What is the remainder when $ n^n $ is divided by $ (n-1) $?












1












$begingroup$


I want to find out the remainder if $ n^n $ is divided by $(n-1)$ . I have tried investigating a number of examples for small $n$:





  • $ 2^2 $ is divided by $1$ gives remainder $0 $


  • $ 3^3 $ is divided by $2$ gives remainder $1$


  • $ 4^4 $ is divided by $3$ gives remainder $1$


  • $ 5^5 $ is divided by $4$ gives remainder $1$


  • $ 6^6 $ is divided by $5$ gives remainder $1$


But I can't identify any pattern to find out the remainder. So please help me to find out the remainder if $ n^n $ is divided by $ n-1 $.






elementary-number-theory







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    $4^4$ divided by $3$ gives $1$. The answer is always $1$
    $endgroup$
    – Mohammad Zuhair Khan
    Dec 30 '18 at 9:58










  • $begingroup$
    That is a huge mistake . Many many thanks
    $endgroup$
    – Standard Equation
    Dec 30 '18 at 10:00


















1












$begingroup$


I want to find out the remainder if $ n^n $ is divided by $(n-1)$ . I have tried investigating a number of examples for small $n$:





  • $ 2^2 $ is divided by $1$ gives remainder $0 $


  • $ 3^3 $ is divided by $2$ gives remainder $1$


  • $ 4^4 $ is divided by $3$ gives remainder $1$


  • $ 5^5 $ is divided by $4$ gives remainder $1$


  • $ 6^6 $ is divided by $5$ gives remainder $1$


But I can't identify any pattern to find out the remainder. So please help me to find out the remainder if $ n^n $ is divided by $ n-1 $.






elementary-number-theory







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    $4^4$ divided by $3$ gives $1$. The answer is always $1$
    $endgroup$
    – Mohammad Zuhair Khan
    Dec 30 '18 at 9:58










  • $begingroup$
    That is a huge mistake . Many many thanks
    $endgroup$
    – Standard Equation
    Dec 30 '18 at 10:00
















1












1








1





$begingroup$


I want to find out the remainder if $ n^n $ is divided by $(n-1)$ . I have tried investigating a number of examples for small $n$:





  • $ 2^2 $ is divided by $1$ gives remainder $0 $


  • $ 3^3 $ is divided by $2$ gives remainder $1$


  • $ 4^4 $ is divided by $3$ gives remainder $1$


  • $ 5^5 $ is divided by $4$ gives remainder $1$


  • $ 6^6 $ is divided by $5$ gives remainder $1$


But I can't identify any pattern to find out the remainder. So please help me to find out the remainder if $ n^n $ is divided by $ n-1 $.






elementary-number-theory







share|cite|improve this question











$endgroup$




I want to find out the remainder if $ n^n $ is divided by $(n-1)$ . I have tried investigating a number of examples for small $n$:





  • $ 2^2 $ is divided by $1$ gives remainder $0 $


  • $ 3^3 $ is divided by $2$ gives remainder $1$


  • $ 4^4 $ is divided by $3$ gives remainder $1$


  • $ 5^5 $ is divided by $4$ gives remainder $1$


  • $ 6^6 $ is divided by $5$ gives remainder $1$


But I can't identify any pattern to find out the remainder. So please help me to find out the remainder if $ n^n $ is divided by $ n-1 $.






elementary-number-theory




elementary-number-theory






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 30 '18 at 15:32









Bill Dubuque

214k29198660




214k29198660










asked Dec 30 '18 at 9:56









Standard EquationStandard Equation

247113




247113








  • 1




    $begingroup$
    $4^4$ divided by $3$ gives $1$. The answer is always $1$
    $endgroup$
    – Mohammad Zuhair Khan
    Dec 30 '18 at 9:58










  • $begingroup$
    That is a huge mistake . Many many thanks
    $endgroup$
    – Standard Equation
    Dec 30 '18 at 10:00
















  • 1




    $begingroup$
    $4^4$ divided by $3$ gives $1$. The answer is always $1$
    $endgroup$
    – Mohammad Zuhair Khan
    Dec 30 '18 at 9:58










  • $begingroup$
    That is a huge mistake . Many many thanks
    $endgroup$
    – Standard Equation
    Dec 30 '18 at 10:00










1




1




$begingroup$
$4^4$ divided by $3$ gives $1$. The answer is always $1$
$endgroup$
– Mohammad Zuhair Khan
Dec 30 '18 at 9:58




$begingroup$
$4^4$ divided by $3$ gives $1$. The answer is always $1$
$endgroup$
– Mohammad Zuhair Khan
Dec 30 '18 at 9:58












$begingroup$
That is a huge mistake . Many many thanks
$endgroup$
– Standard Equation
Dec 30 '18 at 10:00






$begingroup$
That is a huge mistake . Many many thanks
$endgroup$
– Standard Equation
Dec 30 '18 at 10:00












4 Answers
4






active

oldest

votes


















5












$begingroup$

The other answers are all correct. But since you posted such an elementary problem, I guess it would be better if you learnt how to attack this problem rather than only the solution of it.



See, you need to find out $n^n pmod{n-1}$ where $n>2$. Whenever you need to find mod of such products, the first thing to try is finding out the mod of individual terms (the others being some manipulation or Euler's/Fermat's little theorem if it's a power etc.). Now, in this case, it is evident that $n equiv (n-1)+1 equiv 1 pmod{n-1}$.



We know that, taking mod is multiplicative. So, $n^n equiv (1)^n pmod{n-1} equiv 1 pmod{n-1}$.



Please comment if you need further clarifications.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    I think it would have better if you used equiv $equiv$ instead of =
    $endgroup$
    – Mohammad Zuhair Khan
    Dec 30 '18 at 11:12










  • $begingroup$
    @MohammadZuhairKhan You're right. Edited.
    $endgroup$
    – SinTan1729
    Dec 30 '18 at 15:24



















7












$begingroup$

$$n^n equiv ((n-1)+1)^n equiv 1^n equiv 1 pmod {n-1}$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    It would probably more helpful to the OP if you edited in where this equation comes from. (I understand it, but OP need not necessarily do so, it's mostly just an equation completely out of context. >_>)
    $endgroup$
    – Eevee Trainer
    Dec 30 '18 at 10:01






  • 1




    $begingroup$
    I cant understand your solution . Can you please make elaboration of this solution ?
    $endgroup$
    – Standard Equation
    Dec 30 '18 at 10:04










  • $begingroup$
    @StandardEquation Do you know modular arithmetic / congruences? It helps to add such context to your questions. You accepted an answer using modular arithmetic, so how could you understand that but not this? (they are equivalent)
    $endgroup$
    – Bill Dubuque
    Dec 30 '18 at 15:43





















7












$begingroup$

$n^n = (n^{n-1} + n^{n-2} +ldots + 1)(n-1) + 1$. Its a telescope sum.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    It would probably more helpful to the OP if you edited in where this equation comes from. (I understand it, but OP need not necessarily do so, it's mostly just an equation completely out of context. >_>)
    $endgroup$
    – Eevee Trainer
    Dec 30 '18 at 10:01






  • 1




    $begingroup$
    This solution make sense to me . Many many thanks
    $endgroup$
    – Standard Equation
    Dec 30 '18 at 10:05



















4












$begingroup$

For n>2:



Binomial expansion:



$(1+(n-1))^n=$



$sum_{k=1}^{n}binom{n}{k}(n-1)^k+1;$



The first term is divisible by $(n-1)$, the remainder is $1$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Except when $,|n-1| = 1,$ when the remainder $,r= 0neq 1,,$ but $,0equiv 1pmod{!pm!1},,$ i.e. $, requiv 1pmod{n!-!1} $ is always true.
    $endgroup$
    – Bill Dubuque
    Dec 30 '18 at 15:37












  • $begingroup$
    Bill.Oops. Thanks for your comment. n>2. Shall fix it.
    $endgroup$
    – Peter Szilas
    Dec 30 '18 at 17:41












Your Answer








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',
autoActivateHeartbeat: false,
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%2f3056668%2fwhat-is-the-remainder-when-nn-is-divided-by-n-1%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























4 Answers
4






active

oldest

votes








4 Answers
4






active

oldest

votes









active

oldest

votes






active

oldest

votes









5












$begingroup$

The other answers are all correct. But since you posted such an elementary problem, I guess it would be better if you learnt how to attack this problem rather than only the solution of it.



See, you need to find out $n^n pmod{n-1}$ where $n>2$. Whenever you need to find mod of such products, the first thing to try is finding out the mod of individual terms (the others being some manipulation or Euler's/Fermat's little theorem if it's a power etc.). Now, in this case, it is evident that $n equiv (n-1)+1 equiv 1 pmod{n-1}$.



We know that, taking mod is multiplicative. So, $n^n equiv (1)^n pmod{n-1} equiv 1 pmod{n-1}$.



Please comment if you need further clarifications.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    I think it would have better if you used equiv $equiv$ instead of =
    $endgroup$
    – Mohammad Zuhair Khan
    Dec 30 '18 at 11:12










  • $begingroup$
    @MohammadZuhairKhan You're right. Edited.
    $endgroup$
    – SinTan1729
    Dec 30 '18 at 15:24
















5












$begingroup$

The other answers are all correct. But since you posted such an elementary problem, I guess it would be better if you learnt how to attack this problem rather than only the solution of it.



See, you need to find out $n^n pmod{n-1}$ where $n>2$. Whenever you need to find mod of such products, the first thing to try is finding out the mod of individual terms (the others being some manipulation or Euler's/Fermat's little theorem if it's a power etc.). Now, in this case, it is evident that $n equiv (n-1)+1 equiv 1 pmod{n-1}$.



We know that, taking mod is multiplicative. So, $n^n equiv (1)^n pmod{n-1} equiv 1 pmod{n-1}$.



Please comment if you need further clarifications.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    I think it would have better if you used equiv $equiv$ instead of =
    $endgroup$
    – Mohammad Zuhair Khan
    Dec 30 '18 at 11:12










  • $begingroup$
    @MohammadZuhairKhan You're right. Edited.
    $endgroup$
    – SinTan1729
    Dec 30 '18 at 15:24














5












5








5





$begingroup$

The other answers are all correct. But since you posted such an elementary problem, I guess it would be better if you learnt how to attack this problem rather than only the solution of it.



See, you need to find out $n^n pmod{n-1}$ where $n>2$. Whenever you need to find mod of such products, the first thing to try is finding out the mod of individual terms (the others being some manipulation or Euler's/Fermat's little theorem if it's a power etc.). Now, in this case, it is evident that $n equiv (n-1)+1 equiv 1 pmod{n-1}$.



We know that, taking mod is multiplicative. So, $n^n equiv (1)^n pmod{n-1} equiv 1 pmod{n-1}$.



Please comment if you need further clarifications.






share|cite|improve this answer











$endgroup$



The other answers are all correct. But since you posted such an elementary problem, I guess it would be better if you learnt how to attack this problem rather than only the solution of it.



See, you need to find out $n^n pmod{n-1}$ where $n>2$. Whenever you need to find mod of such products, the first thing to try is finding out the mod of individual terms (the others being some manipulation or Euler's/Fermat's little theorem if it's a power etc.). Now, in this case, it is evident that $n equiv (n-1)+1 equiv 1 pmod{n-1}$.



We know that, taking mod is multiplicative. So, $n^n equiv (1)^n pmod{n-1} equiv 1 pmod{n-1}$.



Please comment if you need further clarifications.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 30 '18 at 15:24

























answered Dec 30 '18 at 10:10









SinTan1729SinTan1729

2,689724




2,689724








  • 1




    $begingroup$
    I think it would have better if you used equiv $equiv$ instead of =
    $endgroup$
    – Mohammad Zuhair Khan
    Dec 30 '18 at 11:12










  • $begingroup$
    @MohammadZuhairKhan You're right. Edited.
    $endgroup$
    – SinTan1729
    Dec 30 '18 at 15:24














  • 1




    $begingroup$
    I think it would have better if you used equiv $equiv$ instead of =
    $endgroup$
    – Mohammad Zuhair Khan
    Dec 30 '18 at 11:12










  • $begingroup$
    @MohammadZuhairKhan You're right. Edited.
    $endgroup$
    – SinTan1729
    Dec 30 '18 at 15:24








1




1




$begingroup$
I think it would have better if you used equiv $equiv$ instead of =
$endgroup$
– Mohammad Zuhair Khan
Dec 30 '18 at 11:12




$begingroup$
I think it would have better if you used equiv $equiv$ instead of =
$endgroup$
– Mohammad Zuhair Khan
Dec 30 '18 at 11:12












$begingroup$
@MohammadZuhairKhan You're right. Edited.
$endgroup$
– SinTan1729
Dec 30 '18 at 15:24




$begingroup$
@MohammadZuhairKhan You're right. Edited.
$endgroup$
– SinTan1729
Dec 30 '18 at 15:24











7












$begingroup$

$$n^n equiv ((n-1)+1)^n equiv 1^n equiv 1 pmod {n-1}$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    It would probably more helpful to the OP if you edited in where this equation comes from. (I understand it, but OP need not necessarily do so, it's mostly just an equation completely out of context. >_>)
    $endgroup$
    – Eevee Trainer
    Dec 30 '18 at 10:01






  • 1




    $begingroup$
    I cant understand your solution . Can you please make elaboration of this solution ?
    $endgroup$
    – Standard Equation
    Dec 30 '18 at 10:04










  • $begingroup$
    @StandardEquation Do you know modular arithmetic / congruences? It helps to add such context to your questions. You accepted an answer using modular arithmetic, so how could you understand that but not this? (they are equivalent)
    $endgroup$
    – Bill Dubuque
    Dec 30 '18 at 15:43


















7












$begingroup$

$$n^n equiv ((n-1)+1)^n equiv 1^n equiv 1 pmod {n-1}$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    It would probably more helpful to the OP if you edited in where this equation comes from. (I understand it, but OP need not necessarily do so, it's mostly just an equation completely out of context. >_>)
    $endgroup$
    – Eevee Trainer
    Dec 30 '18 at 10:01






  • 1




    $begingroup$
    I cant understand your solution . Can you please make elaboration of this solution ?
    $endgroup$
    – Standard Equation
    Dec 30 '18 at 10:04










  • $begingroup$
    @StandardEquation Do you know modular arithmetic / congruences? It helps to add such context to your questions. You accepted an answer using modular arithmetic, so how could you understand that but not this? (they are equivalent)
    $endgroup$
    – Bill Dubuque
    Dec 30 '18 at 15:43
















7












7








7





$begingroup$

$$n^n equiv ((n-1)+1)^n equiv 1^n equiv 1 pmod {n-1}$$






share|cite|improve this answer









$endgroup$



$$n^n equiv ((n-1)+1)^n equiv 1^n equiv 1 pmod {n-1}$$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 30 '18 at 9:58









Kenny LauKenny Lau

20.1k2260




20.1k2260












  • $begingroup$
    It would probably more helpful to the OP if you edited in where this equation comes from. (I understand it, but OP need not necessarily do so, it's mostly just an equation completely out of context. >_>)
    $endgroup$
    – Eevee Trainer
    Dec 30 '18 at 10:01






  • 1




    $begingroup$
    I cant understand your solution . Can you please make elaboration of this solution ?
    $endgroup$
    – Standard Equation
    Dec 30 '18 at 10:04










  • $begingroup$
    @StandardEquation Do you know modular arithmetic / congruences? It helps to add such context to your questions. You accepted an answer using modular arithmetic, so how could you understand that but not this? (they are equivalent)
    $endgroup$
    – Bill Dubuque
    Dec 30 '18 at 15:43




















  • $begingroup$
    It would probably more helpful to the OP if you edited in where this equation comes from. (I understand it, but OP need not necessarily do so, it's mostly just an equation completely out of context. >_>)
    $endgroup$
    – Eevee Trainer
    Dec 30 '18 at 10:01






  • 1




    $begingroup$
    I cant understand your solution . Can you please make elaboration of this solution ?
    $endgroup$
    – Standard Equation
    Dec 30 '18 at 10:04










  • $begingroup$
    @StandardEquation Do you know modular arithmetic / congruences? It helps to add such context to your questions. You accepted an answer using modular arithmetic, so how could you understand that but not this? (they are equivalent)
    $endgroup$
    – Bill Dubuque
    Dec 30 '18 at 15:43


















$begingroup$
It would probably more helpful to the OP if you edited in where this equation comes from. (I understand it, but OP need not necessarily do so, it's mostly just an equation completely out of context. >_>)
$endgroup$
– Eevee Trainer
Dec 30 '18 at 10:01




$begingroup$
It would probably more helpful to the OP if you edited in where this equation comes from. (I understand it, but OP need not necessarily do so, it's mostly just an equation completely out of context. >_>)
$endgroup$
– Eevee Trainer
Dec 30 '18 at 10:01




1




1




$begingroup$
I cant understand your solution . Can you please make elaboration of this solution ?
$endgroup$
– Standard Equation
Dec 30 '18 at 10:04




$begingroup$
I cant understand your solution . Can you please make elaboration of this solution ?
$endgroup$
– Standard Equation
Dec 30 '18 at 10:04












$begingroup$
@StandardEquation Do you know modular arithmetic / congruences? It helps to add such context to your questions. You accepted an answer using modular arithmetic, so how could you understand that but not this? (they are equivalent)
$endgroup$
– Bill Dubuque
Dec 30 '18 at 15:43






$begingroup$
@StandardEquation Do you know modular arithmetic / congruences? It helps to add such context to your questions. You accepted an answer using modular arithmetic, so how could you understand that but not this? (they are equivalent)
$endgroup$
– Bill Dubuque
Dec 30 '18 at 15:43













7












$begingroup$

$n^n = (n^{n-1} + n^{n-2} +ldots + 1)(n-1) + 1$. Its a telescope sum.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    It would probably more helpful to the OP if you edited in where this equation comes from. (I understand it, but OP need not necessarily do so, it's mostly just an equation completely out of context. >_>)
    $endgroup$
    – Eevee Trainer
    Dec 30 '18 at 10:01






  • 1




    $begingroup$
    This solution make sense to me . Many many thanks
    $endgroup$
    – Standard Equation
    Dec 30 '18 at 10:05
















7












$begingroup$

$n^n = (n^{n-1} + n^{n-2} +ldots + 1)(n-1) + 1$. Its a telescope sum.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    It would probably more helpful to the OP if you edited in where this equation comes from. (I understand it, but OP need not necessarily do so, it's mostly just an equation completely out of context. >_>)
    $endgroup$
    – Eevee Trainer
    Dec 30 '18 at 10:01






  • 1




    $begingroup$
    This solution make sense to me . Many many thanks
    $endgroup$
    – Standard Equation
    Dec 30 '18 at 10:05














7












7








7





$begingroup$

$n^n = (n^{n-1} + n^{n-2} +ldots + 1)(n-1) + 1$. Its a telescope sum.






share|cite|improve this answer











$endgroup$



$n^n = (n^{n-1} + n^{n-2} +ldots + 1)(n-1) + 1$. Its a telescope sum.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 30 '18 at 17:23

























answered Dec 30 '18 at 10:00









WuestenfuxWuestenfux

5,5911513




5,5911513








  • 1




    $begingroup$
    It would probably more helpful to the OP if you edited in where this equation comes from. (I understand it, but OP need not necessarily do so, it's mostly just an equation completely out of context. >_>)
    $endgroup$
    – Eevee Trainer
    Dec 30 '18 at 10:01






  • 1




    $begingroup$
    This solution make sense to me . Many many thanks
    $endgroup$
    – Standard Equation
    Dec 30 '18 at 10:05














  • 1




    $begingroup$
    It would probably more helpful to the OP if you edited in where this equation comes from. (I understand it, but OP need not necessarily do so, it's mostly just an equation completely out of context. >_>)
    $endgroup$
    – Eevee Trainer
    Dec 30 '18 at 10:01






  • 1




    $begingroup$
    This solution make sense to me . Many many thanks
    $endgroup$
    – Standard Equation
    Dec 30 '18 at 10:05








1




1




$begingroup$
It would probably more helpful to the OP if you edited in where this equation comes from. (I understand it, but OP need not necessarily do so, it's mostly just an equation completely out of context. >_>)
$endgroup$
– Eevee Trainer
Dec 30 '18 at 10:01




$begingroup$
It would probably more helpful to the OP if you edited in where this equation comes from. (I understand it, but OP need not necessarily do so, it's mostly just an equation completely out of context. >_>)
$endgroup$
– Eevee Trainer
Dec 30 '18 at 10:01




1




1




$begingroup$
This solution make sense to me . Many many thanks
$endgroup$
– Standard Equation
Dec 30 '18 at 10:05




$begingroup$
This solution make sense to me . Many many thanks
$endgroup$
– Standard Equation
Dec 30 '18 at 10:05











4












$begingroup$

For n>2:



Binomial expansion:



$(1+(n-1))^n=$



$sum_{k=1}^{n}binom{n}{k}(n-1)^k+1;$



The first term is divisible by $(n-1)$, the remainder is $1$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Except when $,|n-1| = 1,$ when the remainder $,r= 0neq 1,,$ but $,0equiv 1pmod{!pm!1},,$ i.e. $, requiv 1pmod{n!-!1} $ is always true.
    $endgroup$
    – Bill Dubuque
    Dec 30 '18 at 15:37












  • $begingroup$
    Bill.Oops. Thanks for your comment. n>2. Shall fix it.
    $endgroup$
    – Peter Szilas
    Dec 30 '18 at 17:41
















4












$begingroup$

For n>2:



Binomial expansion:



$(1+(n-1))^n=$



$sum_{k=1}^{n}binom{n}{k}(n-1)^k+1;$



The first term is divisible by $(n-1)$, the remainder is $1$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Except when $,|n-1| = 1,$ when the remainder $,r= 0neq 1,,$ but $,0equiv 1pmod{!pm!1},,$ i.e. $, requiv 1pmod{n!-!1} $ is always true.
    $endgroup$
    – Bill Dubuque
    Dec 30 '18 at 15:37












  • $begingroup$
    Bill.Oops. Thanks for your comment. n>2. Shall fix it.
    $endgroup$
    – Peter Szilas
    Dec 30 '18 at 17:41














4












4








4





$begingroup$

For n>2:



Binomial expansion:



$(1+(n-1))^n=$



$sum_{k=1}^{n}binom{n}{k}(n-1)^k+1;$



The first term is divisible by $(n-1)$, the remainder is $1$.






share|cite|improve this answer











$endgroup$



For n>2:



Binomial expansion:



$(1+(n-1))^n=$



$sum_{k=1}^{n}binom{n}{k}(n-1)^k+1;$



The first term is divisible by $(n-1)$, the remainder is $1$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 30 '18 at 17:42

























answered Dec 30 '18 at 10:06









Peter SzilasPeter Szilas

12k2822




12k2822












  • $begingroup$
    Except when $,|n-1| = 1,$ when the remainder $,r= 0neq 1,,$ but $,0equiv 1pmod{!pm!1},,$ i.e. $, requiv 1pmod{n!-!1} $ is always true.
    $endgroup$
    – Bill Dubuque
    Dec 30 '18 at 15:37












  • $begingroup$
    Bill.Oops. Thanks for your comment. n>2. Shall fix it.
    $endgroup$
    – Peter Szilas
    Dec 30 '18 at 17:41


















  • $begingroup$
    Except when $,|n-1| = 1,$ when the remainder $,r= 0neq 1,,$ but $,0equiv 1pmod{!pm!1},,$ i.e. $, requiv 1pmod{n!-!1} $ is always true.
    $endgroup$
    – Bill Dubuque
    Dec 30 '18 at 15:37












  • $begingroup$
    Bill.Oops. Thanks for your comment. n>2. Shall fix it.
    $endgroup$
    – Peter Szilas
    Dec 30 '18 at 17:41
















$begingroup$
Except when $,|n-1| = 1,$ when the remainder $,r= 0neq 1,,$ but $,0equiv 1pmod{!pm!1},,$ i.e. $, requiv 1pmod{n!-!1} $ is always true.
$endgroup$
– Bill Dubuque
Dec 30 '18 at 15:37






$begingroup$
Except when $,|n-1| = 1,$ when the remainder $,r= 0neq 1,,$ but $,0equiv 1pmod{!pm!1},,$ i.e. $, requiv 1pmod{n!-!1} $ is always true.
$endgroup$
– Bill Dubuque
Dec 30 '18 at 15:37














$begingroup$
Bill.Oops. Thanks for your comment. n>2. Shall fix it.
$endgroup$
– Peter Szilas
Dec 30 '18 at 17:41




$begingroup$
Bill.Oops. Thanks for your comment. n>2. Shall fix it.
$endgroup$
– Peter Szilas
Dec 30 '18 at 17:41


















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.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3056668%2fwhat-is-the-remainder-when-nn-is-divided-by-n-1%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

Bundesstraße 106

Image
Dieser Artikel beschreibt die Bundesstraße 106 in Deutschland. Zur gleichnamigen Straße in Österreich siehe Mölltal Straße. Vorlage:Infobox hochrangige Straße/Wartung/DE-B Bundesstraße 106 in Deutschland Karte Basisdaten Betreiber: Deutschland  Bundesrepublik Deutschland Straßenbeginn: Wismar ( 53° 54′  N , 11° 24′  O 53.900536 11.396875 ) Straßenende: Schwerin ( 53° 36′  N , 11° 24′  O 53.592785 11.404892 ) Gesamtlänge: 40,7 km Bundesland: Mecklenburg-Vorpommern Ausbauzustand: zweistreifig [1] Bundesstraße 106 in Schwerin (Umgehungsstraße) Straßenverlauf Land Mecklenburg-Vorpommern Landkreis Nordwestmecklenburg Ortsumgehung Wismar Wallensteingraben über   Dorf Mecklenburg Wallensteingraben Groß Stieten Bad Kleinen OT Niendorf Bahnstrecke Lübeck–Bad Kleinen Zickhusen Lübstorf Klein Trebbow OT Kirch Stück
Read more

Verónica Boquete

Image
Verónica Boquete im Trikot der Nationalmannschaft (2013) Personalia Name Verónica Boquete Giadans Geburtstag 9. April 1987 Geburtsort Santiago de Compostela, Spanien Größe 159 cm Position Mittelfeld / Sturm Juniorinnen Jahre Station 2001–2005 Xuventú Aguiño Frauen Jahre Station Spiele (Tore) 1 2005–2008 Prainsa Zaragoza 83 (40) 2008–2010 Espanyol Barcelona 53 (30) 2010 Buffalo Flash 9 0 (9) 2010 Chicago Red Stars 3 0 (1) 2010–2011 Espanyol Barcelona 26 (41) 2011 Philadelphia Independence 13 0 (5) 2011 Energija Woronesch 5 0 (1) 2012–2014 Tyresö FF 39 (14) 2014 Portland Thorns FC 15 0 (4) 2014–2015 1. FFC Frankfurt 23 0 (7) 2015–2016 FC Bayern München 9 0 (0) 2016–2018 Paris Saint-Germain 31 0 (8) 2018 Peking BG Phoenix 2019– Utah Royals FC 0 0 (0) Nationalmannschaft Jahre Auswahl
Read more

Ida-Boy-Ed-Garten

Image
Die Lage des Ida-Boy-Ed-Gartens, rot markiert Der Ida-Boy-Ed-Garten Der Ida-Boy-Ed-Garten, Blick von der Wakenitzmauer durch das Tor in der Stadtmauer Der Ida-Boy-Ed-Garten ist eine Straße der Lübecker Altstadt. Inhaltsverzeichnis 1 Lage 2 Geschichte 3 Literatur 4 Weblinks Lage | Der etwa 90 Meter lange Ida-Boy-Ed-Garten befindet sich an der nördlichen Spitze der Altstadtinsel, vor dem Burgtor. Die in einem Bogen durch eine Grünanlage hangaufwärts führende Straße verbindet die Wakenitzmauer mit der Burgtorbrücke. Geschichte | Der heutige Ida-Boy-Ed-Garten wurde ursprünglich als Verlängerung der Wakenitzmauer angelegt. 1903 wurde in den erhaltenen Teil der mittelalterlichen Stadtmauer, an dem die Wakenitzmauer bei der Einmündung der Kaiserstraße begann, ein Tor mit einem großen Bogen für die Durchführung der Straße und zwei kleineren seitlichen Öffnungen für Fußgänger gebrochen, um für die neu errichteten Wohnhäuser in diesem Bereich
Read more