What's the remainder if $99{,}999^{99}$ is divided by $999{,}999$?











up vote
4
down vote

favorite
1












What's the remainder if $99{,}999^{99}$ is divided by $999{,}999$?



Is there any formula or trick method to achieve this?
Also kindly ignore the improper use of tag as i don't know which tag to choose










share|cite|improve this question
























  • Hint: relabel $999999$ as $x$. What's the remainder after division?
    – John Douma
    Nov 21 at 20:58








  • 1




    Computer says 123579, but I don't see elegant trick besides laboriously doing square-and-multiply
    – Hagen von Eitzen
    Nov 21 at 21:07










  • @JohnDouma: I don't get it.
    – TonyK
    Nov 21 at 23:21










  • @Hagen Your hardware is consistent with my wetware - see my answer.
    – Bill Dubuque
    Nov 21 at 23:34










  • @TonyK I misread the question. I thought both numbers consisted of six $9$s.
    – John Douma
    Nov 21 at 23:39















up vote
4
down vote

favorite
1












What's the remainder if $99{,}999^{99}$ is divided by $999{,}999$?



Is there any formula or trick method to achieve this?
Also kindly ignore the improper use of tag as i don't know which tag to choose










share|cite|improve this question
























  • Hint: relabel $999999$ as $x$. What's the remainder after division?
    – John Douma
    Nov 21 at 20:58








  • 1




    Computer says 123579, but I don't see elegant trick besides laboriously doing square-and-multiply
    – Hagen von Eitzen
    Nov 21 at 21:07










  • @JohnDouma: I don't get it.
    – TonyK
    Nov 21 at 23:21










  • @Hagen Your hardware is consistent with my wetware - see my answer.
    – Bill Dubuque
    Nov 21 at 23:34










  • @TonyK I misread the question. I thought both numbers consisted of six $9$s.
    – John Douma
    Nov 21 at 23:39













up vote
4
down vote

favorite
1









up vote
4
down vote

favorite
1






1





What's the remainder if $99{,}999^{99}$ is divided by $999{,}999$?



Is there any formula or trick method to achieve this?
Also kindly ignore the improper use of tag as i don't know which tag to choose










share|cite|improve this question















What's the remainder if $99{,}999^{99}$ is divided by $999{,}999$?



Is there any formula or trick method to achieve this?
Also kindly ignore the improper use of tag as i don't know which tag to choose







elementary-number-theory modular-arithmetic






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 22 at 0:01









Barry Cipra

58.6k653122




58.6k653122










asked Nov 21 at 20:56









Girish venkata

212




212












  • Hint: relabel $999999$ as $x$. What's the remainder after division?
    – John Douma
    Nov 21 at 20:58








  • 1




    Computer says 123579, but I don't see elegant trick besides laboriously doing square-and-multiply
    – Hagen von Eitzen
    Nov 21 at 21:07










  • @JohnDouma: I don't get it.
    – TonyK
    Nov 21 at 23:21










  • @Hagen Your hardware is consistent with my wetware - see my answer.
    – Bill Dubuque
    Nov 21 at 23:34










  • @TonyK I misread the question. I thought both numbers consisted of six $9$s.
    – John Douma
    Nov 21 at 23:39


















  • Hint: relabel $999999$ as $x$. What's the remainder after division?
    – John Douma
    Nov 21 at 20:58








  • 1




    Computer says 123579, but I don't see elegant trick besides laboriously doing square-and-multiply
    – Hagen von Eitzen
    Nov 21 at 21:07










  • @JohnDouma: I don't get it.
    – TonyK
    Nov 21 at 23:21










  • @Hagen Your hardware is consistent with my wetware - see my answer.
    – Bill Dubuque
    Nov 21 at 23:34










  • @TonyK I misread the question. I thought both numbers consisted of six $9$s.
    – John Douma
    Nov 21 at 23:39
















Hint: relabel $999999$ as $x$. What's the remainder after division?
– John Douma
Nov 21 at 20:58






Hint: relabel $999999$ as $x$. What's the remainder after division?
– John Douma
Nov 21 at 20:58






1




1




Computer says 123579, but I don't see elegant trick besides laboriously doing square-and-multiply
– Hagen von Eitzen
Nov 21 at 21:07




Computer says 123579, but I don't see elegant trick besides laboriously doing square-and-multiply
– Hagen von Eitzen
Nov 21 at 21:07












@JohnDouma: I don't get it.
– TonyK
Nov 21 at 23:21




@JohnDouma: I don't get it.
– TonyK
Nov 21 at 23:21












@Hagen Your hardware is consistent with my wetware - see my answer.
– Bill Dubuque
Nov 21 at 23:34




@Hagen Your hardware is consistent with my wetware - see my answer.
– Bill Dubuque
Nov 21 at 23:34












@TonyK I misread the question. I thought both numbers consisted of six $9$s.
– John Douma
Nov 21 at 23:39




@TonyK I misread the question. I thought both numbers consisted of six $9$s.
– John Douma
Nov 21 at 23:39










2 Answers
2






active

oldest

votes

















up vote
2
down vote













$!bmod 999999!:, 10cdot99999equiv -9 $ so $ 99999 equiv -9/10$



Thus $,99999^{large99}equiv -9^{large 99}/10^{large 99}equiv -9^{large 99}/10^{large 3}equiv -10^{large 3}cdot 9^{large 99} $ via $ 10^{large
6}equiv 1$



$n := 999999/27 = 7cdot 11cdot 13cdot 37.,$ mod each $p,,$ $9,$ has order $,3,5,3,9,$ so $,9^{large color{#c00}{45}}equiv 1pmod{!n}$



so $ 9^{large 99}!bmod 999999 = 27(3cdot 9^{large color{#c00}{97}}!bmod n) = 27(3cdot 9^{large 7}!bmod n) equiv 9^{large 9}!pmod{!999999}$



Hence we conclude $ 99999^{large 99}equiv -10^{large 3}cdot 9^{large 99}equiv {-}10^{large 3}cdot 9^{large 9}equiv 123579,pmod{!999999}$






share|cite|improve this answer























  • Due to $999999$ isn't a prime number you should say that $10$ has an inverse because it's coprime with $999999$.
    – P De Donato
    Nov 21 at 22:51










  • Then you can't arbitrarly pass from mod $999999$ to mod $n$.
    – P De Donato
    Nov 21 at 22:52










  • He's using the Chinese remainder theorem there. It's clear that $99999^{99} equiv 0 bmod 27$.
    – Qiaochu Yuan
    Nov 21 at 22:53










  • @PDeDonato That's obvious: $bmod 10^6-1!:, 10cdot 10^5equiv 1 $ so $10^{-1}equiv 10^5 $
    – Bill Dubuque
    Nov 21 at 22:54












  • I know it, but the answer isn't so clear.
    – P De Donato
    Nov 21 at 22:56


















up vote
2
down vote













We work in the ring $R=Bbb Z/N$, with operation modulo $N=999999=10^6-1$.
(Equalities below address computations in $R$.)
Then
$$
begin{aligned}
99999^{99}
&=(99999-999999)^{99} =(-900000)^{99}=-9^{99}cdot (10^5)^{99}\
&=-3^{198}cdot 10^{495}=-3^{198}cdot {underbrace{(10^6)}_{=1}}^{82}cdot 10^3=-3^{198}cdot 1000 .
end{aligned}
$$

Now the order of (the unit) $3$ in the ring



$Bbb Z/37037
=Bbb Z/(7cdot 11cdot 13cdot 37)
cong
(Bbb Z/7)times
(Bbb Z/11)times
(Bbb Z/13)times
(Bbb Z/37)
$



is $90$, but we need only the simpler information that $3^{180}$ is one modulo $37037$. This is so because $180$ is a multiple of $(7-1)$, $(11-1)$, $(13-1)$, and $(37-1)$. So instead of $3^{198}$ we write the smaller power $3^{18}=387420489$.



We finally search for a number which is $0$ modulo $27$, and also $-387420489000$ modulo $37037$, which is $12468$. After rearrangements modulo $3,9,27$ we get the result
$$123579 .$$





Note: A computer algebra system like sage delivers immediately



sage: Zmod(999999)(99999)^99
123579





share|cite|improve this answer





















  • This is essentially the same as my answer, except I used the mod distributive law form of CRT (which makes it simpler).
    – Bill Dubuque
    Nov 22 at 0:45












  • I committed it now after typing in the train... Yes, same computation.
    – dan_fulea
    Nov 22 at 0:50










  • Thank you very much, it took some time but i eventually understood
    – Girish venkata
    Nov 26 at 18:12











Your Answer





StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3008349%2fwhats-the-remainder-if-99-99999-is-divided-by-999-999%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
2
down vote













$!bmod 999999!:, 10cdot99999equiv -9 $ so $ 99999 equiv -9/10$



Thus $,99999^{large99}equiv -9^{large 99}/10^{large 99}equiv -9^{large 99}/10^{large 3}equiv -10^{large 3}cdot 9^{large 99} $ via $ 10^{large
6}equiv 1$



$n := 999999/27 = 7cdot 11cdot 13cdot 37.,$ mod each $p,,$ $9,$ has order $,3,5,3,9,$ so $,9^{large color{#c00}{45}}equiv 1pmod{!n}$



so $ 9^{large 99}!bmod 999999 = 27(3cdot 9^{large color{#c00}{97}}!bmod n) = 27(3cdot 9^{large 7}!bmod n) equiv 9^{large 9}!pmod{!999999}$



Hence we conclude $ 99999^{large 99}equiv -10^{large 3}cdot 9^{large 99}equiv {-}10^{large 3}cdot 9^{large 9}equiv 123579,pmod{!999999}$






share|cite|improve this answer























  • Due to $999999$ isn't a prime number you should say that $10$ has an inverse because it's coprime with $999999$.
    – P De Donato
    Nov 21 at 22:51










  • Then you can't arbitrarly pass from mod $999999$ to mod $n$.
    – P De Donato
    Nov 21 at 22:52










  • He's using the Chinese remainder theorem there. It's clear that $99999^{99} equiv 0 bmod 27$.
    – Qiaochu Yuan
    Nov 21 at 22:53










  • @PDeDonato That's obvious: $bmod 10^6-1!:, 10cdot 10^5equiv 1 $ so $10^{-1}equiv 10^5 $
    – Bill Dubuque
    Nov 21 at 22:54












  • I know it, but the answer isn't so clear.
    – P De Donato
    Nov 21 at 22:56















up vote
2
down vote













$!bmod 999999!:, 10cdot99999equiv -9 $ so $ 99999 equiv -9/10$



Thus $,99999^{large99}equiv -9^{large 99}/10^{large 99}equiv -9^{large 99}/10^{large 3}equiv -10^{large 3}cdot 9^{large 99} $ via $ 10^{large
6}equiv 1$



$n := 999999/27 = 7cdot 11cdot 13cdot 37.,$ mod each $p,,$ $9,$ has order $,3,5,3,9,$ so $,9^{large color{#c00}{45}}equiv 1pmod{!n}$



so $ 9^{large 99}!bmod 999999 = 27(3cdot 9^{large color{#c00}{97}}!bmod n) = 27(3cdot 9^{large 7}!bmod n) equiv 9^{large 9}!pmod{!999999}$



Hence we conclude $ 99999^{large 99}equiv -10^{large 3}cdot 9^{large 99}equiv {-}10^{large 3}cdot 9^{large 9}equiv 123579,pmod{!999999}$






share|cite|improve this answer























  • Due to $999999$ isn't a prime number you should say that $10$ has an inverse because it's coprime with $999999$.
    – P De Donato
    Nov 21 at 22:51










  • Then you can't arbitrarly pass from mod $999999$ to mod $n$.
    – P De Donato
    Nov 21 at 22:52










  • He's using the Chinese remainder theorem there. It's clear that $99999^{99} equiv 0 bmod 27$.
    – Qiaochu Yuan
    Nov 21 at 22:53










  • @PDeDonato That's obvious: $bmod 10^6-1!:, 10cdot 10^5equiv 1 $ so $10^{-1}equiv 10^5 $
    – Bill Dubuque
    Nov 21 at 22:54












  • I know it, but the answer isn't so clear.
    – P De Donato
    Nov 21 at 22:56













up vote
2
down vote










up vote
2
down vote









$!bmod 999999!:, 10cdot99999equiv -9 $ so $ 99999 equiv -9/10$



Thus $,99999^{large99}equiv -9^{large 99}/10^{large 99}equiv -9^{large 99}/10^{large 3}equiv -10^{large 3}cdot 9^{large 99} $ via $ 10^{large
6}equiv 1$



$n := 999999/27 = 7cdot 11cdot 13cdot 37.,$ mod each $p,,$ $9,$ has order $,3,5,3,9,$ so $,9^{large color{#c00}{45}}equiv 1pmod{!n}$



so $ 9^{large 99}!bmod 999999 = 27(3cdot 9^{large color{#c00}{97}}!bmod n) = 27(3cdot 9^{large 7}!bmod n) equiv 9^{large 9}!pmod{!999999}$



Hence we conclude $ 99999^{large 99}equiv -10^{large 3}cdot 9^{large 99}equiv {-}10^{large 3}cdot 9^{large 9}equiv 123579,pmod{!999999}$






share|cite|improve this answer














$!bmod 999999!:, 10cdot99999equiv -9 $ so $ 99999 equiv -9/10$



Thus $,99999^{large99}equiv -9^{large 99}/10^{large 99}equiv -9^{large 99}/10^{large 3}equiv -10^{large 3}cdot 9^{large 99} $ via $ 10^{large
6}equiv 1$



$n := 999999/27 = 7cdot 11cdot 13cdot 37.,$ mod each $p,,$ $9,$ has order $,3,5,3,9,$ so $,9^{large color{#c00}{45}}equiv 1pmod{!n}$



so $ 9^{large 99}!bmod 999999 = 27(3cdot 9^{large color{#c00}{97}}!bmod n) = 27(3cdot 9^{large 7}!bmod n) equiv 9^{large 9}!pmod{!999999}$



Hence we conclude $ 99999^{large 99}equiv -10^{large 3}cdot 9^{large 99}equiv {-}10^{large 3}cdot 9^{large 9}equiv 123579,pmod{!999999}$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Nov 22 at 0:39

























answered Nov 21 at 22:35









Bill Dubuque

207k29189624




207k29189624












  • Due to $999999$ isn't a prime number you should say that $10$ has an inverse because it's coprime with $999999$.
    – P De Donato
    Nov 21 at 22:51










  • Then you can't arbitrarly pass from mod $999999$ to mod $n$.
    – P De Donato
    Nov 21 at 22:52










  • He's using the Chinese remainder theorem there. It's clear that $99999^{99} equiv 0 bmod 27$.
    – Qiaochu Yuan
    Nov 21 at 22:53










  • @PDeDonato That's obvious: $bmod 10^6-1!:, 10cdot 10^5equiv 1 $ so $10^{-1}equiv 10^5 $
    – Bill Dubuque
    Nov 21 at 22:54












  • I know it, but the answer isn't so clear.
    – P De Donato
    Nov 21 at 22:56


















  • Due to $999999$ isn't a prime number you should say that $10$ has an inverse because it's coprime with $999999$.
    – P De Donato
    Nov 21 at 22:51










  • Then you can't arbitrarly pass from mod $999999$ to mod $n$.
    – P De Donato
    Nov 21 at 22:52










  • He's using the Chinese remainder theorem there. It's clear that $99999^{99} equiv 0 bmod 27$.
    – Qiaochu Yuan
    Nov 21 at 22:53










  • @PDeDonato That's obvious: $bmod 10^6-1!:, 10cdot 10^5equiv 1 $ so $10^{-1}equiv 10^5 $
    – Bill Dubuque
    Nov 21 at 22:54












  • I know it, but the answer isn't so clear.
    – P De Donato
    Nov 21 at 22:56
















Due to $999999$ isn't a prime number you should say that $10$ has an inverse because it's coprime with $999999$.
– P De Donato
Nov 21 at 22:51




Due to $999999$ isn't a prime number you should say that $10$ has an inverse because it's coprime with $999999$.
– P De Donato
Nov 21 at 22:51












Then you can't arbitrarly pass from mod $999999$ to mod $n$.
– P De Donato
Nov 21 at 22:52




Then you can't arbitrarly pass from mod $999999$ to mod $n$.
– P De Donato
Nov 21 at 22:52












He's using the Chinese remainder theorem there. It's clear that $99999^{99} equiv 0 bmod 27$.
– Qiaochu Yuan
Nov 21 at 22:53




He's using the Chinese remainder theorem there. It's clear that $99999^{99} equiv 0 bmod 27$.
– Qiaochu Yuan
Nov 21 at 22:53












@PDeDonato That's obvious: $bmod 10^6-1!:, 10cdot 10^5equiv 1 $ so $10^{-1}equiv 10^5 $
– Bill Dubuque
Nov 21 at 22:54






@PDeDonato That's obvious: $bmod 10^6-1!:, 10cdot 10^5equiv 1 $ so $10^{-1}equiv 10^5 $
– Bill Dubuque
Nov 21 at 22:54














I know it, but the answer isn't so clear.
– P De Donato
Nov 21 at 22:56




I know it, but the answer isn't so clear.
– P De Donato
Nov 21 at 22:56










up vote
2
down vote













We work in the ring $R=Bbb Z/N$, with operation modulo $N=999999=10^6-1$.
(Equalities below address computations in $R$.)
Then
$$
begin{aligned}
99999^{99}
&=(99999-999999)^{99} =(-900000)^{99}=-9^{99}cdot (10^5)^{99}\
&=-3^{198}cdot 10^{495}=-3^{198}cdot {underbrace{(10^6)}_{=1}}^{82}cdot 10^3=-3^{198}cdot 1000 .
end{aligned}
$$

Now the order of (the unit) $3$ in the ring



$Bbb Z/37037
=Bbb Z/(7cdot 11cdot 13cdot 37)
cong
(Bbb Z/7)times
(Bbb Z/11)times
(Bbb Z/13)times
(Bbb Z/37)
$



is $90$, but we need only the simpler information that $3^{180}$ is one modulo $37037$. This is so because $180$ is a multiple of $(7-1)$, $(11-1)$, $(13-1)$, and $(37-1)$. So instead of $3^{198}$ we write the smaller power $3^{18}=387420489$.



We finally search for a number which is $0$ modulo $27$, and also $-387420489000$ modulo $37037$, which is $12468$. After rearrangements modulo $3,9,27$ we get the result
$$123579 .$$





Note: A computer algebra system like sage delivers immediately



sage: Zmod(999999)(99999)^99
123579





share|cite|improve this answer





















  • This is essentially the same as my answer, except I used the mod distributive law form of CRT (which makes it simpler).
    – Bill Dubuque
    Nov 22 at 0:45












  • I committed it now after typing in the train... Yes, same computation.
    – dan_fulea
    Nov 22 at 0:50










  • Thank you very much, it took some time but i eventually understood
    – Girish venkata
    Nov 26 at 18:12















up vote
2
down vote













We work in the ring $R=Bbb Z/N$, with operation modulo $N=999999=10^6-1$.
(Equalities below address computations in $R$.)
Then
$$
begin{aligned}
99999^{99}
&=(99999-999999)^{99} =(-900000)^{99}=-9^{99}cdot (10^5)^{99}\
&=-3^{198}cdot 10^{495}=-3^{198}cdot {underbrace{(10^6)}_{=1}}^{82}cdot 10^3=-3^{198}cdot 1000 .
end{aligned}
$$

Now the order of (the unit) $3$ in the ring



$Bbb Z/37037
=Bbb Z/(7cdot 11cdot 13cdot 37)
cong
(Bbb Z/7)times
(Bbb Z/11)times
(Bbb Z/13)times
(Bbb Z/37)
$



is $90$, but we need only the simpler information that $3^{180}$ is one modulo $37037$. This is so because $180$ is a multiple of $(7-1)$, $(11-1)$, $(13-1)$, and $(37-1)$. So instead of $3^{198}$ we write the smaller power $3^{18}=387420489$.



We finally search for a number which is $0$ modulo $27$, and also $-387420489000$ modulo $37037$, which is $12468$. After rearrangements modulo $3,9,27$ we get the result
$$123579 .$$





Note: A computer algebra system like sage delivers immediately



sage: Zmod(999999)(99999)^99
123579





share|cite|improve this answer





















  • This is essentially the same as my answer, except I used the mod distributive law form of CRT (which makes it simpler).
    – Bill Dubuque
    Nov 22 at 0:45












  • I committed it now after typing in the train... Yes, same computation.
    – dan_fulea
    Nov 22 at 0:50










  • Thank you very much, it took some time but i eventually understood
    – Girish venkata
    Nov 26 at 18:12













up vote
2
down vote










up vote
2
down vote









We work in the ring $R=Bbb Z/N$, with operation modulo $N=999999=10^6-1$.
(Equalities below address computations in $R$.)
Then
$$
begin{aligned}
99999^{99}
&=(99999-999999)^{99} =(-900000)^{99}=-9^{99}cdot (10^5)^{99}\
&=-3^{198}cdot 10^{495}=-3^{198}cdot {underbrace{(10^6)}_{=1}}^{82}cdot 10^3=-3^{198}cdot 1000 .
end{aligned}
$$

Now the order of (the unit) $3$ in the ring



$Bbb Z/37037
=Bbb Z/(7cdot 11cdot 13cdot 37)
cong
(Bbb Z/7)times
(Bbb Z/11)times
(Bbb Z/13)times
(Bbb Z/37)
$



is $90$, but we need only the simpler information that $3^{180}$ is one modulo $37037$. This is so because $180$ is a multiple of $(7-1)$, $(11-1)$, $(13-1)$, and $(37-1)$. So instead of $3^{198}$ we write the smaller power $3^{18}=387420489$.



We finally search for a number which is $0$ modulo $27$, and also $-387420489000$ modulo $37037$, which is $12468$. After rearrangements modulo $3,9,27$ we get the result
$$123579 .$$





Note: A computer algebra system like sage delivers immediately



sage: Zmod(999999)(99999)^99
123579





share|cite|improve this answer












We work in the ring $R=Bbb Z/N$, with operation modulo $N=999999=10^6-1$.
(Equalities below address computations in $R$.)
Then
$$
begin{aligned}
99999^{99}
&=(99999-999999)^{99} =(-900000)^{99}=-9^{99}cdot (10^5)^{99}\
&=-3^{198}cdot 10^{495}=-3^{198}cdot {underbrace{(10^6)}_{=1}}^{82}cdot 10^3=-3^{198}cdot 1000 .
end{aligned}
$$

Now the order of (the unit) $3$ in the ring



$Bbb Z/37037
=Bbb Z/(7cdot 11cdot 13cdot 37)
cong
(Bbb Z/7)times
(Bbb Z/11)times
(Bbb Z/13)times
(Bbb Z/37)
$



is $90$, but we need only the simpler information that $3^{180}$ is one modulo $37037$. This is so because $180$ is a multiple of $(7-1)$, $(11-1)$, $(13-1)$, and $(37-1)$. So instead of $3^{198}$ we write the smaller power $3^{18}=387420489$.



We finally search for a number which is $0$ modulo $27$, and also $-387420489000$ modulo $37037$, which is $12468$. After rearrangements modulo $3,9,27$ we get the result
$$123579 .$$





Note: A computer algebra system like sage delivers immediately



sage: Zmod(999999)(99999)^99
123579






share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Nov 22 at 0:42









dan_fulea

6,2101312




6,2101312












  • This is essentially the same as my answer, except I used the mod distributive law form of CRT (which makes it simpler).
    – Bill Dubuque
    Nov 22 at 0:45












  • I committed it now after typing in the train... Yes, same computation.
    – dan_fulea
    Nov 22 at 0:50










  • Thank you very much, it took some time but i eventually understood
    – Girish venkata
    Nov 26 at 18:12


















  • This is essentially the same as my answer, except I used the mod distributive law form of CRT (which makes it simpler).
    – Bill Dubuque
    Nov 22 at 0:45












  • I committed it now after typing in the train... Yes, same computation.
    – dan_fulea
    Nov 22 at 0:50










  • Thank you very much, it took some time but i eventually understood
    – Girish venkata
    Nov 26 at 18:12
















This is essentially the same as my answer, except I used the mod distributive law form of CRT (which makes it simpler).
– Bill Dubuque
Nov 22 at 0:45






This is essentially the same as my answer, except I used the mod distributive law form of CRT (which makes it simpler).
– Bill Dubuque
Nov 22 at 0:45














I committed it now after typing in the train... Yes, same computation.
– dan_fulea
Nov 22 at 0:50




I committed it now after typing in the train... Yes, same computation.
– dan_fulea
Nov 22 at 0:50












Thank you very much, it took some time but i eventually understood
– Girish venkata
Nov 26 at 18:12




Thank you very much, it took some time but i eventually understood
– Girish venkata
Nov 26 at 18:12


















draft saved

draft discarded




















































Thanks for contributing an answer to Mathematics Stack Exchange!


  • Please be sure to answer the question. Provide details and share your research!

But avoid



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


Use MathJax to format equations. MathJax reference.


To learn more, see our tips on writing great answers.





Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


Please pay close attention to the following guidance:


  • Please be sure to answer the question. Provide details and share your research!

But avoid



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


To learn more, see our tips on writing great answers.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3008349%2fwhats-the-remainder-if-99-99999-is-divided-by-999-999%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown





















































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown

































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown







Popular posts from this blog

Probability when a professor distributes a quiz and homework assignment to a class of n students.

Aardman Animations

Are they similar matrix