The Most Powerful Numberwhats the Missing Number?Making 24 - Most Complicated WayOthello - Most number of legal moves in a given turnThe Most Intelligent PrinceFind a Strobogrammatic number, so if we square it, the result is a pandigit numberConsecutive Numbers Sum of a numberThe most mines that can be swept!The most probable keyMost efficient way to travel home for the Holidays?Solve for the Number in the number..square..cube relationship
Are runways booked by airlines to land their planes?
To exponential digit growth and beyond!
Why does Bran want to find Drogon?
What is the recommended procedure to land a taildragger in a crosswind?
Freedom of Speech and Assembly in China
Fixie: how to learn to ride: step by step
Did this character show any indication of wanting to rule before S8E6?
Can we assume that a hash function with high collision resistance also means highly uniform distribution?
How to teach an undergraduate course without having taken that course formally before?
Shorten or merge multiple lines of `&> /dev/null &`
Who knighted this character?
Expected maximum number of unpaired socks
How did NASA Langley end up with the first 737?
How can I properly write this equation in Latex?
Can a UK national work as a paid shop assistant in the USA?
Why sampling a periodic signal doesn't yield a periodic discrete signal?
Testing using real data of the customer
Can you still travel to America on the ESTA waiver program if you have been to Iran in transit?
How to melt snow without fire or using body heat?
Best shape for a necromancer's undead minions for battle?
Why is this integration method not valid?
Removing the last element of a list
Why did it take so long for Germany to allow electric scooters / e-rollers on the roads?
Is this homebrew "Cactus Grenade" cantrip balanced?
The Most Powerful Number
whats the Missing Number?Making 24 - Most Complicated WayOthello - Most number of legal moves in a given turnThe Most Intelligent PrinceFind a Strobogrammatic number, so if we square it, the result is a pandigit numberConsecutive Numbers Sum of a numberThe most mines that can be swept!The most probable keyMost efficient way to travel home for the Holidays?Solve for the Number in the number..square..cube relationship
$begingroup$
A power-full number is defined as a number with $n$ amount of non-zero digits $D_x$.
The number takes the form $D_n$&$...$&$D_2$&$D_1$. The digits in a power-full number satisfies the rule
$D_x+1$&$D_x=k_x^x+1$ where $k_x$ are positive integers. This rule applies for $1le xlt n$.
Essentially a power-full number has paired digits that equal some perfect power corresponding to its position in the chain + 1.
Given that the magnitude of a power-full number and $n$ determines how powerful it is (the bigger the better), what is the most powerful number?
mathematics logical-deduction no-computers
asked 9 hours ago
AdamAdam
4331221
$endgroup$
|
show 2 more comments
$begingroup$
A power-full number is defined as a number with $n$ amount of non-zero digits $D_x$.
The number takes the form $D_n$&$...$&$D_2$&$D_1$. The digits in a power-full number satisfies the rule
$D_x+1$&$D_x=k_x^x+1$ where $k_x$ are positive integers. This rule applies for $1le xlt n$.
Essentially a power-full number has paired digits that equal some perfect power corresponding to its position in the chain + 1.
Given that the magnitude of a power-full number and $n$ determines how powerful it is (the bigger the better), what is the most powerful number?
mathematics logical-deduction no-computers
asked 9 hours ago
AdamAdam
4331221
$endgroup$
$begingroup$
Feel free to help with formatting if there is an easier way to convey this puzzle
$endgroup$
– Adam
9 hours ago
1
$begingroup$
Do you mean kx are positive integers or is there a single k?
$endgroup$
– Jonathan Allan
9 hours ago
$begingroup$
Oops, fixed! @JonathanAllan
$endgroup$
– Adam
9 hours ago
$begingroup$
My edits were to make the spelling consistent; now you have both 'power-full' and 'powerful' in your title and body. Pedants may think the definitions are different ...
$endgroup$
– Glorfindel
8 hours ago
$begingroup$
@Veedrac $1le xlt n$ shouldn't include $n$. I thought it would be more confusing to have $1le xle n-1$
$endgroup$
– Adam
7 hours ago
|
show 2 more comments
$begingroup$
A power-full number is defined as a number with $n$ amount of non-zero digits $D_x$.
The number takes the form $D_n$&$...$&$D_2$&$D_1$. The digits in a power-full number satisfies the rule
$D_x+1$&$D_x=k_x^x+1$ where $k_x$ are positive integers. This rule applies for $1le xlt n$.
Essentially a power-full number has paired digits that equal some perfect power corresponding to its position in the chain + 1.
Given that the magnitude of a power-full number and $n$ determines how powerful it is (the bigger the better), what is the most powerful number?
mathematics logical-deduction no-computers
asked 9 hours ago
AdamAdam
4331221
$endgroup$
A power-full number is defined as a number with $n$ amount of non-zero digits $D_x$.
The number takes the form $D_n$&$...$&$D_2$&$D_1$. The digits in a power-full number satisfies the rule
$D_x+1$&$D_x=k_x^x+1$ where $k_x$ are positive integers. This rule applies for $1le xlt n$.
Essentially a power-full number has paired digits that equal some perfect power corresponding to its position in the chain + 1.
Given that the magnitude of a power-full number and $n$ determines how powerful it is (the bigger the better), what is the most powerful number?
mathematics logical-deduction no-computers
mathematics logical-deduction no-computers
asked 9 hours ago
AdamAdam
4331221
asked 9 hours ago
AdamAdam
4331221
edited 9 hours ago
Adam
asked 9 hours ago
AdamAdam
4331221
asked 9 hours ago
AdamAdam
4331221
asked 9 hours ago
AdamAdam
4331221
4331221
$begingroup$
Feel free to help with formatting if there is an easier way to convey this puzzle
$endgroup$
– Adam
9 hours ago
1
$begingroup$
Do you mean kx are positive integers or is there a single k?
$endgroup$
– Jonathan Allan
9 hours ago
$begingroup$
Oops, fixed! @JonathanAllan
$endgroup$
– Adam
9 hours ago
$begingroup$
My edits were to make the spelling consistent; now you have both 'power-full' and 'powerful' in your title and body. Pedants may think the definitions are different ...
$endgroup$
– Glorfindel
8 hours ago
$begingroup$
@Veedrac $1le xlt n$ shouldn't include $n$. I thought it would be more confusing to have $1le xle n-1$
$endgroup$
– Adam
7 hours ago
|
show 2 more comments
$begingroup$
Feel free to help with formatting if there is an easier way to convey this puzzle
$endgroup$
– Adam
9 hours ago
1
$begingroup$
Do you mean kx are positive integers or is there a single k?
$endgroup$
– Jonathan Allan
9 hours ago
$begingroup$
Oops, fixed! @JonathanAllan
$endgroup$
– Adam
9 hours ago
$begingroup$
My edits were to make the spelling consistent; now you have both 'power-full' and 'powerful' in your title and body. Pedants may think the definitions are different ...
$endgroup$
– Glorfindel
8 hours ago
$begingroup$
@Veedrac $1le xlt n$ shouldn't include $n$. I thought it would be more confusing to have $1le xle n-1$
$endgroup$
– Adam
7 hours ago
$begingroup$
Feel free to help with formatting if there is an easier way to convey this puzzle
$endgroup$
– Adam
9 hours ago
$begingroup$
Feel free to help with formatting if there is an easier way to convey this puzzle
$endgroup$
– Adam
9 hours ago
1
1
$begingroup$
Do you mean kx are positive integers or is there a single k?
$endgroup$
– Jonathan Allan
9 hours ago
$begingroup$
Do you mean kx are positive integers or is there a single k?
$endgroup$
– Jonathan Allan
9 hours ago
$begingroup$
Oops, fixed! @JonathanAllan
$endgroup$
– Adam
9 hours ago
$begingroup$
Oops, fixed! @JonathanAllan
$endgroup$
– Adam
9 hours ago
$begingroup$
My edits were to make the spelling consistent; now you have both 'power-full' and 'powerful' in your title and body. Pedants may think the definitions are different ...
$endgroup$
– Glorfindel
8 hours ago
$begingroup$
My edits were to make the spelling consistent; now you have both 'power-full' and 'powerful' in your title and body. Pedants may think the definitions are different ...
$endgroup$
– Glorfindel
8 hours ago
$begingroup$
@Veedrac $1le xlt n$ shouldn't include $n$. I thought it would be more confusing to have $1le xle n-1$
$endgroup$
– Adam
7 hours ago
$begingroup$
@Veedrac $1le xlt n$ shouldn't include $n$. I thought it would be more confusing to have $1le xle n-1$
$endgroup$
– Adam
7 hours ago
|
show 2 more comments
1 Answer
1
active
oldest
votes
$begingroup$
I think the most powerful number is
1649, $n=4$ (or, if zeros are allowed: 01649, $n=5$)
It's powerful since
49 is a square, 64 a third power, 16 a fourth power (and 01 a fifth power).
In order to have a more powerful number
with $n=6$, it needs to start with two digits forming a sixth power, so either 01 or 64. However, there is no two-digit fifth power starting with 1 or 4; we have only 01 and 32.
Another $n=5$ powerful number without zeros should start with 32, but there's no two-digit fourth power starting with 2, so that's impossible too.
Another $n=4$ powerful number could start with 81, but there are no two-digit third powers starting with one. Similarly, there are no other two-digit third powers starting with 6 than 64, so no improvement can be made there either.
answered 9 hours ago
GlorfindelGlorfindel
15.4k45890
$endgroup$
1
$begingroup$
To complete the demonstration, you should add why you can't have the same n but the leading digit being the number of letters of the 5th word of your answer. (the reason is similar to what you already said)
$endgroup$
– Evargalo
8 hours ago
1
$begingroup$
@Evargalo you're right, thanks; updated
$endgroup$
– Glorfindel
8 hours ago
add a comment |
Your Answer
StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "559"
;
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: false,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: null,
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%2fpuzzling.stackexchange.com%2fquestions%2f84194%2fthe-most-powerful-number%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
$begingroup$
I think the most powerful number is
1649, $n=4$ (or, if zeros are allowed: 01649, $n=5$)
It's powerful since
49 is a square, 64 a third power, 16 a fourth power (and 01 a fifth power).
In order to have a more powerful number
with $n=6$, it needs to start with two digits forming a sixth power, so either 01 or 64. However, there is no two-digit fifth power starting with 1 or 4; we have only 01 and 32.
Another $n=5$ powerful number without zeros should start with 32, but there's no two-digit fourth power starting with 2, so that's impossible too.
Another $n=4$ powerful number could start with 81, but there are no two-digit third powers starting with one. Similarly, there are no other two-digit third powers starting with 6 than 64, so no improvement can be made there either.
answered 9 hours ago
GlorfindelGlorfindel
15.4k45890
$endgroup$
1
$begingroup$
To complete the demonstration, you should add why you can't have the same n but the leading digit being the number of letters of the 5th word of your answer. (the reason is similar to what you already said)
$endgroup$
– Evargalo
8 hours ago
1
$begingroup$
@Evargalo you're right, thanks; updated
$endgroup$
– Glorfindel
8 hours ago
add a comment |
$begingroup$
I think the most powerful number is
1649, $n=4$ (or, if zeros are allowed: 01649, $n=5$)
It's powerful since
49 is a square, 64 a third power, 16 a fourth power (and 01 a fifth power).
In order to have a more powerful number
with $n=6$, it needs to start with two digits forming a sixth power, so either 01 or 64. However, there is no two-digit fifth power starting with 1 or 4; we have only 01 and 32.
Another $n=5$ powerful number without zeros should start with 32, but there's no two-digit fourth power starting with 2, so that's impossible too.
Another $n=4$ powerful number could start with 81, but there are no two-digit third powers starting with one. Similarly, there are no other two-digit third powers starting with 6 than 64, so no improvement can be made there either.
answered 9 hours ago
GlorfindelGlorfindel
15.4k45890
$endgroup$
1
$begingroup$
To complete the demonstration, you should add why you can't have the same n but the leading digit being the number of letters of the 5th word of your answer. (the reason is similar to what you already said)
$endgroup$
– Evargalo
8 hours ago
1
$begingroup$
@Evargalo you're right, thanks; updated
$endgroup$
– Glorfindel
8 hours ago
add a comment |
$begingroup$
I think the most powerful number is
1649, $n=4$ (or, if zeros are allowed: 01649, $n=5$)
It's powerful since
49 is a square, 64 a third power, 16 a fourth power (and 01 a fifth power).
In order to have a more powerful number
with $n=6$, it needs to start with two digits forming a sixth power, so either 01 or 64. However, there is no two-digit fifth power starting with 1 or 4; we have only 01 and 32.
Another $n=5$ powerful number without zeros should start with 32, but there's no two-digit fourth power starting with 2, so that's impossible too.
Another $n=4$ powerful number could start with 81, but there are no two-digit third powers starting with one. Similarly, there are no other two-digit third powers starting with 6 than 64, so no improvement can be made there either.
answered 9 hours ago
GlorfindelGlorfindel
15.4k45890
$endgroup$
I think the most powerful number is
1649, $n=4$ (or, if zeros are allowed: 01649, $n=5$)
It's powerful since
49 is a square, 64 a third power, 16 a fourth power (and 01 a fifth power).
In order to have a more powerful number
with $n=6$, it needs to start with two digits forming a sixth power, so either 01 or 64. However, there is no two-digit fifth power starting with 1 or 4; we have only 01 and 32.
Another $n=5$ powerful number without zeros should start with 32, but there's no two-digit fourth power starting with 2, so that's impossible too.
Another $n=4$ powerful number could start with 81, but there are no two-digit third powers starting with one. Similarly, there are no other two-digit third powers starting with 6 than 64, so no improvement can be made there either.
answered 9 hours ago
GlorfindelGlorfindel
15.4k45890
edited 8 hours ago
answered 9 hours ago
GlorfindelGlorfindel
15.4k45890
answered 9 hours ago
GlorfindelGlorfindel
15.4k45890
answered 9 hours ago
GlorfindelGlorfindel
15.4k45890
15.4k45890
1
$begingroup$
To complete the demonstration, you should add why you can't have the same n but the leading digit being the number of letters of the 5th word of your answer. (the reason is similar to what you already said)
$endgroup$
– Evargalo
8 hours ago
1
$begingroup$
@Evargalo you're right, thanks; updated
$endgroup$
– Glorfindel
8 hours ago
add a comment |
1
$begingroup$
To complete the demonstration, you should add why you can't have the same n but the leading digit being the number of letters of the 5th word of your answer. (the reason is similar to what you already said)
$endgroup$
– Evargalo
8 hours ago
1
$begingroup$
@Evargalo you're right, thanks; updated
$endgroup$
– Glorfindel
8 hours ago
1
1
$begingroup$
To complete the demonstration, you should add why you can't have the same n but the leading digit being the number of letters of the 5th word of your answer. (the reason is similar to what you already said)
$endgroup$
– Evargalo
8 hours ago
$begingroup$
To complete the demonstration, you should add why you can't have the same n but the leading digit being the number of letters of the 5th word of your answer. (the reason is similar to what you already said)
$endgroup$
– Evargalo
8 hours ago
1
1
$begingroup$
@Evargalo you're right, thanks; updated
$endgroup$
– Glorfindel
8 hours ago
$begingroup$
@Evargalo you're right, thanks; updated
$endgroup$
– Glorfindel
8 hours ago
add a comment |
Thanks for contributing an answer to Puzzling 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%2fpuzzling.stackexchange.com%2fquestions%2f84194%2fthe-most-powerful-number%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
$begingroup$
Feel free to help with formatting if there is an easier way to convey this puzzle
$endgroup$
– Adam
9 hours ago
1
$begingroup$
Do you mean kx are positive integers or is there a single k?
$endgroup$
– Jonathan Allan
9 hours ago
$begingroup$
Oops, fixed! @JonathanAllan
$endgroup$
– Adam
9 hours ago
$begingroup$
My edits were to make the spelling consistent; now you have both 'power-full' and 'powerful' in your title and body. Pedants may think the definitions are different ...
$endgroup$
– Glorfindel
8 hours ago
$begingroup$
@Veedrac $1le xlt n$ shouldn't include $n$. I thought it would be more confusing to have $1le xle n-1$
$endgroup$
– Adam
7 hours ago