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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?

Teodorico Levoff
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2 Answers2

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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 $b-1$.

Martin R
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I would say no... For instance, consider the multiple of $10$ to be $20$, then sum of digits of $20$ will be $2$ which indeed in base $11$ is also $2$. But $2$ is not a multiple of $10$.

Anand
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  • wrong in base 11, 20 base 10, is 19 sum of digits is 10 a multiple of ten. this follows from the fact that a base bk+1 number needs it's sum of digits to be a multiple of b to actually represent a multiple of b. –  Jul 11 '19 at 12:19