a-conjecture-of-mine

An exercise on polyglossy: the same problem solved on multiple languages

proof.tex (1393B)

 1 \documentclass{article}
 2 \usepackage{amsmath, amssymb, amsthm}
 3 
 4 \newtheorem*{conjecture}{Conjecture}
 5 \renewcommand{\qedsymbol}{$\blacksquare$}
 6 \newcommand{\Mod}{\;(\textup{mod} \; 9)}
 7 
 8 \title{A Conjecture of Mine}
 9 \author{Pablo Emilio Escobar Gaviria}
10 \date{August 15, 2020}
11 
12 \begin{document}
13 \maketitle
14 
15 \begin{conjecture}
16   Let \(S : \mathbb{N} \rightarrow \mathbb{N}\) be the sum of the digits of a
17   natural number. Then \(S(n + m) \equiv S(n) + S(m) \Mod\) for all natural 
18   numbers \(n\) and \(m\).
19 \end{conjecture}
20 
21 \begin{proof}
22   If \(n = n_0 \cdot 10^0 + \cdots + n_k \cdot 10^k \in \mathbb{N}\) then:
23 
24   \[
25     \begin{split}
26       n
27       & = n_0 \cdot 10^0 + \cdots + n_k \cdot 10^k \\
28       & = n_0 \cdot (10^0 - 1 + 1) + \cdots + n_k \cdot (10^k - 1 + 1) \\
29       & = (n_0 \cdot (10^0 - 1) + \cdots + n_k \cdot (10^k - 1))
30         + (n_0 + \cdots + n_k) \\
31       & = S(n) + n_0 \cdot (10^0 - 1) + \cdots + n_k \cdot (10^k - 1)
32     \end{split}
33   \]
34 
35   Therefore:
36 
37   \[
38     \begin{split}
39       n 
40       & \equiv S(n) + n_0 \cdot (10^0 - 1) + \cdots + n_k \cdot (10^k - 1) \\
41       & \equiv S(n) + 0 + \cdots + 0 \\
42       & \equiv S(n) \Mod
43     \end{split}
44   \]
45 
46   Similarly, if \(m \in \mathbb{N}\) then \(m \equiv S(m) \Mod\). Finally:
47 
48   \[
49     \begin{split}
50       S(n + m)
51       & \equiv n + m \\
52       & \equiv S(n) + S(m) \Mod
53     \end{split}
54   \]
55 \end{proof}
56 \end{document}
57