Prove that $1982cdot 1983|underbrace{22 ldots2}_{1980}$
Prove that $1982cdot 1983|underbrace{22 ldots2}_{1980}$
Since $underbrace{22 ldots2}_{1980}=frac{2(10^{1980}-1)}{10-1}$ then I need to show that $9cdot 1982cdot 1983| 2(10^{1980}-1)$, or $9cdot 991cdot 1983|(10^{1980}-1)$ since $(9,991,1983)=1$ if I show that $9|(10^{1980}-1)$, $991|(10^{1980}-1)$, $1983|(10^{1980}-1)$ then I finish the task.
Since $10equiv 1 (mod{9})$ then $10^{1980}equiv 1(mod{9})$ so $10^{1980}-1equiv 0 (mod{9})$ so we show that $9|(10^{1980}-1)$.
In $991|(10^{1980}-1)$ we can use Euler's function since $(991,10)=1$ so $10^{varphi (991)}equiv 1 (mod 991)$ so $10^{990}equiv 1 (mod{9})$ then if I square $10^{2cdot 990}equiv 1 (mod{991})$ so $10^{1980}-1equiv 0 (mod{991})$.
Then $1983|(10^{1980}-1)$ $1983=3cdot 661$ $3|10^{1980}-1$ it is trivial, and we can you Euler's theorem then $10^{660}equiv 1(mod{661})$ and then $10^{3cdot 660}equiv 1 (mod{661})$ so $10^{1980}-1equiv 0 (mod{661})$.
But I just not so sure when you show something like this and number $x,y,z in mathbb N$ (x,y,z)=1 and if say (y,x)=3 then that not change anything in task, if we want to show that x|n y|n and z|n? If you understand what I ask, here $(9,1983)=3$, but $(9,1982,1983)=1$
discrete-mathematics modular-arithmetic divisibility
add a comment |
Prove that $1982cdot 1983|underbrace{22 ldots2}_{1980}$
Since $underbrace{22 ldots2}_{1980}=frac{2(10^{1980}-1)}{10-1}$ then I need to show that $9cdot 1982cdot 1983| 2(10^{1980}-1)$, or $9cdot 991cdot 1983|(10^{1980}-1)$ since $(9,991,1983)=1$ if I show that $9|(10^{1980}-1)$, $991|(10^{1980}-1)$, $1983|(10^{1980}-1)$ then I finish the task.
Since $10equiv 1 (mod{9})$ then $10^{1980}equiv 1(mod{9})$ so $10^{1980}-1equiv 0 (mod{9})$ so we show that $9|(10^{1980}-1)$.
In $991|(10^{1980}-1)$ we can use Euler's function since $(991,10)=1$ so $10^{varphi (991)}equiv 1 (mod 991)$ so $10^{990}equiv 1 (mod{9})$ then if I square $10^{2cdot 990}equiv 1 (mod{991})$ so $10^{1980}-1equiv 0 (mod{991})$.
Then $1983|(10^{1980}-1)$ $1983=3cdot 661$ $3|10^{1980}-1$ it is trivial, and we can you Euler's theorem then $10^{660}equiv 1(mod{661})$ and then $10^{3cdot 660}equiv 1 (mod{661})$ so $10^{1980}-1equiv 0 (mod{661})$.
But I just not so sure when you show something like this and number $x,y,z in mathbb N$ (x,y,z)=1 and if say (y,x)=3 then that not change anything in task, if we want to show that x|n y|n and z|n? If you understand what I ask, here $(9,1983)=3$, but $(9,1982,1983)=1$
discrete-mathematics modular-arithmetic divisibility
1
You need a backslash before "underbrace".
– Shaun
Nov 23 at 6:58
add a comment |
Prove that $1982cdot 1983|underbrace{22 ldots2}_{1980}$
Since $underbrace{22 ldots2}_{1980}=frac{2(10^{1980}-1)}{10-1}$ then I need to show that $9cdot 1982cdot 1983| 2(10^{1980}-1)$, or $9cdot 991cdot 1983|(10^{1980}-1)$ since $(9,991,1983)=1$ if I show that $9|(10^{1980}-1)$, $991|(10^{1980}-1)$, $1983|(10^{1980}-1)$ then I finish the task.
Since $10equiv 1 (mod{9})$ then $10^{1980}equiv 1(mod{9})$ so $10^{1980}-1equiv 0 (mod{9})$ so we show that $9|(10^{1980}-1)$.
In $991|(10^{1980}-1)$ we can use Euler's function since $(991,10)=1$ so $10^{varphi (991)}equiv 1 (mod 991)$ so $10^{990}equiv 1 (mod{9})$ then if I square $10^{2cdot 990}equiv 1 (mod{991})$ so $10^{1980}-1equiv 0 (mod{991})$.
Then $1983|(10^{1980}-1)$ $1983=3cdot 661$ $3|10^{1980}-1$ it is trivial, and we can you Euler's theorem then $10^{660}equiv 1(mod{661})$ and then $10^{3cdot 660}equiv 1 (mod{661})$ so $10^{1980}-1equiv 0 (mod{661})$.
But I just not so sure when you show something like this and number $x,y,z in mathbb N$ (x,y,z)=1 and if say (y,x)=3 then that not change anything in task, if we want to show that x|n y|n and z|n? If you understand what I ask, here $(9,1983)=3$, but $(9,1982,1983)=1$
discrete-mathematics modular-arithmetic divisibility
Prove that $1982cdot 1983|underbrace{22 ldots2}_{1980}$
Since $underbrace{22 ldots2}_{1980}=frac{2(10^{1980}-1)}{10-1}$ then I need to show that $9cdot 1982cdot 1983| 2(10^{1980}-1)$, or $9cdot 991cdot 1983|(10^{1980}-1)$ since $(9,991,1983)=1$ if I show that $9|(10^{1980}-1)$, $991|(10^{1980}-1)$, $1983|(10^{1980}-1)$ then I finish the task.
Since $10equiv 1 (mod{9})$ then $10^{1980}equiv 1(mod{9})$ so $10^{1980}-1equiv 0 (mod{9})$ so we show that $9|(10^{1980}-1)$.
In $991|(10^{1980}-1)$ we can use Euler's function since $(991,10)=1$ so $10^{varphi (991)}equiv 1 (mod 991)$ so $10^{990}equiv 1 (mod{9})$ then if I square $10^{2cdot 990}equiv 1 (mod{991})$ so $10^{1980}-1equiv 0 (mod{991})$.
Then $1983|(10^{1980}-1)$ $1983=3cdot 661$ $3|10^{1980}-1$ it is trivial, and we can you Euler's theorem then $10^{660}equiv 1(mod{661})$ and then $10^{3cdot 660}equiv 1 (mod{661})$ so $10^{1980}-1equiv 0 (mod{661})$.
But I just not so sure when you show something like this and number $x,y,z in mathbb N$ (x,y,z)=1 and if say (y,x)=3 then that not change anything in task, if we want to show that x|n y|n and z|n? If you understand what I ask, here $(9,1983)=3$, but $(9,1982,1983)=1$
discrete-mathematics modular-arithmetic divisibility
discrete-mathematics modular-arithmetic divisibility
edited Nov 23 at 7:04
asked Nov 23 at 6:57
Marko Škorić
70310
70310
1
You need a backslash before "underbrace".
– Shaun
Nov 23 at 6:58
add a comment |
1
You need a backslash before "underbrace".
– Shaun
Nov 23 at 6:58
1
1
You need a backslash before "underbrace".
– Shaun
Nov 23 at 6:58
You need a backslash before "underbrace".
– Shaun
Nov 23 at 6:58
add a comment |
2 Answers
2
active
oldest
votes
While $gcd(9,1982,1983)=1$, you have $gcd(9,1983)ne 1$. You should rather factor
$ 9cdot 1982cdot 1983$ as $27cdot 661cdot 991$.
add a comment |
$$1982cdot 1983|underbrace{22 ldots2}_{1980}$$
$$2cdot991cdot 1983|underbrace{left(11 ldots1right)}_{1980}cdot2$$
This implies that $gcd geq 2$ and therefore it is $1982cdot 1983|underbrace{22 ldots2}_{1980}$
add a comment |
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',
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
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3010051%2fprove-that-1982-cdot-1983-underbrace22-ldots2-1980%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
While $gcd(9,1982,1983)=1$, you have $gcd(9,1983)ne 1$. You should rather factor
$ 9cdot 1982cdot 1983$ as $27cdot 661cdot 991$.
add a comment |
While $gcd(9,1982,1983)=1$, you have $gcd(9,1983)ne 1$. You should rather factor
$ 9cdot 1982cdot 1983$ as $27cdot 661cdot 991$.
add a comment |
While $gcd(9,1982,1983)=1$, you have $gcd(9,1983)ne 1$. You should rather factor
$ 9cdot 1982cdot 1983$ as $27cdot 661cdot 991$.
While $gcd(9,1982,1983)=1$, you have $gcd(9,1983)ne 1$. You should rather factor
$ 9cdot 1982cdot 1983$ as $27cdot 661cdot 991$.
answered Nov 23 at 7:16
Hagen von Eitzen
275k21268495
275k21268495
add a comment |
add a comment |
$$1982cdot 1983|underbrace{22 ldots2}_{1980}$$
$$2cdot991cdot 1983|underbrace{left(11 ldots1right)}_{1980}cdot2$$
This implies that $gcd geq 2$ and therefore it is $1982cdot 1983|underbrace{22 ldots2}_{1980}$
add a comment |
$$1982cdot 1983|underbrace{22 ldots2}_{1980}$$
$$2cdot991cdot 1983|underbrace{left(11 ldots1right)}_{1980}cdot2$$
This implies that $gcd geq 2$ and therefore it is $1982cdot 1983|underbrace{22 ldots2}_{1980}$
add a comment |
$$1982cdot 1983|underbrace{22 ldots2}_{1980}$$
$$2cdot991cdot 1983|underbrace{left(11 ldots1right)}_{1980}cdot2$$
This implies that $gcd geq 2$ and therefore it is $1982cdot 1983|underbrace{22 ldots2}_{1980}$
$$1982cdot 1983|underbrace{22 ldots2}_{1980}$$
$$2cdot991cdot 1983|underbrace{left(11 ldots1right)}_{1980}cdot2$$
This implies that $gcd geq 2$ and therefore it is $1982cdot 1983|underbrace{22 ldots2}_{1980}$
edited Nov 23 at 8:48
answered Nov 23 at 7:37
Darío A. Gutiérrez
2,59141530
2,59141530
add a comment |
add a comment |
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.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3010051%2fprove-that-1982-cdot-1983-underbrace22-ldots2-1980%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
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
1
You need a backslash before "underbrace".
– Shaun
Nov 23 at 6:58