Q:

The sum of digits of a multiple of 10 converted to base 11 is still a multiple of 10?

Is it true that the sum of digits of a multiple of 10 if converted to it's base 11 form is also a multiple of 10. Is there a formal proof on this? 

A:

If 
$$
m = a_0 +  a_1 11  + a_2 11^2 + \ldots a_n 11^n
$$
then the base-$11$ digit sum of $m$ is $s = a_0 + a_1 + \ldots + a_n$ and
$$
 m - s = a_1 (11- 1) + a_2 (11^2-1) + \ldots a_n (11^n -1 ) \\
 = 10 \left ( a_ 1 + a_2 (11+1) + \ldots + a_n (11^{n-1} + \ldots + 1)\right)
$$
is a multiple of 10. Therefore the base-$11$ digit sum is a multiple of $10$ if and only if the number itself is a multiple of $10$.

Generally, if 
$$
m = a_0 +  a_1 b  + a_2 b^2 + \ldots a_n b^n
$$
is the base-$b$ representation of $m$ and $s = a_0 + a_1 + \ldots + a_n$ the base-$b$ digit sum then
$$
m - s = P(b) - P(1)
$$
where $P(x) = a_0 + \ldots + a_n x^n$ is a polynomial with integer coefficients. It follows (see for example Show that $a - b \mid f(a) - f(b)$) that
$$
b - 1 \mid P(b) - P(1) = m-s 
$$
so that the base-$b$ digit sum of $m$ is a multiple of $b-1$ if and only if $m$ is a multiple of $b-1$.

