find a number $aabb$ such that is full square [duplicate]











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  • Find all digits $a,b$ such that the number $overline{aabb}$ is a perfect square

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I know that I can write $aabb=1000a+100a+10b+b$, $a,bin {0,1,2,3,4,5,6,7,8,9}, aneq 0$, so $aabb=1100a+11b=m^2$, where $min mathbb N$, I notice that $m^2=11(100a+b)$ so I need to find some number that $11mid m^2$ and $11mid m$,$1100<m^2<9999$ or $33<m<99$ so I just put number in calculator and 88 is answer, so $aabb=7744$, but is there any other better solution?










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marked as duplicate by Ng Chung Tak, Jyrki Lahtonen, amWhy, Leucippus, jgon Nov 16 at 2:56


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.



















    up vote
    0
    down vote

    favorite
    2













    This question already has an answer here:




    • Find all digits $a,b$ such that the number $overline{aabb}$ is a perfect square

      5 answers




    I know that I can write $aabb=1000a+100a+10b+b$, $a,bin {0,1,2,3,4,5,6,7,8,9}, aneq 0$, so $aabb=1100a+11b=m^2$, where $min mathbb N$, I notice that $m^2=11(100a+b)$ so I need to find some number that $11mid m^2$ and $11mid m$,$1100<m^2<9999$ or $33<m<99$ so I just put number in calculator and 88 is answer, so $aabb=7744$, but is there any other better solution?










    share|cite|improve this question















    marked as duplicate by Ng Chung Tak, Jyrki Lahtonen, amWhy, Leucippus, jgon Nov 16 at 2:56


    This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

















      up vote
      0
      down vote

      favorite
      2









      up vote
      0
      down vote

      favorite
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      2






      This question already has an answer here:




      • Find all digits $a,b$ such that the number $overline{aabb}$ is a perfect square

        5 answers




      I know that I can write $aabb=1000a+100a+10b+b$, $a,bin {0,1,2,3,4,5,6,7,8,9}, aneq 0$, so $aabb=1100a+11b=m^2$, where $min mathbb N$, I notice that $m^2=11(100a+b)$ so I need to find some number that $11mid m^2$ and $11mid m$,$1100<m^2<9999$ or $33<m<99$ so I just put number in calculator and 88 is answer, so $aabb=7744$, but is there any other better solution?










      share|cite|improve this question
















      This question already has an answer here:




      • Find all digits $a,b$ such that the number $overline{aabb}$ is a perfect square

        5 answers




      I know that I can write $aabb=1000a+100a+10b+b$, $a,bin {0,1,2,3,4,5,6,7,8,9}, aneq 0$, so $aabb=1100a+11b=m^2$, where $min mathbb N$, I notice that $m^2=11(100a+b)$ so I need to find some number that $11mid m^2$ and $11mid m$,$1100<m^2<9999$ or $33<m<99$ so I just put number in calculator and 88 is answer, so $aabb=7744$, but is there any other better solution?





      This question already has an answer here:




      • Find all digits $a,b$ such that the number $overline{aabb}$ is a perfect square

        5 answers








      divisibility






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      edited Nov 15 at 21:40









      amWhy

      191k27223437




      191k27223437










      asked Nov 15 at 18:23









      Marko Škorić

      58110




      58110




      marked as duplicate by Ng Chung Tak, Jyrki Lahtonen, amWhy, Leucippus, jgon Nov 16 at 2:56


      This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.






      marked as duplicate by Ng Chung Tak, Jyrki Lahtonen, amWhy, Leucippus, jgon Nov 16 at 2:56


      This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
























          1 Answer
          1






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          up vote
          1
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          We have that



          $$11mid (100a+b) implies (a+b)equiv 0 pmod{11},$$



          and then try with $(a,b)$ such that $a+b=11$, excluding the cases $b=2,3,7,8$.






          share|cite|improve this answer























          • It's not immediately clear why you exclude cases of b = 2,3,7,8?
            – IanF1
            Nov 15 at 19:32






          • 1




            @IanF1 Because $n^2mod 10$ can’t take those values.
            – gimusi
            Nov 15 at 19:34










          • of course - dumb of me. Might be worth including in the answer for the benefit of other dumb people.
            – IanF1
            Nov 15 at 19:35










          • "a divides b": a mid b which renders $amid b$, when formatted in mathjax. Don't use a|b
            – amWhy
            Nov 15 at 21:38












          • @amWhy Thanks for the editing and the suggestion! Bye
            – gimusi
            Nov 15 at 21:49


















          1 Answer
          1






          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes








          up vote
          1
          down vote













          We have that



          $$11mid (100a+b) implies (a+b)equiv 0 pmod{11},$$



          and then try with $(a,b)$ such that $a+b=11$, excluding the cases $b=2,3,7,8$.






          share|cite|improve this answer























          • It's not immediately clear why you exclude cases of b = 2,3,7,8?
            – IanF1
            Nov 15 at 19:32






          • 1




            @IanF1 Because $n^2mod 10$ can’t take those values.
            – gimusi
            Nov 15 at 19:34










          • of course - dumb of me. Might be worth including in the answer for the benefit of other dumb people.
            – IanF1
            Nov 15 at 19:35










          • "a divides b": a mid b which renders $amid b$, when formatted in mathjax. Don't use a|b
            – amWhy
            Nov 15 at 21:38












          • @amWhy Thanks for the editing and the suggestion! Bye
            – gimusi
            Nov 15 at 21:49















          up vote
          1
          down vote













          We have that



          $$11mid (100a+b) implies (a+b)equiv 0 pmod{11},$$



          and then try with $(a,b)$ such that $a+b=11$, excluding the cases $b=2,3,7,8$.






          share|cite|improve this answer























          • It's not immediately clear why you exclude cases of b = 2,3,7,8?
            – IanF1
            Nov 15 at 19:32






          • 1




            @IanF1 Because $n^2mod 10$ can’t take those values.
            – gimusi
            Nov 15 at 19:34










          • of course - dumb of me. Might be worth including in the answer for the benefit of other dumb people.
            – IanF1
            Nov 15 at 19:35










          • "a divides b": a mid b which renders $amid b$, when formatted in mathjax. Don't use a|b
            – amWhy
            Nov 15 at 21:38












          • @amWhy Thanks for the editing and the suggestion! Bye
            – gimusi
            Nov 15 at 21:49













          up vote
          1
          down vote










          up vote
          1
          down vote









          We have that



          $$11mid (100a+b) implies (a+b)equiv 0 pmod{11},$$



          and then try with $(a,b)$ such that $a+b=11$, excluding the cases $b=2,3,7,8$.






          share|cite|improve this answer














          We have that



          $$11mid (100a+b) implies (a+b)equiv 0 pmod{11},$$



          and then try with $(a,b)$ such that $a+b=11$, excluding the cases $b=2,3,7,8$.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Nov 15 at 21:37









          amWhy

          191k27223437




          191k27223437










          answered Nov 15 at 18:27









          gimusi

          85.8k74294




          85.8k74294












          • It's not immediately clear why you exclude cases of b = 2,3,7,8?
            – IanF1
            Nov 15 at 19:32






          • 1




            @IanF1 Because $n^2mod 10$ can’t take those values.
            – gimusi
            Nov 15 at 19:34










          • of course - dumb of me. Might be worth including in the answer for the benefit of other dumb people.
            – IanF1
            Nov 15 at 19:35










          • "a divides b": a mid b which renders $amid b$, when formatted in mathjax. Don't use a|b
            – amWhy
            Nov 15 at 21:38












          • @amWhy Thanks for the editing and the suggestion! Bye
            – gimusi
            Nov 15 at 21:49


















          • It's not immediately clear why you exclude cases of b = 2,3,7,8?
            – IanF1
            Nov 15 at 19:32






          • 1




            @IanF1 Because $n^2mod 10$ can’t take those values.
            – gimusi
            Nov 15 at 19:34










          • of course - dumb of me. Might be worth including in the answer for the benefit of other dumb people.
            – IanF1
            Nov 15 at 19:35










          • "a divides b": a mid b which renders $amid b$, when formatted in mathjax. Don't use a|b
            – amWhy
            Nov 15 at 21:38












          • @amWhy Thanks for the editing and the suggestion! Bye
            – gimusi
            Nov 15 at 21:49
















          It's not immediately clear why you exclude cases of b = 2,3,7,8?
          – IanF1
          Nov 15 at 19:32




          It's not immediately clear why you exclude cases of b = 2,3,7,8?
          – IanF1
          Nov 15 at 19:32




          1




          1




          @IanF1 Because $n^2mod 10$ can’t take those values.
          – gimusi
          Nov 15 at 19:34




          @IanF1 Because $n^2mod 10$ can’t take those values.
          – gimusi
          Nov 15 at 19:34












          of course - dumb of me. Might be worth including in the answer for the benefit of other dumb people.
          – IanF1
          Nov 15 at 19:35




          of course - dumb of me. Might be worth including in the answer for the benefit of other dumb people.
          – IanF1
          Nov 15 at 19:35












          "a divides b": a mid b which renders $amid b$, when formatted in mathjax. Don't use a|b
          – amWhy
          Nov 15 at 21:38






          "a divides b": a mid b which renders $amid b$, when formatted in mathjax. Don't use a|b
          – amWhy
          Nov 15 at 21:38














          @amWhy Thanks for the editing and the suggestion! Bye
          – gimusi
          Nov 15 at 21:49




          @amWhy Thanks for the editing and the suggestion! Bye
          – gimusi
          Nov 15 at 21:49



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