What is the remainder of $3^{13} div 25$ by hand using Euler's theorem?











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Well, I know how to solve this problem by using some basic stuff:



$3^3 equiv 2 rightarrow 3^{12} equiv 2^4 rightarrow 3^{13} equiv 48equiv 23pmod{25}$



But what I'm trying to do is to solve this ONLY using Euler's theorem.
What I tried was (all congruences are in mod 25):



$phi(25)=20, space rightarrow 3^{13}equiv3^{13mod{phi(25)}}equiv3^{13mod20}$
which still gives me $3^{13}$ and there's no way I could calculate that.
So I tried to raise both sides of the starting equation to the power of two so I get something easier to calculate:



$3^{26}equiv x^2 equiv 3^{26mod20}equiv3^6$ and then $3^6equiv(3^{3})^2equiv27^2equiv2^2equiv4$ therefore $x^2=4 rightarrow x=2,-2$, hence $3^{13} equiv 2$ or $3^{13}equiv23$ (And we know that the correct answer is 23)



We can verify which one of $2$ and $23$ is the correct answer using Euler's theorem $a^{phi(n)}equiv1pmod{n}$ but I'm trying to find a shorter, possibly smarter answer which DOES use Euler's theorem. Any thoughts?










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  • $!!bmod 25!: 3^{large 20}equiv 1,Rightarrow,3^{large 10}equiv pm1,,$ must be $-1$ to be correct $!bmod 5. $ Equivalently, check your $2$ cases $!bmod 5 $
    – Bill Dubuque
    Nov 20 at 16:57















up vote
1
down vote

favorite
2












Well, I know how to solve this problem by using some basic stuff:



$3^3 equiv 2 rightarrow 3^{12} equiv 2^4 rightarrow 3^{13} equiv 48equiv 23pmod{25}$



But what I'm trying to do is to solve this ONLY using Euler's theorem.
What I tried was (all congruences are in mod 25):



$phi(25)=20, space rightarrow 3^{13}equiv3^{13mod{phi(25)}}equiv3^{13mod20}$
which still gives me $3^{13}$ and there's no way I could calculate that.
So I tried to raise both sides of the starting equation to the power of two so I get something easier to calculate:



$3^{26}equiv x^2 equiv 3^{26mod20}equiv3^6$ and then $3^6equiv(3^{3})^2equiv27^2equiv2^2equiv4$ therefore $x^2=4 rightarrow x=2,-2$, hence $3^{13} equiv 2$ or $3^{13}equiv23$ (And we know that the correct answer is 23)



We can verify which one of $2$ and $23$ is the correct answer using Euler's theorem $a^{phi(n)}equiv1pmod{n}$ but I'm trying to find a shorter, possibly smarter answer which DOES use Euler's theorem. Any thoughts?










share|cite|improve this question
























  • $!!bmod 25!: 3^{large 20}equiv 1,Rightarrow,3^{large 10}equiv pm1,,$ must be $-1$ to be correct $!bmod 5. $ Equivalently, check your $2$ cases $!bmod 5 $
    – Bill Dubuque
    Nov 20 at 16:57













up vote
1
down vote

favorite
2









up vote
1
down vote

favorite
2






2





Well, I know how to solve this problem by using some basic stuff:



$3^3 equiv 2 rightarrow 3^{12} equiv 2^4 rightarrow 3^{13} equiv 48equiv 23pmod{25}$



But what I'm trying to do is to solve this ONLY using Euler's theorem.
What I tried was (all congruences are in mod 25):



$phi(25)=20, space rightarrow 3^{13}equiv3^{13mod{phi(25)}}equiv3^{13mod20}$
which still gives me $3^{13}$ and there's no way I could calculate that.
So I tried to raise both sides of the starting equation to the power of two so I get something easier to calculate:



$3^{26}equiv x^2 equiv 3^{26mod20}equiv3^6$ and then $3^6equiv(3^{3})^2equiv27^2equiv2^2equiv4$ therefore $x^2=4 rightarrow x=2,-2$, hence $3^{13} equiv 2$ or $3^{13}equiv23$ (And we know that the correct answer is 23)



We can verify which one of $2$ and $23$ is the correct answer using Euler's theorem $a^{phi(n)}equiv1pmod{n}$ but I'm trying to find a shorter, possibly smarter answer which DOES use Euler's theorem. Any thoughts?










share|cite|improve this question















Well, I know how to solve this problem by using some basic stuff:



$3^3 equiv 2 rightarrow 3^{12} equiv 2^4 rightarrow 3^{13} equiv 48equiv 23pmod{25}$



But what I'm trying to do is to solve this ONLY using Euler's theorem.
What I tried was (all congruences are in mod 25):



$phi(25)=20, space rightarrow 3^{13}equiv3^{13mod{phi(25)}}equiv3^{13mod20}$
which still gives me $3^{13}$ and there's no way I could calculate that.
So I tried to raise both sides of the starting equation to the power of two so I get something easier to calculate:



$3^{26}equiv x^2 equiv 3^{26mod20}equiv3^6$ and then $3^6equiv(3^{3})^2equiv27^2equiv2^2equiv4$ therefore $x^2=4 rightarrow x=2,-2$, hence $3^{13} equiv 2$ or $3^{13}equiv23$ (And we know that the correct answer is 23)



We can verify which one of $2$ and $23$ is the correct answer using Euler's theorem $a^{phi(n)}equiv1pmod{n}$ but I'm trying to find a shorter, possibly smarter answer which DOES use Euler's theorem. Any thoughts?







modular-arithmetic totient-function






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edited Nov 20 at 16:03









amWhy

191k27223439




191k27223439










asked Nov 20 at 15:51









Sean Goudarzi

1375




1375












  • $!!bmod 25!: 3^{large 20}equiv 1,Rightarrow,3^{large 10}equiv pm1,,$ must be $-1$ to be correct $!bmod 5. $ Equivalently, check your $2$ cases $!bmod 5 $
    – Bill Dubuque
    Nov 20 at 16:57


















  • $!!bmod 25!: 3^{large 20}equiv 1,Rightarrow,3^{large 10}equiv pm1,,$ must be $-1$ to be correct $!bmod 5. $ Equivalently, check your $2$ cases $!bmod 5 $
    – Bill Dubuque
    Nov 20 at 16:57
















$!!bmod 25!: 3^{large 20}equiv 1,Rightarrow,3^{large 10}equiv pm1,,$ must be $-1$ to be correct $!bmod 5. $ Equivalently, check your $2$ cases $!bmod 5 $
– Bill Dubuque
Nov 20 at 16:57




$!!bmod 25!: 3^{large 20}equiv 1,Rightarrow,3^{large 10}equiv pm1,,$ must be $-1$ to be correct $!bmod 5. $ Equivalently, check your $2$ cases $!bmod 5 $
– Bill Dubuque
Nov 20 at 16:57










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You've already used Euler's theorem to show that $3^{13}equiv2pmod{25}$ or $3^{13}equiv23pmod{25}$, as
$$(3^{13})^2equiv3^{26}equiv3^{26mod{phi(25)}}equiv3^{26mod{20}}equiv3^6equiv4pmod{25}.$$
You can use Euler's theorem mod $5$ to finish the proof; you know that
$$3^{13}equiv3^{13mod{phi(5)}}equiv3^{13mod{4}}equiv3^1pmod{5},$$
and hence $3^{13}equiv23pmod{25}$.






share|cite|improve this answer



















  • 1




    I don't understand the second part, $3^{13}equiv23pmod{25}$, is there any property that I'm missing in modular arithmetics? What I conclude from the first expression is that $3^{13}$ divided by 25 could be any number from {3, 8, 13, 18, 23}
    – Sean Goudarzi
    Nov 20 at 16:07












  • I was assuming the work you showed in your question already; let me expand my answer to clarify this.
    – Servaes
    Nov 20 at 16:26













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You've already used Euler's theorem to show that $3^{13}equiv2pmod{25}$ or $3^{13}equiv23pmod{25}$, as
$$(3^{13})^2equiv3^{26}equiv3^{26mod{phi(25)}}equiv3^{26mod{20}}equiv3^6equiv4pmod{25}.$$
You can use Euler's theorem mod $5$ to finish the proof; you know that
$$3^{13}equiv3^{13mod{phi(5)}}equiv3^{13mod{4}}equiv3^1pmod{5},$$
and hence $3^{13}equiv23pmod{25}$.






share|cite|improve this answer



















  • 1




    I don't understand the second part, $3^{13}equiv23pmod{25}$, is there any property that I'm missing in modular arithmetics? What I conclude from the first expression is that $3^{13}$ divided by 25 could be any number from {3, 8, 13, 18, 23}
    – Sean Goudarzi
    Nov 20 at 16:07












  • I was assuming the work you showed in your question already; let me expand my answer to clarify this.
    – Servaes
    Nov 20 at 16:26

















up vote
1
down vote



accepted










You've already used Euler's theorem to show that $3^{13}equiv2pmod{25}$ or $3^{13}equiv23pmod{25}$, as
$$(3^{13})^2equiv3^{26}equiv3^{26mod{phi(25)}}equiv3^{26mod{20}}equiv3^6equiv4pmod{25}.$$
You can use Euler's theorem mod $5$ to finish the proof; you know that
$$3^{13}equiv3^{13mod{phi(5)}}equiv3^{13mod{4}}equiv3^1pmod{5},$$
and hence $3^{13}equiv23pmod{25}$.






share|cite|improve this answer



















  • 1




    I don't understand the second part, $3^{13}equiv23pmod{25}$, is there any property that I'm missing in modular arithmetics? What I conclude from the first expression is that $3^{13}$ divided by 25 could be any number from {3, 8, 13, 18, 23}
    – Sean Goudarzi
    Nov 20 at 16:07












  • I was assuming the work you showed in your question already; let me expand my answer to clarify this.
    – Servaes
    Nov 20 at 16:26















up vote
1
down vote



accepted







up vote
1
down vote



accepted






You've already used Euler's theorem to show that $3^{13}equiv2pmod{25}$ or $3^{13}equiv23pmod{25}$, as
$$(3^{13})^2equiv3^{26}equiv3^{26mod{phi(25)}}equiv3^{26mod{20}}equiv3^6equiv4pmod{25}.$$
You can use Euler's theorem mod $5$ to finish the proof; you know that
$$3^{13}equiv3^{13mod{phi(5)}}equiv3^{13mod{4}}equiv3^1pmod{5},$$
and hence $3^{13}equiv23pmod{25}$.






share|cite|improve this answer














You've already used Euler's theorem to show that $3^{13}equiv2pmod{25}$ or $3^{13}equiv23pmod{25}$, as
$$(3^{13})^2equiv3^{26}equiv3^{26mod{phi(25)}}equiv3^{26mod{20}}equiv3^6equiv4pmod{25}.$$
You can use Euler's theorem mod $5$ to finish the proof; you know that
$$3^{13}equiv3^{13mod{phi(5)}}equiv3^{13mod{4}}equiv3^1pmod{5},$$
and hence $3^{13}equiv23pmod{25}$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Nov 20 at 16:28

























answered Nov 20 at 15:54









Servaes

21.8k33792




21.8k33792








  • 1




    I don't understand the second part, $3^{13}equiv23pmod{25}$, is there any property that I'm missing in modular arithmetics? What I conclude from the first expression is that $3^{13}$ divided by 25 could be any number from {3, 8, 13, 18, 23}
    – Sean Goudarzi
    Nov 20 at 16:07












  • I was assuming the work you showed in your question already; let me expand my answer to clarify this.
    – Servaes
    Nov 20 at 16:26
















  • 1




    I don't understand the second part, $3^{13}equiv23pmod{25}$, is there any property that I'm missing in modular arithmetics? What I conclude from the first expression is that $3^{13}$ divided by 25 could be any number from {3, 8, 13, 18, 23}
    – Sean Goudarzi
    Nov 20 at 16:07












  • I was assuming the work you showed in your question already; let me expand my answer to clarify this.
    – Servaes
    Nov 20 at 16:26










1




1




I don't understand the second part, $3^{13}equiv23pmod{25}$, is there any property that I'm missing in modular arithmetics? What I conclude from the first expression is that $3^{13}$ divided by 25 could be any number from {3, 8, 13, 18, 23}
– Sean Goudarzi
Nov 20 at 16:07






I don't understand the second part, $3^{13}equiv23pmod{25}$, is there any property that I'm missing in modular arithmetics? What I conclude from the first expression is that $3^{13}$ divided by 25 could be any number from {3, 8, 13, 18, 23}
– Sean Goudarzi
Nov 20 at 16:07














I was assuming the work you showed in your question already; let me expand my answer to clarify this.
– Servaes
Nov 20 at 16:26






I was assuming the work you showed in your question already; let me expand my answer to clarify this.
– Servaes
Nov 20 at 16:26




















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