$n$-th term of the series 1 27 125 1000$n$-th term of the series 9 81 961 9801…General term of a series that subtracts the square root of every square.Given a series how do I create a function to describe that seriescalculate the intersection of two number seriesSimplify Square Root Expression $sqrt125 - sqrt5$Relationship between perfect squares and infinite series (zeta function)Perfect square with negative constant termPerfect Square and its multipleHow To find the 1991-th number of this seriesHow many perfect powers are there amoung the first 1000 positive integersnth term of the series 1, 16, 24, 1024
What does the Digital Threat scope actually do?
What can I do if someone tampers with my SSH public key?
How can I portion out frozen cookie dough?
When an outsider describes family relationships, which point of view are they using?
What should I do when a paper is published similar to my PhD thesis without citation?
Trocar background-image com delay via jQuery
Difference between `nmap local-IP-address` and `nmap localhost`
The preposition for the verb (avenge) - avenge sb/sth (on OR from) sb
Automaton recognizing ambiguously accepted words of another automaton
Short scifi story where reproductive organs are converted to produce "materials", pregnant protagonist is "found fit" to be a mother
Giving a career talk in my old university, how prominently should I tell students my salary?
Strange opamp's output impedance in spice
What is better: yes / no radio, or simple checkbox?
How to write a chaotic neutral protagonist and prevent my readers from thinking they are evil?
If nine coins are tossed, what is the probability that the number of heads is even?
How can a demon take control of a human body during REM sleep?
How do spaceships determine each other's mass in space?
Why restrict private health insurance?
Would those living in a "perfect society" not understand satire
Do Paladin Auras of Differing Oaths Stack?
Is divide-by-zero a security vulnerability?
Smooth vector fields on a surface modulo diffeomorphisms
Writing text next to a table
Are small insurances worth it?
$n$-th term of the series 1 27 125 1000
$n$-th term of the series 9 81 961 9801…General term of a series that subtracts the square root of every square.Given a series how do I create a function to describe that seriescalculate the intersection of two number seriesSimplify Square Root Expression $sqrt125 - sqrt5$Relationship between perfect squares and infinite series (zeta function)Perfect square with negative constant termPerfect Square and its multipleHow To find the 1991-th number of this seriesHow many perfect powers are there amoung the first 1000 positive integersnth term of the series 1, 16, 24, 1024
$begingroup$
What will be the nth term of the series 1 27 125 1000
for $n = 1$, it is 1
for $n = 2$, it is 27
for $n = 3$, it is 125
for $n = 4$, it is 1000
square-numbers perfect-numbers perfect-powers
edited yesterday
Bernard
122k741116
asked yesterday
sroxsrox
1
$endgroup$
add a comment |
$begingroup$
What will be the nth term of the series 1 27 125 1000
for $n = 1$, it is 1
for $n = 2$, it is 27
for $n = 3$, it is 125
for $n = 4$, it is 1000
square-numbers perfect-numbers perfect-powers
edited yesterday
Bernard
122k741116
asked yesterday
sroxsrox
1
$endgroup$
1
$begingroup$
Obviously $10^3n$ is the smallest cube of $3n+1$ digits.
$endgroup$
– Peter Foreman
yesterday
$begingroup$
I suggest reverting to the original title, "Smallest perfect cube of $n$ digits", and the original form of the question. The pre-existing answers make less sense as written with the newer, less specific title and question.
$endgroup$
– Travis
yesterday
add a comment |
$begingroup$
What will be the nth term of the series 1 27 125 1000
for $n = 1$, it is 1
for $n = 2$, it is 27
for $n = 3$, it is 125
for $n = 4$, it is 1000
square-numbers perfect-numbers perfect-powers
edited yesterday
Bernard
122k741116
asked yesterday
sroxsrox
1
$endgroup$
What will be the nth term of the series 1 27 125 1000
for $n = 1$, it is 1
for $n = 2$, it is 27
for $n = 3$, it is 125
for $n = 4$, it is 1000
square-numbers perfect-numbers perfect-powers
square-numbers perfect-numbers perfect-powers
edited yesterday
Bernard
122k741116
asked yesterday
sroxsrox
1
edited yesterday
Bernard
122k741116
asked yesterday
sroxsrox
1
edited yesterday
Bernard
122k741116
edited yesterday
Bernard
122k741116
edited yesterday
Bernard
122k741116
122k741116
asked yesterday
sroxsrox
1
asked yesterday
sroxsrox
1
asked yesterday
sroxsrox
1
1
1
$begingroup$
Obviously $10^3n$ is the smallest cube of $3n+1$ digits.
$endgroup$
– Peter Foreman
yesterday
$begingroup$
I suggest reverting to the original title, "Smallest perfect cube of $n$ digits", and the original form of the question. The pre-existing answers make less sense as written with the newer, less specific title and question.
$endgroup$
– Travis
yesterday
add a comment |
1
$begingroup$
Obviously $10^3n$ is the smallest cube of $3n+1$ digits.
$endgroup$
– Peter Foreman
yesterday
$begingroup$
I suggest reverting to the original title, "Smallest perfect cube of $n$ digits", and the original form of the question. The pre-existing answers make less sense as written with the newer, less specific title and question.
$endgroup$
– Travis
yesterday
1
1
$begingroup$
Obviously $10^3n$ is the smallest cube of $3n+1$ digits.
$endgroup$
– Peter Foreman
yesterday
$begingroup$
Obviously $10^3n$ is the smallest cube of $3n+1$ digits.
$endgroup$
– Peter Foreman
yesterday
$begingroup$
I suggest reverting to the original title, "Smallest perfect cube of $n$ digits", and the original form of the question. The pre-existing answers make less sense as written with the newer, less specific title and question.
$endgroup$
– Travis
yesterday
$begingroup$
I suggest reverting to the original title, "Smallest perfect cube of $n$ digits", and the original form of the question. The pre-existing answers make less sense as written with the newer, less specific title and question.
$endgroup$
– Travis
yesterday
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
For all residue classes modulo $3$, the smallest cube with $n$ digits is the smallest perfect cube $geq 10^n - 1$, so its cube root is the smallest integer $geq 10^(n - 1) / 3$, namely $lceil 10^(n - 1) / 3 rceil$. So, the smallest cube with $n$ digits is $$boxedlceil 10^(n - 1) / 3 rceil^3 .$$
If particular, if $n equiv 1 pmod 3$, then $10^n - 1 = 1 underbrace0 cdots 0_n$, which is the smallest number with $3 n + 1$ digits.
For $n equiv 2 pmod 3$, so, for $n = 3 m + 2$, we have $10^(n - 1) / 3 = lceil 10^m cdot 10^1 / 3 rceil,$ so to produce the smallest integer whose cube has $n$ digits, we take $10^1 / 3 = 2.15443ldots$, move the decimal place rightward $m$ times, and round up, giving $3, 22, 216, ldots$; their respective cubes are $27, 10648, 10077696, ldots$.
A similar argument shows that for $n equiv 0 pmod 3$ we consider $10^2 / 3 = 4.64158ldots$, so the smallest integers are $5, 47, 465, ldots$, and the corresponding cubes are $125, 103823, 100544625, ldots$.
answered yesterday
TravisTravis
62.8k767149
$endgroup$
$begingroup$
Perhaps the downvoter would explain their objection?
$endgroup$
– Travis
yesterday
2
$begingroup$
Probably not. They usually don't, in my experience.
$endgroup$
– Brian Tung
yesterday
1
$begingroup$
I voted down the question (not any of the answers) as this is one of a sequence of several homework questions where the asker is getting others to do their work for them. I take the old fashioned view that the asker will learn nothing from copying out answers posted here that are not understood.
$endgroup$
– Martin Hansen
yesterday
$begingroup$
I agree, and now that I'm aware of OP's broader behavior (and now that OP has, strangely, changed the original question here for the worse) I would have done the same if I hadn't already run out of votes for the day.
$endgroup$
– Travis
yesterday
add a comment |
$begingroup$
Suppose $nge 1$
If $10^n-1$ is not already a cube , take $$m=lceil (10^n-1)^frac13rceil $$ Then, $m^3$ is the smallest cube with $n$ digits.
answered yesterday
PeterPeter
48.5k1139135
$endgroup$
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%2f3140407%2fn-th-term-of-the-series-1-27-125-1000%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
$begingroup$
For all residue classes modulo $3$, the smallest cube with $n$ digits is the smallest perfect cube $geq 10^n - 1$, so its cube root is the smallest integer $geq 10^(n - 1) / 3$, namely $lceil 10^(n - 1) / 3 rceil$. So, the smallest cube with $n$ digits is $$boxedlceil 10^(n - 1) / 3 rceil^3 .$$
If particular, if $n equiv 1 pmod 3$, then $10^n - 1 = 1 underbrace0 cdots 0_n$, which is the smallest number with $3 n + 1$ digits.
For $n equiv 2 pmod 3$, so, for $n = 3 m + 2$, we have $10^(n - 1) / 3 = lceil 10^m cdot 10^1 / 3 rceil,$ so to produce the smallest integer whose cube has $n$ digits, we take $10^1 / 3 = 2.15443ldots$, move the decimal place rightward $m$ times, and round up, giving $3, 22, 216, ldots$; their respective cubes are $27, 10648, 10077696, ldots$.
A similar argument shows that for $n equiv 0 pmod 3$ we consider $10^2 / 3 = 4.64158ldots$, so the smallest integers are $5, 47, 465, ldots$, and the corresponding cubes are $125, 103823, 100544625, ldots$.
answered yesterday
TravisTravis
62.8k767149
$endgroup$
$begingroup$
Perhaps the downvoter would explain their objection?
$endgroup$
– Travis
yesterday
2
$begingroup$
Probably not. They usually don't, in my experience.
$endgroup$
– Brian Tung
yesterday
1
$begingroup$
I voted down the question (not any of the answers) as this is one of a sequence of several homework questions where the asker is getting others to do their work for them. I take the old fashioned view that the asker will learn nothing from copying out answers posted here that are not understood.
$endgroup$
– Martin Hansen
yesterday
$begingroup$
I agree, and now that I'm aware of OP's broader behavior (and now that OP has, strangely, changed the original question here for the worse) I would have done the same if I hadn't already run out of votes for the day.
$endgroup$
– Travis
yesterday
add a comment |
$begingroup$
For all residue classes modulo $3$, the smallest cube with $n$ digits is the smallest perfect cube $geq 10^n - 1$, so its cube root is the smallest integer $geq 10^(n - 1) / 3$, namely $lceil 10^(n - 1) / 3 rceil$. So, the smallest cube with $n$ digits is $$boxedlceil 10^(n - 1) / 3 rceil^3 .$$
If particular, if $n equiv 1 pmod 3$, then $10^n - 1 = 1 underbrace0 cdots 0_n$, which is the smallest number with $3 n + 1$ digits.
For $n equiv 2 pmod 3$, so, for $n = 3 m + 2$, we have $10^(n - 1) / 3 = lceil 10^m cdot 10^1 / 3 rceil,$ so to produce the smallest integer whose cube has $n$ digits, we take $10^1 / 3 = 2.15443ldots$, move the decimal place rightward $m$ times, and round up, giving $3, 22, 216, ldots$; their respective cubes are $27, 10648, 10077696, ldots$.
A similar argument shows that for $n equiv 0 pmod 3$ we consider $10^2 / 3 = 4.64158ldots$, so the smallest integers are $5, 47, 465, ldots$, and the corresponding cubes are $125, 103823, 100544625, ldots$.
answered yesterday
TravisTravis
62.8k767149
$endgroup$
$begingroup$
Perhaps the downvoter would explain their objection?
$endgroup$
– Travis
yesterday
2
$begingroup$
Probably not. They usually don't, in my experience.
$endgroup$
– Brian Tung
yesterday
1
$begingroup$
I voted down the question (not any of the answers) as this is one of a sequence of several homework questions where the asker is getting others to do their work for them. I take the old fashioned view that the asker will learn nothing from copying out answers posted here that are not understood.
$endgroup$
– Martin Hansen
yesterday
$begingroup$
I agree, and now that I'm aware of OP's broader behavior (and now that OP has, strangely, changed the original question here for the worse) I would have done the same if I hadn't already run out of votes for the day.
$endgroup$
– Travis
yesterday
add a comment |
$begingroup$
For all residue classes modulo $3$, the smallest cube with $n$ digits is the smallest perfect cube $geq 10^n - 1$, so its cube root is the smallest integer $geq 10^(n - 1) / 3$, namely $lceil 10^(n - 1) / 3 rceil$. So, the smallest cube with $n$ digits is $$boxedlceil 10^(n - 1) / 3 rceil^3 .$$
If particular, if $n equiv 1 pmod 3$, then $10^n - 1 = 1 underbrace0 cdots 0_n$, which is the smallest number with $3 n + 1$ digits.
For $n equiv 2 pmod 3$, so, for $n = 3 m + 2$, we have $10^(n - 1) / 3 = lceil 10^m cdot 10^1 / 3 rceil,$ so to produce the smallest integer whose cube has $n$ digits, we take $10^1 / 3 = 2.15443ldots$, move the decimal place rightward $m$ times, and round up, giving $3, 22, 216, ldots$; their respective cubes are $27, 10648, 10077696, ldots$.
A similar argument shows that for $n equiv 0 pmod 3$ we consider $10^2 / 3 = 4.64158ldots$, so the smallest integers are $5, 47, 465, ldots$, and the corresponding cubes are $125, 103823, 100544625, ldots$.
answered yesterday
TravisTravis
62.8k767149
$endgroup$
For all residue classes modulo $3$, the smallest cube with $n$ digits is the smallest perfect cube $geq 10^n - 1$, so its cube root is the smallest integer $geq 10^(n - 1) / 3$, namely $lceil 10^(n - 1) / 3 rceil$. So, the smallest cube with $n$ digits is $$boxedlceil 10^(n - 1) / 3 rceil^3 .$$
If particular, if $n equiv 1 pmod 3$, then $10^n - 1 = 1 underbrace0 cdots 0_n$, which is the smallest number with $3 n + 1$ digits.
For $n equiv 2 pmod 3$, so, for $n = 3 m + 2$, we have $10^(n - 1) / 3 = lceil 10^m cdot 10^1 / 3 rceil,$ so to produce the smallest integer whose cube has $n$ digits, we take $10^1 / 3 = 2.15443ldots$, move the decimal place rightward $m$ times, and round up, giving $3, 22, 216, ldots$; their respective cubes are $27, 10648, 10077696, ldots$.
A similar argument shows that for $n equiv 0 pmod 3$ we consider $10^2 / 3 = 4.64158ldots$, so the smallest integers are $5, 47, 465, ldots$, and the corresponding cubes are $125, 103823, 100544625, ldots$.
answered yesterday
TravisTravis
62.8k767149
answered yesterday
TravisTravis
62.8k767149
answered yesterday
TravisTravis
62.8k767149
answered yesterday
TravisTravis
62.8k767149
62.8k767149
$begingroup$
Perhaps the downvoter would explain their objection?
$endgroup$
– Travis
yesterday
2
$begingroup$
Probably not. They usually don't, in my experience.
$endgroup$
– Brian Tung
yesterday
1
$begingroup$
I voted down the question (not any of the answers) as this is one of a sequence of several homework questions where the asker is getting others to do their work for them. I take the old fashioned view that the asker will learn nothing from copying out answers posted here that are not understood.
$endgroup$
– Martin Hansen
yesterday
$begingroup$
I agree, and now that I'm aware of OP's broader behavior (and now that OP has, strangely, changed the original question here for the worse) I would have done the same if I hadn't already run out of votes for the day.
$endgroup$
– Travis
yesterday
add a comment |
$begingroup$
Perhaps the downvoter would explain their objection?
$endgroup$
– Travis
yesterday
2
$begingroup$
Probably not. They usually don't, in my experience.
$endgroup$
– Brian Tung
yesterday
1
$begingroup$
I voted down the question (not any of the answers) as this is one of a sequence of several homework questions where the asker is getting others to do their work for them. I take the old fashioned view that the asker will learn nothing from copying out answers posted here that are not understood.
$endgroup$
– Martin Hansen
yesterday
$begingroup$
I agree, and now that I'm aware of OP's broader behavior (and now that OP has, strangely, changed the original question here for the worse) I would have done the same if I hadn't already run out of votes for the day.
$endgroup$
– Travis
yesterday
$begingroup$
Perhaps the downvoter would explain their objection?
$endgroup$
– Travis
yesterday
$begingroup$
Perhaps the downvoter would explain their objection?
$endgroup$
– Travis
yesterday
2
2
$begingroup$
Probably not. They usually don't, in my experience.
$endgroup$
– Brian Tung
yesterday
$begingroup$
Probably not. They usually don't, in my experience.
$endgroup$
– Brian Tung
yesterday
1
1
$begingroup$
I voted down the question (not any of the answers) as this is one of a sequence of several homework questions where the asker is getting others to do their work for them. I take the old fashioned view that the asker will learn nothing from copying out answers posted here that are not understood.
$endgroup$
– Martin Hansen
yesterday
$begingroup$
I voted down the question (not any of the answers) as this is one of a sequence of several homework questions where the asker is getting others to do their work for them. I take the old fashioned view that the asker will learn nothing from copying out answers posted here that are not understood.
$endgroup$
– Martin Hansen
yesterday
$begingroup$
I agree, and now that I'm aware of OP's broader behavior (and now that OP has, strangely, changed the original question here for the worse) I would have done the same if I hadn't already run out of votes for the day.
$endgroup$
– Travis
yesterday
$begingroup$
I agree, and now that I'm aware of OP's broader behavior (and now that OP has, strangely, changed the original question here for the worse) I would have done the same if I hadn't already run out of votes for the day.
$endgroup$
– Travis
yesterday
add a comment |
$begingroup$
Suppose $nge 1$
If $10^n-1$ is not already a cube , take $$m=lceil (10^n-1)^frac13rceil $$ Then, $m^3$ is the smallest cube with $n$ digits.
answered yesterday
PeterPeter
48.5k1139135
$endgroup$
add a comment |
$begingroup$
Suppose $nge 1$
If $10^n-1$ is not already a cube , take $$m=lceil (10^n-1)^frac13rceil $$ Then, $m^3$ is the smallest cube with $n$ digits.
answered yesterday
PeterPeter
48.5k1139135
$endgroup$
add a comment |
$begingroup$
Suppose $nge 1$
If $10^n-1$ is not already a cube , take $$m=lceil (10^n-1)^frac13rceil $$ Then, $m^3$ is the smallest cube with $n$ digits.
answered yesterday
PeterPeter
48.5k1139135
$endgroup$
Suppose $nge 1$
If $10^n-1$ is not already a cube , take $$m=lceil (10^n-1)^frac13rceil $$ Then, $m^3$ is the smallest cube with $n$ digits.
answered yesterday
PeterPeter
48.5k1139135
answered yesterday
PeterPeter
48.5k1139135
answered yesterday
PeterPeter
48.5k1139135
answered yesterday
PeterPeter
48.5k1139135
48.5k1139135
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.
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%2f3140407%2fn-th-term-of-the-series-1-27-125-1000%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
$begingroup$
Obviously $10^3n$ is the smallest cube of $3n+1$ digits.
$endgroup$
– Peter Foreman
yesterday
$begingroup$
I suggest reverting to the original title, "Smallest perfect cube of $n$ digits", and the original form of the question. The pre-existing answers make less sense as written with the newer, less specific title and question.
$endgroup$
– Travis
yesterday