Prove $ n equiv s(n) ($mod$ 3)$ using the fact that $ [10^n] = [1]$. [duplicate]












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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.






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 Dec 8 '18 at 21:26


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    This question already has an answer here:




    • Why is $9$ special in testing divisiblity by $9$ by summing decimal digits? (casting out nines)

      6 answers




    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.






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     Dec 8 '18 at 21:26


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      This question already has an answer here:




      • Why is $9$ special in testing divisiblity by $9$ by summing decimal digits? (casting out nines)

        6 answers




      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.






      modular-arithmetic







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      This question already has an answer here:




      • Why is $9$ special in testing divisiblity by $9$ by summing decimal digits? (casting out nines)

        6 answers




      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.





      This question already has an answer here:




      • Why is $9$ special in testing divisiblity by $9$ by summing decimal digits? (casting out nines)

        6 answers







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      edited Dec 8 '18 at 20:08







      Claire

















      asked Dec 8 '18 at 18:48









      ClaireClaire

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


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      marked as duplicate by Bill Dubuque modular-arithmetic
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          $begingroup$

          Hint: Let $$z_n=a_n10^n+a_{n-1}10^{n-1}+...+a_1cdot 10+a_0$$ with your hint we get
          $$z_nequiv a_n+a_{n-1}+...+a_1+a_0mod 3$$ since $$10^iequiv 1mod 3$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            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


















          1 Answer
          1






          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          1












          $begingroup$

          Hint: Let $$z_n=a_n10^n+a_{n-1}10^{n-1}+...+a_1cdot 10+a_0$$ with your hint we get
          $$z_nequiv a_n+a_{n-1}+...+a_1+a_0mod 3$$ since $$10^iequiv 1mod 3$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            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
















          1












          $begingroup$

          Hint: Let $$z_n=a_n10^n+a_{n-1}10^{n-1}+...+a_1cdot 10+a_0$$ with your hint we get
          $$z_nequiv a_n+a_{n-1}+...+a_1+a_0mod 3$$ since $$10^iequiv 1mod 3$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            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














          1












          1








          1





          $begingroup$

          Hint: Let $$z_n=a_n10^n+a_{n-1}10^{n-1}+...+a_1cdot 10+a_0$$ with your hint we get
          $$z_nequiv a_n+a_{n-1}+...+a_1+a_0mod 3$$ since $$10^iequiv 1mod 3$$






          share|cite|improve this answer









          $endgroup$



          Hint: Let $$z_n=a_n10^n+a_{n-1}10^{n-1}+...+a_1cdot 10+a_0$$ with your hint we get
          $$z_nequiv a_n+a_{n-1}+...+a_1+a_0mod 3$$ since $$10^iequiv 1mod 3$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 8 '18 at 19:02









          Dr. Sonnhard GraubnerDr. Sonnhard Graubner

          74.8k42865




          74.8k42865












          • $begingroup$
            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


















          • $begingroup$
            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
















          $begingroup$
          Okay, that makes sense to me now.
          $endgroup$
          – Claire
          Dec 8 '18 at 19:22




          $begingroup$
          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




          $begingroup$
          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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