Congruence involving CRT












2














I was working on a problem, I arrived at the point at which I have to find $17^{{{17}^{17}}^{17}} pmod {25}$



My attempt: $$ 17^{{{17}^{17}}^{17}}equiv 17^{{{{17}^{17}}^{17}} pmod{phi(25)}} pmod {25} $$$$17^{{{17}^{17}}}equiv 17^{{{{17}^{17}}} pmod{phi(20)}} pmod{20}$$$$17^{17}equiv1^{17}equiv1pmod{phi(20)=8} $$ Thus:$$17^{{{17}^{17}}^{17}}equiv 17^{17}equiv17^{-8}equiv(17^{-1})^{8}pmod{25}$$
The inverse of $17$ modulo $25$ its $3$ since $17cdot3=51$, so:$$17^{{{17}^{17}}^{17}}equiv3^{8} equiv3^3cdot3^3cdot3^2equiv2cdot2cdot9equiv36equiv11pmod{25}$$
I checked but the solutions says that it is actually congruent to $2$ and not $11$, everything before the inverse it's fine for sure since OP follows the same procedure, what did I do wrong?






elementary-number-theory modular-arithmetic chinese-remainder-theorem







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    2














    I was working on a problem, I arrived at the point at which I have to find $17^{{{17}^{17}}^{17}} pmod {25}$



    My attempt: $$ 17^{{{17}^{17}}^{17}}equiv 17^{{{{17}^{17}}^{17}} pmod{phi(25)}} pmod {25} $$$$17^{{{17}^{17}}}equiv 17^{{{{17}^{17}}} pmod{phi(20)}} pmod{20}$$$$17^{17}equiv1^{17}equiv1pmod{phi(20)=8} $$ Thus:$$17^{{{17}^{17}}^{17}}equiv 17^{17}equiv17^{-8}equiv(17^{-1})^{8}pmod{25}$$
    The inverse of $17$ modulo $25$ its $3$ since $17cdot3=51$, so:$$17^{{{17}^{17}}^{17}}equiv3^{8} equiv3^3cdot3^3cdot3^2equiv2cdot2cdot9equiv36equiv11pmod{25}$$
    I checked but the solutions says that it is actually congruent to $2$ and not $11$, everything before the inverse it's fine for sure since OP follows the same procedure, what did I do wrong?






    elementary-number-theory modular-arithmetic chinese-remainder-theorem







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      2












      2








      2







      I was working on a problem, I arrived at the point at which I have to find $17^{{{17}^{17}}^{17}} pmod {25}$



      My attempt: $$ 17^{{{17}^{17}}^{17}}equiv 17^{{{{17}^{17}}^{17}} pmod{phi(25)}} pmod {25} $$$$17^{{{17}^{17}}}equiv 17^{{{{17}^{17}}} pmod{phi(20)}} pmod{20}$$$$17^{17}equiv1^{17}equiv1pmod{phi(20)=8} $$ Thus:$$17^{{{17}^{17}}^{17}}equiv 17^{17}equiv17^{-8}equiv(17^{-1})^{8}pmod{25}$$
      The inverse of $17$ modulo $25$ its $3$ since $17cdot3=51$, so:$$17^{{{17}^{17}}^{17}}equiv3^{8} equiv3^3cdot3^3cdot3^2equiv2cdot2cdot9equiv36equiv11pmod{25}$$
      I checked but the solutions says that it is actually congruent to $2$ and not $11$, everything before the inverse it's fine for sure since OP follows the same procedure, what did I do wrong?






      elementary-number-theory modular-arithmetic chinese-remainder-theorem







      share|cite|improve this question













      I was working on a problem, I arrived at the point at which I have to find $17^{{{17}^{17}}^{17}} pmod {25}$



      My attempt: $$ 17^{{{17}^{17}}^{17}}equiv 17^{{{{17}^{17}}^{17}} pmod{phi(25)}} pmod {25} $$$$17^{{{17}^{17}}}equiv 17^{{{{17}^{17}}} pmod{phi(20)}} pmod{20}$$$$17^{17}equiv1^{17}equiv1pmod{phi(20)=8} $$ Thus:$$17^{{{17}^{17}}^{17}}equiv 17^{17}equiv17^{-8}equiv(17^{-1})^{8}pmod{25}$$
      The inverse of $17$ modulo $25$ its $3$ since $17cdot3=51$, so:$$17^{{{17}^{17}}^{17}}equiv3^{8} equiv3^3cdot3^3cdot3^2equiv2cdot2cdot9equiv36equiv11pmod{25}$$
      I checked but the solutions says that it is actually congruent to $2$ and not $11$, everything before the inverse it's fine for sure since OP follows the same procedure, what did I do wrong?






      elementary-number-theory modular-arithmetic chinese-remainder-theorem




      elementary-number-theory modular-arithmetic chinese-remainder-theorem






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      asked Dec 9 at 11:53









      Spasoje Durovic

      19410




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          There is a mistake.
          $$17^{{{17}^{17}}^{17}}equiv 17^{17}equiv17^{color{red}{-8}}equiv(17^{-1})^{color{red}{8}}pmod{25}$$
          should be
          $$17^{{{17}^{17}}^{17}}equiv 17^{17}equiv17^{color{red}{-3}}equiv(17^{-1})^{color{Red}{3}}pmod{25}$$






          share|cite|improve this answer

















          • 1




            Wait, I know this might be a dumb question but... $17^{17} neq 17^{17 pmod {25}} pmod{25}$ that's the error right?
            – Spasoje Durovic
            Dec 9 at 12:25










          • How silly of me!... I had used the same logic more than once, then I completely messed it up at the end, Thanks!
            – Spasoje Durovic
            Dec 9 at 12:30










          • Haha happens sometimes :) I don't see how CRT applies here though, you've just used lil fermat
            – rsadhvika
            Dec 9 at 12:32












          • Yeah, CRT applied in a different part of the problem since i was evaluating that ugly $17^{17^{17^{17}}}$ thing in $pmod{100}$ so I broke it down into two congruences, one modulo $4$ and the other modulo $25$
            – Spasoje Durovic
            Dec 9 at 14:26











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          There is a mistake.
          $$17^{{{17}^{17}}^{17}}equiv 17^{17}equiv17^{color{red}{-8}}equiv(17^{-1})^{color{red}{8}}pmod{25}$$
          should be
          $$17^{{{17}^{17}}^{17}}equiv 17^{17}equiv17^{color{red}{-3}}equiv(17^{-1})^{color{Red}{3}}pmod{25}$$






          share|cite|improve this answer

















          • 1




            Wait, I know this might be a dumb question but... $17^{17} neq 17^{17 pmod {25}} pmod{25}$ that's the error right?
            – Spasoje Durovic
            Dec 9 at 12:25










          • How silly of me!... I had used the same logic more than once, then I completely messed it up at the end, Thanks!
            – Spasoje Durovic
            Dec 9 at 12:30










          • Haha happens sometimes :) I don't see how CRT applies here though, you've just used lil fermat
            – rsadhvika
            Dec 9 at 12:32












          • Yeah, CRT applied in a different part of the problem since i was evaluating that ugly $17^{17^{17^{17}}}$ thing in $pmod{100}$ so I broke it down into two congruences, one modulo $4$ and the other modulo $25$
            – Spasoje Durovic
            Dec 9 at 14:26
















          2














          There is a mistake.
          $$17^{{{17}^{17}}^{17}}equiv 17^{17}equiv17^{color{red}{-8}}equiv(17^{-1})^{color{red}{8}}pmod{25}$$
          should be
          $$17^{{{17}^{17}}^{17}}equiv 17^{17}equiv17^{color{red}{-3}}equiv(17^{-1})^{color{Red}{3}}pmod{25}$$






          share|cite|improve this answer

















          • 1




            Wait, I know this might be a dumb question but... $17^{17} neq 17^{17 pmod {25}} pmod{25}$ that's the error right?
            – Spasoje Durovic
            Dec 9 at 12:25










          • How silly of me!... I had used the same logic more than once, then I completely messed it up at the end, Thanks!
            – Spasoje Durovic
            Dec 9 at 12:30










          • Haha happens sometimes :) I don't see how CRT applies here though, you've just used lil fermat
            – rsadhvika
            Dec 9 at 12:32












          • Yeah, CRT applied in a different part of the problem since i was evaluating that ugly $17^{17^{17^{17}}}$ thing in $pmod{100}$ so I broke it down into two congruences, one modulo $4$ and the other modulo $25$
            – Spasoje Durovic
            Dec 9 at 14:26














          2












          2








          2






          There is a mistake.
          $$17^{{{17}^{17}}^{17}}equiv 17^{17}equiv17^{color{red}{-8}}equiv(17^{-1})^{color{red}{8}}pmod{25}$$
          should be
          $$17^{{{17}^{17}}^{17}}equiv 17^{17}equiv17^{color{red}{-3}}equiv(17^{-1})^{color{Red}{3}}pmod{25}$$






          share|cite|improve this answer












          There is a mistake.
          $$17^{{{17}^{17}}^{17}}equiv 17^{17}equiv17^{color{red}{-8}}equiv(17^{-1})^{color{red}{8}}pmod{25}$$
          should be
          $$17^{{{17}^{17}}^{17}}equiv 17^{17}equiv17^{color{red}{-3}}equiv(17^{-1})^{color{Red}{3}}pmod{25}$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 9 at 12:11









          rsadhvika

          1,6631228




          1,6631228








          • 1




            Wait, I know this might be a dumb question but... $17^{17} neq 17^{17 pmod {25}} pmod{25}$ that's the error right?
            – Spasoje Durovic
            Dec 9 at 12:25










          • How silly of me!... I had used the same logic more than once, then I completely messed it up at the end, Thanks!
            – Spasoje Durovic
            Dec 9 at 12:30










          • Haha happens sometimes :) I don't see how CRT applies here though, you've just used lil fermat
            – rsadhvika
            Dec 9 at 12:32












          • Yeah, CRT applied in a different part of the problem since i was evaluating that ugly $17^{17^{17^{17}}}$ thing in $pmod{100}$ so I broke it down into two congruences, one modulo $4$ and the other modulo $25$
            – Spasoje Durovic
            Dec 9 at 14:26














          • 1




            Wait, I know this might be a dumb question but... $17^{17} neq 17^{17 pmod {25}} pmod{25}$ that's the error right?
            – Spasoje Durovic
            Dec 9 at 12:25










          • How silly of me!... I had used the same logic more than once, then I completely messed it up at the end, Thanks!
            – Spasoje Durovic
            Dec 9 at 12:30










          • Haha happens sometimes :) I don't see how CRT applies here though, you've just used lil fermat
            – rsadhvika
            Dec 9 at 12:32












          • Yeah, CRT applied in a different part of the problem since i was evaluating that ugly $17^{17^{17^{17}}}$ thing in $pmod{100}$ so I broke it down into two congruences, one modulo $4$ and the other modulo $25$
            – Spasoje Durovic
            Dec 9 at 14:26








          1




          1




          Wait, I know this might be a dumb question but... $17^{17} neq 17^{17 pmod {25}} pmod{25}$ that's the error right?
          – Spasoje Durovic
          Dec 9 at 12:25




          Wait, I know this might be a dumb question but... $17^{17} neq 17^{17 pmod {25}} pmod{25}$ that's the error right?
          – Spasoje Durovic
          Dec 9 at 12:25












          How silly of me!... I had used the same logic more than once, then I completely messed it up at the end, Thanks!
          – Spasoje Durovic
          Dec 9 at 12:30




          How silly of me!... I had used the same logic more than once, then I completely messed it up at the end, Thanks!
          – Spasoje Durovic
          Dec 9 at 12:30












          Haha happens sometimes :) I don't see how CRT applies here though, you've just used lil fermat
          – rsadhvika
          Dec 9 at 12:32






          Haha happens sometimes :) I don't see how CRT applies here though, you've just used lil fermat
          – rsadhvika
          Dec 9 at 12:32














          Yeah, CRT applied in a different part of the problem since i was evaluating that ugly $17^{17^{17^{17}}}$ thing in $pmod{100}$ so I broke it down into two congruences, one modulo $4$ and the other modulo $25$
          – Spasoje Durovic
          Dec 9 at 14:26




          Yeah, CRT applied in a different part of the problem since i was evaluating that ugly $17^{17^{17^{17}}}$ thing in $pmod{100}$ so I broke it down into two congruences, one modulo $4$ and the other modulo $25$
          – Spasoje Durovic
          Dec 9 at 14:26


















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