $k\geq \max \{k_0,b,m\}$ and

$$ \begin{align*}p=\frac{s_b(m)}{d}\cdot k+\frac{s_b(r)}{d}\end{align*} $$

is a prime. Since $p>k\geq \max \{b,m\}$ , we have $p\nmid bm$ . Furthermore, let $\tilde {x}$ be the smallest positive integer such that $\tilde {x}\equiv -m^{-1}r\text { (mod }p\text {)}$ . From Lemma 2.3 and (2.1),

$$ \begin{align*}k\geq s_b(\tilde{x})+(b-2)(s_b(p)-1).\end{align*} $$

By Lemma 2.1, there exist nonnegative integers g and h such that

$$ \begin{align*}k=s_b(\tilde{x})+g(b-1)+h\cdot s_b(p).\end{align*} $$

Let $\omega {\kern-1.3pt}\in{\kern-1.3pt} \mathbb {N}$ be a multiple of $p{\kern-1.3pt}-{\kern-1.3pt}1$ such that $b^\omega{\kern-1.3pt}>{\kern-1.3pt}\max \{m,r\}$ . Note that $b^\omega {\kern-1.3pt}\equiv{\kern-1.3pt} 1\text { (mod }p\text {)}$ by Fermat’s little theorem since $p\nmid b$ . We now define a function $\tau _b:\mathbb {N}\to \mathbb {N}$ as follows. For each fixed $n\in \mathbb {N}$ , let $\sigma _{-1}=0$ and $\sigma _i=\sum _{j=0}^i\nu _b(n,j)$ for $0\leq i\leq \lfloor \log _bn\rfloor $ . Then,

$$ \begin{align*}\tau_b(n)=\sum_{j=1}^{\sigma_{\lfloor\log_bn\rfloor}}b^{j\omega+\ell_j},\end{align*} $$

where $\ell _j=i$ for the unique $i\in \{0,1,2,\dotsc ,\lfloor \log _bn\rfloor \}$ satisfying $\sigma _{i-1}<j\leq \sigma _i$ . It is important to notice that the construction of $\tau _b(n)$ guarantees $s_b(\tau _b(n))=\sigma _{\lfloor \log _bn\rfloor }=s_b(n)$ and $\tau _b(n)\equiv \sum _{j=1}^{\sigma _{\lfloor \log _bn\rfloor }}b^{\ell _j}\equiv \sum _{i=0}^{\lfloor \log _bn\rfloor }\nu _b(n,i)b^i\equiv n\text { (mod }p\text {)}$ .

Let $x_0=\tau _b(\tilde {x})$ and, for each positive integer $t\leq g$ , let

$$ \begin{align*}x_t=x_{t-1}-b^{\lfloor\log_bx_{t-1}\rfloor}+\sum_{\iota=1}^bb^{\iota\omega+\lfloor\log_bx_{t-1}\rfloor-1}.\end{align*} $$

From this construction, $s_b(x_t)=s_b(x_{t-1})+b-1$ and

$$ \begin{align*}x_t\equiv x_{t-1}-b^{\lfloor\log_bx_{t-1}\rfloor}+b\cdot b^{\lfloor\log_bx_{t-1}\rfloor-1}\equiv x_{t-1}\text{ (mod }p\text{)}\end{align*} $$

for all $t\leq g$ . It follows that $s_b(x_g)=s_b(x_0)+g(b-1)=s_b(\tilde {x})+g(b-1)$ and ${x_g\equiv x_0\equiv \tilde {x}\text { (mod }p\text {)}}$ . Lastly, let $\alpha $ and $\beta $ be integers such that $b^{\alpha \omega }>x_g$ and $b^{\beta \omega }>\tau _b(p)$ . We define

$$ \begin{align*}x=x_g+\sum_{\iota=0}^{h-1}\tau_b(p)\cdot b^{(\iota\beta+\alpha)\omega}.\end{align*} $$

Now, $s_b(x)=s_b(x_g)+h\cdot s_b(\tau _b(p))=k$ , $x\equiv x_g+\sum _{\iota =0}^{h-1}p\cdot b^{(\iota \beta +\alpha )\omega }\equiv -m^{-1}r\text { (mod }p\text {)}$ and since every summand of x is a distinct power of b where the powers differ by at least $\omega $ , we have $s_b(mx+r)=s_b(m)\cdot s_b(x)+s_b(r)=s_b(m)\cdot k+s_b(r)=dp$ . Therefore, $mx+r$ is a b-Niven number based on the following observations:

• • $mx+r\equiv m(-m^{-1}r)+r\equiv 0\text { (mod }p\text {)}$ ;

• • $d\mid (mx+r)$ since $d\mid m$ and $d\mid r$ by Lemma 2.2;

• • $\gcd (p,d)=1$ since $p>b$ and $d\mid b-1$ .

Proof. Note that for all nonnegative integers n, if $n=\sum _{i=0}^{\lfloor \log _bn\rfloor }\nu _b(n,i)b^i$ , then $s_b(n)=\sum _{i=0}^{\lfloor \log _bn\rfloor }\nu _b(n,i)\equiv n\text { (mod }b-1\text {)}$ . Hence, $s_b(yn)\equiv yn\equiv ys_b(n)\text { (mod }b-1\text {)}$ .

Proof. Let $i_0$ be the smallest nonnegative integer such that $\nu _b(m,i_0)\neq 0$ . Then there exists a nonnegative integer $a\leq b-1$ such that $\nu _b(am,i_0)=\nu _b(a\cdot \nu _b(m,i_0),0)\geq {b}/{2}$ . Next, if $\nu _b(am,i_0+1)\neq b-1$ , then let $m'=am$ ; otherwise, let $m'=(b+1)am$ so that $\nu _b(m',i_0)=\nu _b(am,i_0)\geq {b}/{2}$ and

$$ \begin{align*}\nu_b(m',i_0+1)\equiv\nu_b(am,i_0+1)+\nu_b(am,i_0)\equiv b-1+\nu_b(am,i_0)\not\equiv {b-1}\, (\mathrm{mod}\ {b}).\end{align*} $$

Furthermore, define $m"$ to be a multiple of $m'$ such that the leading digit of $m"$ in base-b representation is at least ${b}/{2}$ , that is, $\nu _b(m",\lfloor \log _bm"\rfloor )\geq {b}/{2}$ . Let $m^*=b^2m"+m'$ . Then $m^*$ is a multiple of m such that $\nu _b(m^*,i_0)\geq {b}/{2}$ , $\nu _b(m^*,i_0+1)\neq b-1$ and $\nu _b(m^*,\lfloor \log _bm^*\rfloor )\geq {b}/{2}$ .

Let $x,y,z$ be integers such that $xs_b(r)+ys_b(m)+z(b-1)=d$ . Define $y^*$ such that $m^*=y^*m$ and let $z^*$ be an integer such that $s_b(m^*)=y^*s_b(m)+z^*(b-1)$ by Lemma 2.5. Letting $m^{**}=(b^{\lfloor \log _bm^*\rfloor -i_0}+1)m^*$ , we see that $\nu _b(m^{**}, \lfloor \log _bm^*\rfloor )=\nu _b(m^*,\lfloor \log _bm^*\rfloor )+\nu _b(m^*,i_0)-b$ and $\nu _b(m^{**},\lfloor \log _bm^*\rfloor +1)=\nu _b(m^*, i_0+1)+1\leq b-1$ . Hence, $s_b(m^{**})=2s_b(m^*)-(b-1)=2y^*s_b(m)+(2z^*-1)(b-1)$ . By Lemma 2.1, there exist nonnegative integers g and h such that $gz^*+h(2z^*-1)\equiv z\text { (mod }s_b(r)\text {)}$ . Let j be a nonnegative integer such that $gy^*+h(2y^*)+j\equiv y\text { (mod }s_b(r)\text {)}$ . Consider

$$ \begin{align*} \overline{m}=&\sum_{\iota=0}^{g-1}m^*b^{\iota(\lfloor\log_bm^*\rfloor+1)}+\sum_{\iota=0}^{h-1}m^{**}b^{\iota(\lfloor\log_bm^{**}\rfloor+1)+g(\lfloor\log_bm^*\rfloor+1)}\\ &+\sum_{\iota=0}^{j-1}mb^{\iota(\lfloor\log_bm\rfloor+1)+g(\lfloor\log_bm^*\rfloor+1)+h(\lfloor\log_bm^{**}\rfloor+1)}. \end{align*} $$

By construction, $\overline {m}$ is a multiple of m and

$$ \begin{align*} s_b(\overline{m})&=gs_b(m^*)+hs_b(m^{**})+js_b(m)\\ &=g(y^*s_b(m)+z^*(b-1))+h(2y^*s_b(m)+(2z^*-1)(b-1))+js_b(m)\\ &=(gy^*+h(2y^*)+j)s_b(m)+(gz^*+h(2z^*-1))(b-1)\\ &\equiv ys_b(m)+z(b-1)\equiv d\text{ (mod }s_b(r)\text{)}. \end{align*} $$

Note that $d\mid s_b(\overline {m})$ since $d\mid s_b(m)$ and $d\mid b-1$ . Therefore, $\gcd (s_b(\overline {m}),s_b(r))=d$ .

Proof. By Proposition 2.6, there exists a multiple $\overline {m}$ of m such that

$$ \begin{align*}\gcd(s_b(\overline{m}),s_b(r),b-1)=\gcd(s_b(\overline{m}),s_b(r)). \end{align*} $$

Hence, by Proposition 2.4, $\mathcal {S}_{\overline {m},r}$ , and thus $\mathcal {S}_{m,r}$ , contains at least one b-Niven number since $\mathcal {S}_{\overline {m},r}$ is a subset of $\mathcal {S}_{m,r}$ . Let this b-Niven number be $\eta m+r$ for some nonnegative integer $\eta $ . Applying the same argument on the arithmetic progression, $\mathcal {S}_{m,(\eta +1)m+r}$ yields another b-Niven number and our proof is complete by induction.