Find the least nonnegative residue of: $42^{173} modulo 13$












0












$begingroup$


I can across this question:



Find the least nonnegative residue of:



$42^{173} modulo 13$



I have done the following:



$42^{10} ≡ 1 mod 13$



$42^{173} = 42^{10 (17) +3}$



$ 42^{173} ≡ 42^{3} mod 13$



$ 42^{3} = 74088$



We can write $74088 = a(13)+r$



so $74088 = 5699(13)+1$



Therefore,



$ 42^{173} ≡ 42^{3}= 74088=5699(13)+1 ≡ 1 mod 13$



Is this the correct way to solve it?






number-theory elementary-number-theory modular-arithmetic quadratic-residues







share|cite|improve this question









$endgroup$








  • 3




    $begingroup$
    Why is $42^{10}equiv 1 pmod{13}$? And have you ever hear of Fermat's Little Theorem?
    $endgroup$
    – fleablood
    Dec 7 '18 at 1:47








  • 1




    $begingroup$
    Your method is correct, but as @fleablood points out you started with a wrong result
    $endgroup$
    – rsadhvika
    Dec 7 '18 at 1:49










  • $begingroup$
    Actually your last step should be helpful : $$42^3equiv 1pmod{13}$$
    $endgroup$
    – rsadhvika
    Dec 7 '18 at 1:51








  • 2




    $begingroup$
    Hint $bmod 13!: 42equiv 3 $ and $,3^{large 3}equiv 1, $ so $,3^{large 173}equiv 3^{large 2},$ by $bmod 3!: 173equiv 1!+!7!+!3equiv 2 $
    $endgroup$
    – Bill Dubuque
    Dec 7 '18 at 1:51












  • $begingroup$
    $42^{10} not equiv 1 mod 13$. You can do $42equiv 3pmod{13}$ and $3^3 equiv 27 equiv 1 pmod {13}$. And do what you did. But better if you know FLT.
    $endgroup$
    – fleablood
    Dec 7 '18 at 1:53
















0












$begingroup$


I can across this question:



Find the least nonnegative residue of:



$42^{173} modulo 13$



I have done the following:



$42^{10} ≡ 1 mod 13$



$42^{173} = 42^{10 (17) +3}$



$ 42^{173} ≡ 42^{3} mod 13$



$ 42^{3} = 74088$



We can write $74088 = a(13)+r$



so $74088 = 5699(13)+1$



Therefore,



$ 42^{173} ≡ 42^{3}= 74088=5699(13)+1 ≡ 1 mod 13$



Is this the correct way to solve it?






number-theory elementary-number-theory modular-arithmetic quadratic-residues







share|cite|improve this question









$endgroup$








  • 3




    $begingroup$
    Why is $42^{10}equiv 1 pmod{13}$? And have you ever hear of Fermat's Little Theorem?
    $endgroup$
    – fleablood
    Dec 7 '18 at 1:47








  • 1




    $begingroup$
    Your method is correct, but as @fleablood points out you started with a wrong result
    $endgroup$
    – rsadhvika
    Dec 7 '18 at 1:49










  • $begingroup$
    Actually your last step should be helpful : $$42^3equiv 1pmod{13}$$
    $endgroup$
    – rsadhvika
    Dec 7 '18 at 1:51








  • 2




    $begingroup$
    Hint $bmod 13!: 42equiv 3 $ and $,3^{large 3}equiv 1, $ so $,3^{large 173}equiv 3^{large 2},$ by $bmod 3!: 173equiv 1!+!7!+!3equiv 2 $
    $endgroup$
    – Bill Dubuque
    Dec 7 '18 at 1:51












  • $begingroup$
    $42^{10} not equiv 1 mod 13$. You can do $42equiv 3pmod{13}$ and $3^3 equiv 27 equiv 1 pmod {13}$. And do what you did. But better if you know FLT.
    $endgroup$
    – fleablood
    Dec 7 '18 at 1:53














0












0








0





$begingroup$


I can across this question:



Find the least nonnegative residue of:



$42^{173} modulo 13$



I have done the following:



$42^{10} ≡ 1 mod 13$



$42^{173} = 42^{10 (17) +3}$



$ 42^{173} ≡ 42^{3} mod 13$



$ 42^{3} = 74088$



We can write $74088 = a(13)+r$



so $74088 = 5699(13)+1$



Therefore,



$ 42^{173} ≡ 42^{3}= 74088=5699(13)+1 ≡ 1 mod 13$



Is this the correct way to solve it?






number-theory elementary-number-theory modular-arithmetic quadratic-residues







share|cite|improve this question









$endgroup$




I can across this question:



Find the least nonnegative residue of:



$42^{173} modulo 13$



I have done the following:



$42^{10} ≡ 1 mod 13$



$42^{173} = 42^{10 (17) +3}$



$ 42^{173} ≡ 42^{3} mod 13$



$ 42^{3} = 74088$



We can write $74088 = a(13)+r$



so $74088 = 5699(13)+1$



Therefore,



$ 42^{173} ≡ 42^{3}= 74088=5699(13)+1 ≡ 1 mod 13$



Is this the correct way to solve it?






number-theory elementary-number-theory modular-arithmetic quadratic-residues




number-theory elementary-number-theory modular-arithmetic quadratic-residues






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 7 '18 at 1:44









HidawHidaw

505624




505624








  • 3




    $begingroup$
    Why is $42^{10}equiv 1 pmod{13}$? And have you ever hear of Fermat's Little Theorem?
    $endgroup$
    – fleablood
    Dec 7 '18 at 1:47








  • 1




    $begingroup$
    Your method is correct, but as @fleablood points out you started with a wrong result
    $endgroup$
    – rsadhvika
    Dec 7 '18 at 1:49










  • $begingroup$
    Actually your last step should be helpful : $$42^3equiv 1pmod{13}$$
    $endgroup$
    – rsadhvika
    Dec 7 '18 at 1:51








  • 2




    $begingroup$
    Hint $bmod 13!: 42equiv 3 $ and $,3^{large 3}equiv 1, $ so $,3^{large 173}equiv 3^{large 2},$ by $bmod 3!: 173equiv 1!+!7!+!3equiv 2 $
    $endgroup$
    – Bill Dubuque
    Dec 7 '18 at 1:51












  • $begingroup$
    $42^{10} not equiv 1 mod 13$. You can do $42equiv 3pmod{13}$ and $3^3 equiv 27 equiv 1 pmod {13}$. And do what you did. But better if you know FLT.
    $endgroup$
    – fleablood
    Dec 7 '18 at 1:53














  • 3




    $begingroup$
    Why is $42^{10}equiv 1 pmod{13}$? And have you ever hear of Fermat's Little Theorem?
    $endgroup$
    – fleablood
    Dec 7 '18 at 1:47








  • 1




    $begingroup$
    Your method is correct, but as @fleablood points out you started with a wrong result
    $endgroup$
    – rsadhvika
    Dec 7 '18 at 1:49










  • $begingroup$
    Actually your last step should be helpful : $$42^3equiv 1pmod{13}$$
    $endgroup$
    – rsadhvika
    Dec 7 '18 at 1:51








  • 2




    $begingroup$
    Hint $bmod 13!: 42equiv 3 $ and $,3^{large 3}equiv 1, $ so $,3^{large 173}equiv 3^{large 2},$ by $bmod 3!: 173equiv 1!+!7!+!3equiv 2 $
    $endgroup$
    – Bill Dubuque
    Dec 7 '18 at 1:51












  • $begingroup$
    $42^{10} not equiv 1 mod 13$. You can do $42equiv 3pmod{13}$ and $3^3 equiv 27 equiv 1 pmod {13}$. And do what you did. But better if you know FLT.
    $endgroup$
    – fleablood
    Dec 7 '18 at 1:53








3




3




$begingroup$
Why is $42^{10}equiv 1 pmod{13}$? And have you ever hear of Fermat's Little Theorem?
$endgroup$
– fleablood
Dec 7 '18 at 1:47






$begingroup$
Why is $42^{10}equiv 1 pmod{13}$? And have you ever hear of Fermat's Little Theorem?
$endgroup$
– fleablood
Dec 7 '18 at 1:47






1




1




$begingroup$
Your method is correct, but as @fleablood points out you started with a wrong result
$endgroup$
– rsadhvika
Dec 7 '18 at 1:49




$begingroup$
Your method is correct, but as @fleablood points out you started with a wrong result
$endgroup$
– rsadhvika
Dec 7 '18 at 1:49












$begingroup$
Actually your last step should be helpful : $$42^3equiv 1pmod{13}$$
$endgroup$
– rsadhvika
Dec 7 '18 at 1:51






$begingroup$
Actually your last step should be helpful : $$42^3equiv 1pmod{13}$$
$endgroup$
– rsadhvika
Dec 7 '18 at 1:51






2




2




$begingroup$
Hint $bmod 13!: 42equiv 3 $ and $,3^{large 3}equiv 1, $ so $,3^{large 173}equiv 3^{large 2},$ by $bmod 3!: 173equiv 1!+!7!+!3equiv 2 $
$endgroup$
– Bill Dubuque
Dec 7 '18 at 1:51






$begingroup$
Hint $bmod 13!: 42equiv 3 $ and $,3^{large 3}equiv 1, $ so $,3^{large 173}equiv 3^{large 2},$ by $bmod 3!: 173equiv 1!+!7!+!3equiv 2 $
$endgroup$
– Bill Dubuque
Dec 7 '18 at 1:51














$begingroup$
$42^{10} not equiv 1 mod 13$. You can do $42equiv 3pmod{13}$ and $3^3 equiv 27 equiv 1 pmod {13}$. And do what you did. But better if you know FLT.
$endgroup$
– fleablood
Dec 7 '18 at 1:53




$begingroup$
$42^{10} not equiv 1 mod 13$. You can do $42equiv 3pmod{13}$ and $3^3 equiv 27 equiv 1 pmod {13}$. And do what you did. But better if you know FLT.
$endgroup$
– fleablood
Dec 7 '18 at 1:53










2 Answers
2






active

oldest

votes


















2












$begingroup$

By Fermat's little theorem, $42^color{blue}{12}cong1pmod{13}$.



So, $42^{173}=({42^{12}})^{14}cdot 42^5cong42^5pmod{13}cong3^5pmod{13}cong243pmod{13}cong9pmod{13}$.






share|cite|improve this answer









$endgroup$





















    2












    $begingroup$

    I don't know why you think $42^{10} equiv 1 pmod{13}$.



    But note: $42=39 + 3 equiv 3pmod {13}$ and $3^3 = 27 equiv 1 pmod {13}$



    So $42^{173} equiv 3^{3*57+2} equiv (3^3)^{57}*3^2 equiv 1*9equiv 9 pmod {13}$.



    By Fermat's little theorem we know $42^{12} equiv 1 mod 13$ and we could do $42^{12*14 +5}equiv 42^5equiv 3^5 = 243 equiv 9pmod{13}$.






    share|cite|improve this answer











    $endgroup$













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      2 Answers
      2






      active

      oldest

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      2 Answers
      2






      active

      oldest

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      active

      oldest

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      active

      oldest

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      2












      $begingroup$

      By Fermat's little theorem, $42^color{blue}{12}cong1pmod{13}$.



      So, $42^{173}=({42^{12}})^{14}cdot 42^5cong42^5pmod{13}cong3^5pmod{13}cong243pmod{13}cong9pmod{13}$.






      share|cite|improve this answer









      $endgroup$


















        2












        $begingroup$

        By Fermat's little theorem, $42^color{blue}{12}cong1pmod{13}$.



        So, $42^{173}=({42^{12}})^{14}cdot 42^5cong42^5pmod{13}cong3^5pmod{13}cong243pmod{13}cong9pmod{13}$.






        share|cite|improve this answer









        $endgroup$
















          2












          2








          2





          $begingroup$

          By Fermat's little theorem, $42^color{blue}{12}cong1pmod{13}$.



          So, $42^{173}=({42^{12}})^{14}cdot 42^5cong42^5pmod{13}cong3^5pmod{13}cong243pmod{13}cong9pmod{13}$.






          share|cite|improve this answer









          $endgroup$



          By Fermat's little theorem, $42^color{blue}{12}cong1pmod{13}$.



          So, $42^{173}=({42^{12}})^{14}cdot 42^5cong42^5pmod{13}cong3^5pmod{13}cong243pmod{13}cong9pmod{13}$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 7 '18 at 2:02









          Chris CusterChris Custer

          11.8k3824




          11.8k3824























              2












              $begingroup$

              I don't know why you think $42^{10} equiv 1 pmod{13}$.



              But note: $42=39 + 3 equiv 3pmod {13}$ and $3^3 = 27 equiv 1 pmod {13}$



              So $42^{173} equiv 3^{3*57+2} equiv (3^3)^{57}*3^2 equiv 1*9equiv 9 pmod {13}$.



              By Fermat's little theorem we know $42^{12} equiv 1 mod 13$ and we could do $42^{12*14 +5}equiv 42^5equiv 3^5 = 243 equiv 9pmod{13}$.






              share|cite|improve this answer











              $endgroup$


















                2












                $begingroup$

                I don't know why you think $42^{10} equiv 1 pmod{13}$.



                But note: $42=39 + 3 equiv 3pmod {13}$ and $3^3 = 27 equiv 1 pmod {13}$



                So $42^{173} equiv 3^{3*57+2} equiv (3^3)^{57}*3^2 equiv 1*9equiv 9 pmod {13}$.



                By Fermat's little theorem we know $42^{12} equiv 1 mod 13$ and we could do $42^{12*14 +5}equiv 42^5equiv 3^5 = 243 equiv 9pmod{13}$.






                share|cite|improve this answer











                $endgroup$
















                  2












                  2








                  2





                  $begingroup$

                  I don't know why you think $42^{10} equiv 1 pmod{13}$.



                  But note: $42=39 + 3 equiv 3pmod {13}$ and $3^3 = 27 equiv 1 pmod {13}$



                  So $42^{173} equiv 3^{3*57+2} equiv (3^3)^{57}*3^2 equiv 1*9equiv 9 pmod {13}$.



                  By Fermat's little theorem we know $42^{12} equiv 1 mod 13$ and we could do $42^{12*14 +5}equiv 42^5equiv 3^5 = 243 equiv 9pmod{13}$.






                  share|cite|improve this answer











                  $endgroup$



                  I don't know why you think $42^{10} equiv 1 pmod{13}$.



                  But note: $42=39 + 3 equiv 3pmod {13}$ and $3^3 = 27 equiv 1 pmod {13}$



                  So $42^{173} equiv 3^{3*57+2} equiv (3^3)^{57}*3^2 equiv 1*9equiv 9 pmod {13}$.



                  By Fermat's little theorem we know $42^{12} equiv 1 mod 13$ and we could do $42^{12*14 +5}equiv 42^5equiv 3^5 = 243 equiv 9pmod{13}$.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Dec 7 '18 at 2:03









                  Andreas Blass

                  49.6k451108




                  49.6k451108










                  answered Dec 7 '18 at 2:00









                  fleabloodfleablood

                  69.4k22685




                  69.4k22685






























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