Prove $ n \equiv s(n)\ (mod\ 3)$ using the fact that $\ [10^n] = [1]$. Let $n = (a_k \times 10^k) + (a_{k-1} \times 10^{k-1}) + \cdots +(a_1 \times 10^1)+ (a_0 \times 10^0)$ and $s(n)=(a_k + a_{k-1}+ \cdots +a_1+a_0)$.

When trying to solve this question, I combined the information given and found that $$n-s(n) = (a_k \times 10^k) + (a_{k-1} \times 10^{k-1}) + \cdots +(a_1 \times 10^1)+ (a_0 \times 10^0)-(a_k + a_{k-1}+ \cdots +a_1+a_0)$$ $$=(a_k \times 10^k-a_k ) + (a_{k-1} \times 10^{k-1}-a_{k-1}) + \cdots +(a_1 \times 10^1-a_1)+ (a_0 \times 10^0-a_0)$$ $$=a_k (10^k-1) + a_{k-1} (10^{k-1}-1) + \cdots +a_1 (10^1-1)+ a_0 (1-1)$$ $$=a_k (10^k-1) + a_{k-1} (10^{k-1}-1) + \cdots +a_1 (9)$$ I'm not sure where to go from here, I thought maybe I could deduce that since, for the case of $\ [10^n] = [1]$ -- since that means that $10^n \equiv 1 (mod \ 3)$ -- I could say that since there exists an integer, call it p, such that $10^n - 1=3p$, and then put that "statement" in the parentheses in the last "=" line that I had above. I have a feeling it doesn't make sense though, and it would be incorrect.


marked as duplicate by Bill Dubuque modular-arithmetic Dec 8 '18 at 21:26

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Hint: Let $$z_n=a_n10^n+a_{n-1}10^{n-1}+...+a_1\cdot 10+a_0$$ with your hint we get $$z_n\equiv a_n+a_{n-1}+...+a_1+a_0\mod 3$$ since $$10^i\equiv 1\mod 3$$

  • $\begingroup$ Okay, that makes sense to me now. $\endgroup$ – Claire Dec 8 '18 at 19:22
  • $\begingroup$ Do you know how I would go about showing that $3|n$ if and only if $3|s(n)$? $\endgroup$ – Claire Dec 8 '18 at 19:22

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