Mixing
How to calculate the division and remainder of a big number manually
$begingroup$
Let $x = 2017^{2017}$. How can I manually determine:
- The remainder of the division of $x$ by $31$
- The least significant digit of $x$
elementary-number-theory discrete-mathematics
asked May 6 '17 at 2:41
shardlshardl
204
$endgroup$
add a comment |
$begingroup$
Let $x = 2017^{2017}$. How can I manually determine:
- The remainder of the division of $x$ by $31$
- The least significant digit of $x$
elementary-number-theory discrete-mathematics
asked May 6 '17 at 2:41
shardlshardl
204
$endgroup$
2
$begingroup$
Your title does not match with the body. Namely, the title mentions calculating the division (I assume you mean quotient), but the body does not require the quotient.
$endgroup$
– Kenny Lau
May 6 '17 at 2:42
add a comment |
$begingroup$
Let $x = 2017^{2017}$. How can I manually determine:
- The remainder of the division of $x$ by $31$
- The least significant digit of $x$
elementary-number-theory discrete-mathematics
asked May 6 '17 at 2:41
shardlshardl
204
$endgroup$
Let $x = 2017^{2017}$. How can I manually determine:
- The remainder of the division of $x$ by $31$
- The least significant digit of $x$
elementary-number-theory discrete-mathematics
elementary-number-theory discrete-mathematics
asked May 6 '17 at 2:41
shardlshardl
204
asked May 6 '17 at 2:41
shardlshardl
204
asked May 6 '17 at 2:41
shardlshardl
204
asked May 6 '17 at 2:41
shardlshardl
204
asked May 6 '17 at 2:41
shardlshardl
204
204
2
$begingroup$
Your title does not match with the body. Namely, the title mentions calculating the division (I assume you mean quotient), but the body does not require the quotient.
$endgroup$
– Kenny Lau
May 6 '17 at 2:42
add a comment |
2
$begingroup$
Your title does not match with the body. Namely, the title mentions calculating the division (I assume you mean quotient), but the body does not require the quotient.
$endgroup$
– Kenny Lau
May 6 '17 at 2:42
2
2
$begingroup$
Your title does not match with the body. Namely, the title mentions calculating the division (I assume you mean quotient), but the body does not require the quotient.
$endgroup$
– Kenny Lau
May 6 '17 at 2:42
$begingroup$
Your title does not match with the body. Namely, the title mentions calculating the division (I assume you mean quotient), but the body does not require the quotient.
$endgroup$
– Kenny Lau
May 6 '17 at 2:42
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
FIRST QUESTION
$$begin{array}{rcll}
2017^{2017} &=& 2017^{30times67+7} \
&=& left(2017^{30}right)^{67} times 2017^7 \
&equiv& 1^{67} times 2^7 & pmod {31} \
&=& 128 \
&equiv& 4 & pmod {31}
end{array}$$
Note that $2017^{30} equiv 1 pmod{31}$ is Fermat's little theorem.
SECOND QUESTION
$$begin{array}{rcll}
2017^{2017} &=& 2017^{4times504+1} \
&=& left(2017^{4}right)^{504} times 2017 \
&equiv& 1^{504} times 7 & pmod {10} \
&=& 7 \
end{array}$$
Note that $2017^4 equiv 1 pmod{10}$ is Euler's totient theorem.
answered May 6 '17 at 2:46
Kenny LauKenny Lau
19.9k2159
$endgroup$
$begingroup$
Why $2017^{{30}^{67}} times 2017^7 equiv 1^{67} times 2^7 pmod {31}$?
$endgroup$
– shardl
May 6 '17 at 3:38
1
$begingroup$
Because $2017^{30}equiv1pmod{31}$ and $2017equiv 2pmod{31}$.
$endgroup$
– Kenny Lau
May 6 '17 at 3:39
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%2f2267991%2fhow-to-calculate-the-division-and-remainder-of-a-big-number-manually%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
FIRST QUESTION
$$begin{array}{rcll}
2017^{2017} &=& 2017^{30times67+7} \
&=& left(2017^{30}right)^{67} times 2017^7 \
&equiv& 1^{67} times 2^7 & pmod {31} \
&=& 128 \
&equiv& 4 & pmod {31}
end{array}$$
Note that $2017^{30} equiv 1 pmod{31}$ is Fermat's little theorem.
SECOND QUESTION
$$begin{array}{rcll}
2017^{2017} &=& 2017^{4times504+1} \
&=& left(2017^{4}right)^{504} times 2017 \
&equiv& 1^{504} times 7 & pmod {10} \
&=& 7 \
end{array}$$
Note that $2017^4 equiv 1 pmod{10}$ is Euler's totient theorem.
answered May 6 '17 at 2:46
Kenny LauKenny Lau
19.9k2159
$endgroup$
$begingroup$
Why $2017^{{30}^{67}} times 2017^7 equiv 1^{67} times 2^7 pmod {31}$?
$endgroup$
– shardl
May 6 '17 at 3:38
1
$begingroup$
Because $2017^{30}equiv1pmod{31}$ and $2017equiv 2pmod{31}$.
$endgroup$
– Kenny Lau
May 6 '17 at 3:39
add a comment |
$begingroup$
FIRST QUESTION
$$begin{array}{rcll}
2017^{2017} &=& 2017^{30times67+7} \
&=& left(2017^{30}right)^{67} times 2017^7 \
&equiv& 1^{67} times 2^7 & pmod {31} \
&=& 128 \
&equiv& 4 & pmod {31}
end{array}$$
Note that $2017^{30} equiv 1 pmod{31}$ is Fermat's little theorem.
SECOND QUESTION
$$begin{array}{rcll}
2017^{2017} &=& 2017^{4times504+1} \
&=& left(2017^{4}right)^{504} times 2017 \
&equiv& 1^{504} times 7 & pmod {10} \
&=& 7 \
end{array}$$
Note that $2017^4 equiv 1 pmod{10}$ is Euler's totient theorem.
answered May 6 '17 at 2:46
Kenny LauKenny Lau
19.9k2159
$endgroup$
$begingroup$
Why $2017^{{30}^{67}} times 2017^7 equiv 1^{67} times 2^7 pmod {31}$?
$endgroup$
– shardl
May 6 '17 at 3:38
1
$begingroup$
Because $2017^{30}equiv1pmod{31}$ and $2017equiv 2pmod{31}$.
$endgroup$
– Kenny Lau
May 6 '17 at 3:39
add a comment |
$begingroup$
FIRST QUESTION
$$begin{array}{rcll}
2017^{2017} &=& 2017^{30times67+7} \
&=& left(2017^{30}right)^{67} times 2017^7 \
&equiv& 1^{67} times 2^7 & pmod {31} \
&=& 128 \
&equiv& 4 & pmod {31}
end{array}$$
Note that $2017^{30} equiv 1 pmod{31}$ is Fermat's little theorem.
SECOND QUESTION
$$begin{array}{rcll}
2017^{2017} &=& 2017^{4times504+1} \
&=& left(2017^{4}right)^{504} times 2017 \
&equiv& 1^{504} times 7 & pmod {10} \
&=& 7 \
end{array}$$
Note that $2017^4 equiv 1 pmod{10}$ is Euler's totient theorem.
answered May 6 '17 at 2:46
Kenny LauKenny Lau
19.9k2159
$endgroup$
FIRST QUESTION
$$begin{array}{rcll}
2017^{2017} &=& 2017^{30times67+7} \
&=& left(2017^{30}right)^{67} times 2017^7 \
&equiv& 1^{67} times 2^7 & pmod {31} \
&=& 128 \
&equiv& 4 & pmod {31}
end{array}$$
Note that $2017^{30} equiv 1 pmod{31}$ is Fermat's little theorem.
SECOND QUESTION
$$begin{array}{rcll}
2017^{2017} &=& 2017^{4times504+1} \
&=& left(2017^{4}right)^{504} times 2017 \
&equiv& 1^{504} times 7 & pmod {10} \
&=& 7 \
end{array}$$
Note that $2017^4 equiv 1 pmod{10}$ is Euler's totient theorem.
answered May 6 '17 at 2:46
Kenny LauKenny Lau
19.9k2159
answered May 6 '17 at 2:46
Kenny LauKenny Lau
19.9k2159
answered May 6 '17 at 2:46
Kenny LauKenny Lau
19.9k2159
answered May 6 '17 at 2:46
Kenny LauKenny Lau
19.9k2159
19.9k2159
$begingroup$
Why $2017^{{30}^{67}} times 2017^7 equiv 1^{67} times 2^7 pmod {31}$?
$endgroup$
– shardl
May 6 '17 at 3:38
1
$begingroup$
Because $2017^{30}equiv1pmod{31}$ and $2017equiv 2pmod{31}$.
$endgroup$
– Kenny Lau
May 6 '17 at 3:39
add a comment |
$begingroup$
Why $2017^{{30}^{67}} times 2017^7 equiv 1^{67} times 2^7 pmod {31}$?
$endgroup$
– shardl
May 6 '17 at 3:38
1
$begingroup$
Because $2017^{30}equiv1pmod{31}$ and $2017equiv 2pmod{31}$.
$endgroup$
– Kenny Lau
May 6 '17 at 3:39
$begingroup$
Why $2017^{{30}^{67}} times 2017^7 equiv 1^{67} times 2^7 pmod {31}$?
$endgroup$
– shardl
May 6 '17 at 3:38
$begingroup$
Why $2017^{{30}^{67}} times 2017^7 equiv 1^{67} times 2^7 pmod {31}$?
$endgroup$
– shardl
May 6 '17 at 3:38
1
1
$begingroup$
Because $2017^{30}equiv1pmod{31}$ and $2017equiv 2pmod{31}$.
$endgroup$
– Kenny Lau
May 6 '17 at 3:39
$begingroup$
Because $2017^{30}equiv1pmod{31}$ and $2017equiv 2pmod{31}$.
$endgroup$
– Kenny Lau
May 6 '17 at 3:39
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.
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%2f2267991%2fhow-to-calculate-the-division-and-remainder-of-a-big-number-manually%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
2
$begingroup$
Your title does not match with the body. Namely, the title mentions calculating the division (I assume you mean quotient), but the body does not require the quotient.
$endgroup$
– Kenny Lau
May 6 '17 at 2:42