37 and Veritasium

Question

In Veritasium's new video about 37 there is brought up something interesting about its multiples.
For any multiple of 37 reverse it and put a 0 between all of its digits and the new number will be a multiple for 37 Example : $37 \rightarrow703 ,703=19*37$
$74\rightarrow407,407=11*37$
Why does this happen can someone prove it?

Answer 1

I’m not sure if this is the simplest way to present a solution, but I think it works.

Start with $k=10^n a_n + 10^{n-1}a_{n-1}+\dots + 10a_1+ a_0$. Then $k’ = 10^{2n}a_0+10^{2n-2} a_1+ \dots + 10^2a_{n-1}+ a_n$ where $k’$ is the integer formed by reversing the digits of $k$ and adding in $0$s between adjacent digits.

We note that $\bmod {37}: 10^3\equiv 1, 10^2\equiv 26, 10\equiv 10$. Then, $$\bmod {37}: k \equiv \underbrace{(a_{0}+a_{3}+a_6+\dots)}_{S_1}+10(\underbrace{a_1+a_4+a_7+\dots}_{S_2})+26(\underbrace{a_2+a_5+a_8+\dots}_{S_3})\equiv 0$$ whereas, $$\mod {37}: k’\equiv (\underbrace{a_n+a_{n-3}+a_{n-6}+\dots}_{L_1})+26(\underbrace{a_{n-1}+a_{n-4}+a_{n-7}+\dots}_{L_2})+10(\underbrace{a_{n-2}+a_{n-5}+a_{n-8}+\dots}_{L_3})$$

Case $1$: Suppose $n$ is a multiple of $3$. Then we get that $S_1=L_1$, $S_2=L_3$ and $S_3=L_2$, thereby giving $\bmod {37}: k\equiv k’$.

Case $2$: $n$ is one more than a multiple of $3$. Then $S_2=L_1$, $L_2=S_1$ and $S_3=L_3$. Then, $\bmod {37}: k’\equiv S_2 +26 S_1+10S_3\implies 10k’ \equiv 10S_2+260S_1+100S_3\equiv S_1+10S_2+26S_3\equiv k \equiv 0$.

Case $3$: $n$ is two more than a multiple of $3$. Then $S_3=L_1$, $ L_3=S_1$, $L_2=S_2$. It follows $\bmod{37}: k’\equiv S_3+26S_2+10S_1\implies 26k’\equiv 26S_3+ 10S_2+S_1\equiv k\equiv 0$.

This shows that $\bmod {37}: k’\equiv 0$ regardless of choice of $n$.

NOTE: we have shown more generally that if $k$ has $3m+1$ digits then $\bmod {37}: k’\equiv k$, if $k$ has $3m+2$ digits then $\bmod {37}: 10k’\equiv k$ and if $k$ has $3m$ digits then $\bmod {37}: 26k’\equiv k$.