Find natural number $0 < n < 30,000$ such that $sqrt[3]{5n}+sqrt{10n}$ is rational
$begingroup$
I was thinking that I could try to make some sort of substitution to convert $sqrt[3]{5n}+sqrt{10n}$ into a polynomial with integer coefficients then use the Rational Roots Theorem to find a rational root. I don't really know if that's going to get me anywhere other than $n=0$.
I would really appreciate some help, or some hints. I don't necessarily want a full solution, but a nudge in the right direction.
irrational-numbers rational-numbers
asked Dec 9 '18 at 15:09
Trevor MasonTrevor Mason
197
$endgroup$
add a comment |
$begingroup$
I was thinking that I could try to make some sort of substitution to convert $sqrt[3]{5n}+sqrt{10n}$ into a polynomial with integer coefficients then use the Rational Roots Theorem to find a rational root. I don't really know if that's going to get me anywhere other than $n=0$.
I would really appreciate some help, or some hints. I don't necessarily want a full solution, but a nudge in the right direction.
irrational-numbers rational-numbers
asked Dec 9 '18 at 15:09
Trevor MasonTrevor Mason
197
$endgroup$
add a comment |
$begingroup$
I was thinking that I could try to make some sort of substitution to convert $sqrt[3]{5n}+sqrt{10n}$ into a polynomial with integer coefficients then use the Rational Roots Theorem to find a rational root. I don't really know if that's going to get me anywhere other than $n=0$.
I would really appreciate some help, or some hints. I don't necessarily want a full solution, but a nudge in the right direction.
irrational-numbers rational-numbers
asked Dec 9 '18 at 15:09
Trevor MasonTrevor Mason
197
$endgroup$
I was thinking that I could try to make some sort of substitution to convert $sqrt[3]{5n}+sqrt{10n}$ into a polynomial with integer coefficients then use the Rational Roots Theorem to find a rational root. I don't really know if that's going to get me anywhere other than $n=0$.
I would really appreciate some help, or some hints. I don't necessarily want a full solution, but a nudge in the right direction.
irrational-numbers rational-numbers
irrational-numbers rational-numbers
asked Dec 9 '18 at 15:09
Trevor MasonTrevor Mason
197
asked Dec 9 '18 at 15:09
Trevor MasonTrevor Mason
197
asked Dec 9 '18 at 15:09
Trevor MasonTrevor Mason
197
asked Dec 9 '18 at 15:09
Trevor MasonTrevor Mason
197
asked Dec 9 '18 at 15:09
Trevor MasonTrevor Mason
197
197
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
You should focus on the prime factorization of $n$. To have a number be a cube, all the primes in its factorization need to come with a power that is a multiple of $3$. Similarly, to be a square the primes need to have an even power.
In your example, all primes except $2$ and $5$ must be sixth powers. Think about what the leading coefficients do for $2$ and $5$.
answered Dec 9 '18 at 15:14
Ross MillikanRoss Millikan
295k23198371
$endgroup$
$begingroup$
by leading coefficients you mean in the prime factorization?
$endgroup$
– Trevor Mason
Dec 9 '18 at 15:31
1
$begingroup$
By leading coefficients I mean the $5$ and $10$ under the radical signs. They mean you don't want the powers of $2$ and $5$ in $n$ to be multiples of $6$.
$endgroup$
– Ross Millikan
Dec 9 '18 at 15:38
$begingroup$
Ah, yes. That makes much more sense. I think I figured it out. Thanks so much for your help!
$endgroup$
– Trevor Mason
Dec 9 '18 at 15:39
$begingroup$
I think the limit of $30,000$ is unfortunately low. I think it should have been chosen large enough for there to be half a dozen solutions or so.
$endgroup$
– Ross Millikan
Dec 9 '18 at 15:56
1
$begingroup$
I just would like to have people recognize that once they find the smallest solution, which is $2^35^5=25,000$ they can multiply it by the sixth power of any number, including $2$ and $5$ and get another solution.
$endgroup$
– Ross Millikan
Dec 9 '18 at 16:16
|
show 1 more 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%2f3032479%2ffind-natural-number-0-n-30-000-such-that-sqrt35n-sqrt10n-is-rati%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
$begingroup$
You should focus on the prime factorization of $n$. To have a number be a cube, all the primes in its factorization need to come with a power that is a multiple of $3$. Similarly, to be a square the primes need to have an even power.
In your example, all primes except $2$ and $5$ must be sixth powers. Think about what the leading coefficients do for $2$ and $5$.
answered Dec 9 '18 at 15:14
Ross MillikanRoss Millikan
295k23198371
$endgroup$
$begingroup$
by leading coefficients you mean in the prime factorization?
$endgroup$
– Trevor Mason
Dec 9 '18 at 15:31
1
$begingroup$
By leading coefficients I mean the $5$ and $10$ under the radical signs. They mean you don't want the powers of $2$ and $5$ in $n$ to be multiples of $6$.
$endgroup$
– Ross Millikan
Dec 9 '18 at 15:38
$begingroup$
Ah, yes. That makes much more sense. I think I figured it out. Thanks so much for your help!
$endgroup$
– Trevor Mason
Dec 9 '18 at 15:39
$begingroup$
I think the limit of $30,000$ is unfortunately low. I think it should have been chosen large enough for there to be half a dozen solutions or so.
$endgroup$
– Ross Millikan
Dec 9 '18 at 15:56
1
$begingroup$
I just would like to have people recognize that once they find the smallest solution, which is $2^35^5=25,000$ they can multiply it by the sixth power of any number, including $2$ and $5$ and get another solution.
$endgroup$
– Ross Millikan
Dec 9 '18 at 16:16
|
show 1 more comment
$begingroup$
You should focus on the prime factorization of $n$. To have a number be a cube, all the primes in its factorization need to come with a power that is a multiple of $3$. Similarly, to be a square the primes need to have an even power.
In your example, all primes except $2$ and $5$ must be sixth powers. Think about what the leading coefficients do for $2$ and $5$.
answered Dec 9 '18 at 15:14
Ross MillikanRoss Millikan
295k23198371
$endgroup$
$begingroup$
by leading coefficients you mean in the prime factorization?
$endgroup$
– Trevor Mason
Dec 9 '18 at 15:31
1
$begingroup$
By leading coefficients I mean the $5$ and $10$ under the radical signs. They mean you don't want the powers of $2$ and $5$ in $n$ to be multiples of $6$.
$endgroup$
– Ross Millikan
Dec 9 '18 at 15:38
$begingroup$
Ah, yes. That makes much more sense. I think I figured it out. Thanks so much for your help!
$endgroup$
– Trevor Mason
Dec 9 '18 at 15:39
$begingroup$
I think the limit of $30,000$ is unfortunately low. I think it should have been chosen large enough for there to be half a dozen solutions or so.
$endgroup$
– Ross Millikan
Dec 9 '18 at 15:56
1
$begingroup$
I just would like to have people recognize that once they find the smallest solution, which is $2^35^5=25,000$ they can multiply it by the sixth power of any number, including $2$ and $5$ and get another solution.
$endgroup$
– Ross Millikan
Dec 9 '18 at 16:16
|
show 1 more comment
$begingroup$
You should focus on the prime factorization of $n$. To have a number be a cube, all the primes in its factorization need to come with a power that is a multiple of $3$. Similarly, to be a square the primes need to have an even power.
In your example, all primes except $2$ and $5$ must be sixth powers. Think about what the leading coefficients do for $2$ and $5$.
answered Dec 9 '18 at 15:14
Ross MillikanRoss Millikan
295k23198371
$endgroup$
You should focus on the prime factorization of $n$. To have a number be a cube, all the primes in its factorization need to come with a power that is a multiple of $3$. Similarly, to be a square the primes need to have an even power.
In your example, all primes except $2$ and $5$ must be sixth powers. Think about what the leading coefficients do for $2$ and $5$.
answered Dec 9 '18 at 15:14
Ross MillikanRoss Millikan
295k23198371
answered Dec 9 '18 at 15:14
Ross MillikanRoss Millikan
295k23198371
answered Dec 9 '18 at 15:14
Ross MillikanRoss Millikan
295k23198371
answered Dec 9 '18 at 15:14
Ross MillikanRoss Millikan
295k23198371
295k23198371
$begingroup$
by leading coefficients you mean in the prime factorization?
$endgroup$
– Trevor Mason
Dec 9 '18 at 15:31
1
$begingroup$
By leading coefficients I mean the $5$ and $10$ under the radical signs. They mean you don't want the powers of $2$ and $5$ in $n$ to be multiples of $6$.
$endgroup$
– Ross Millikan
Dec 9 '18 at 15:38
$begingroup$
Ah, yes. That makes much more sense. I think I figured it out. Thanks so much for your help!
$endgroup$
– Trevor Mason
Dec 9 '18 at 15:39
$begingroup$
I think the limit of $30,000$ is unfortunately low. I think it should have been chosen large enough for there to be half a dozen solutions or so.
$endgroup$
– Ross Millikan
Dec 9 '18 at 15:56
1
$begingroup$
I just would like to have people recognize that once they find the smallest solution, which is $2^35^5=25,000$ they can multiply it by the sixth power of any number, including $2$ and $5$ and get another solution.
$endgroup$
– Ross Millikan
Dec 9 '18 at 16:16
|
show 1 more comment
$begingroup$
by leading coefficients you mean in the prime factorization?
$endgroup$
– Trevor Mason
Dec 9 '18 at 15:31
1
$begingroup$
By leading coefficients I mean the $5$ and $10$ under the radical signs. They mean you don't want the powers of $2$ and $5$ in $n$ to be multiples of $6$.
$endgroup$
– Ross Millikan
Dec 9 '18 at 15:38
$begingroup$
Ah, yes. That makes much more sense. I think I figured it out. Thanks so much for your help!
$endgroup$
– Trevor Mason
Dec 9 '18 at 15:39
$begingroup$
I think the limit of $30,000$ is unfortunately low. I think it should have been chosen large enough for there to be half a dozen solutions or so.
$endgroup$
– Ross Millikan
Dec 9 '18 at 15:56
1
$begingroup$
I just would like to have people recognize that once they find the smallest solution, which is $2^35^5=25,000$ they can multiply it by the sixth power of any number, including $2$ and $5$ and get another solution.
$endgroup$
– Ross Millikan
Dec 9 '18 at 16:16
$begingroup$
by leading coefficients you mean in the prime factorization?
$endgroup$
– Trevor Mason
Dec 9 '18 at 15:31
$begingroup$
by leading coefficients you mean in the prime factorization?
$endgroup$
– Trevor Mason
Dec 9 '18 at 15:31
1
1
$begingroup$
By leading coefficients I mean the $5$ and $10$ under the radical signs. They mean you don't want the powers of $2$ and $5$ in $n$ to be multiples of $6$.
$endgroup$
– Ross Millikan
Dec 9 '18 at 15:38
$begingroup$
By leading coefficients I mean the $5$ and $10$ under the radical signs. They mean you don't want the powers of $2$ and $5$ in $n$ to be multiples of $6$.
$endgroup$
– Ross Millikan
Dec 9 '18 at 15:38
$begingroup$
Ah, yes. That makes much more sense. I think I figured it out. Thanks so much for your help!
$endgroup$
– Trevor Mason
Dec 9 '18 at 15:39
$begingroup$
Ah, yes. That makes much more sense. I think I figured it out. Thanks so much for your help!
$endgroup$
– Trevor Mason
Dec 9 '18 at 15:39
$begingroup$
I think the limit of $30,000$ is unfortunately low. I think it should have been chosen large enough for there to be half a dozen solutions or so.
$endgroup$
– Ross Millikan
Dec 9 '18 at 15:56
$begingroup$
I think the limit of $30,000$ is unfortunately low. I think it should have been chosen large enough for there to be half a dozen solutions or so.
$endgroup$
– Ross Millikan
Dec 9 '18 at 15:56
1
1
$begingroup$
I just would like to have people recognize that once they find the smallest solution, which is $2^35^5=25,000$ they can multiply it by the sixth power of any number, including $2$ and $5$ and get another solution.
$endgroup$
– Ross Millikan
Dec 9 '18 at 16:16
$begingroup$
I just would like to have people recognize that once they find the smallest solution, which is $2^35^5=25,000$ they can multiply it by the sixth power of any number, including $2$ and $5$ and get another solution.
$endgroup$
– Ross Millikan
Dec 9 '18 at 16:16
|
show 1 more 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%2f3032479%2ffind-natural-number-0-n-30-000-such-that-sqrt35n-sqrt10n-is-rati%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
