[Math] A positive integer is divisible by 11 if and only if the sum of the two-digit blocks of its digits is divisible by 11

Question

Prove or disprove

A positive integer is divisible by 11 if and only if the sum of the two-digit blocks of its digits is divisible by 11

"like 10615=11.965 and 01+06+15=22 "
I think it is true statement

I face difficulty to prove it for both side

n=$a_0$ +$a_1$ (100) +……..+$a_k$ (100)$^k$

n≡ 0(mod 11)

any help with that please?

Answer 1

Hint: Note that $11\mid99$ and $100-1\mid100^k-1$.

The latter is because $$ \frac{x^k-1}{x-1}=x^{k-1}+x^{k-2}+\cdots+1 $$ Now write out the number as $$ n=\sum_{k=0}^n\overbrace{\ \ \ \ \ d_k\ \ \ \ \ }^{\text{$2$ digit blocks}}100^k$$ and the sum of the two digit blocks as $$ s=\sum_{k=0}^nd_k$$ then show $99\mid n-s$.