Expanding Factors
$begingroup$
So I recently thought of something. What if you take an integer (like 17 in this case), and find the set of numbers that add together in the closest form, of length 2, but increasing each level. So you would divide 17 into 8 and 9. But then those two numbers would be divided 3 times, producing 2, 3, 3 and 3, 3, 3 respectively. You go any number of levels until the resulting numbers are 0 or 1. Then you count the number of levels. This creates a sort of magnitude of the number. In this case, that magnitude is 3, because it required 3 levels to reach 1's and 0's.
Now the question is, is there any mathematical way of doing this? Any formula that could produce the magnitude of an integer?
mathematics
asked 3 hours ago
RigidityRigidity
483
New contributor
Rigidity is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
So I recently thought of something. What if you take an integer (like 17 in this case), and find the set of numbers that add together in the closest form, of length 2, but increasing each level. So you would divide 17 into 8 and 9. But then those two numbers would be divided 3 times, producing 2, 3, 3 and 3, 3, 3 respectively. You go any number of levels until the resulting numbers are 0 or 1. Then you count the number of levels. This creates a sort of magnitude of the number. In this case, that magnitude is 3, because it required 3 levels to reach 1's and 0's.
Now the question is, is there any mathematical way of doing this? Any formula that could produce the magnitude of an integer?
mathematics
asked 3 hours ago
RigidityRigidity
483
New contributor
Rigidity is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
1
$begingroup$
Probably more suited for math.se.
$endgroup$
– Zimonze
3 hours ago
$begingroup$
Can $17$ be reduced to any two numbers e.g. $6$ and $11$?
$endgroup$
– athin
2 hours ago
$begingroup$
Inverse factorial, rounded down
$endgroup$
– Dr Xorile
1 hour ago
$begingroup$
Actually rounded up and minus 1
$endgroup$
– Dr Xorile
1 hour ago
$begingroup$
@athin It says 'that add together in the closest form'. Bit obscure, but I think it means the set of numbers on the lower level can differ by at most 1.
$endgroup$
– ZanyG
1 hour ago
add a comment |
$begingroup$
So I recently thought of something. What if you take an integer (like 17 in this case), and find the set of numbers that add together in the closest form, of length 2, but increasing each level. So you would divide 17 into 8 and 9. But then those two numbers would be divided 3 times, producing 2, 3, 3 and 3, 3, 3 respectively. You go any number of levels until the resulting numbers are 0 or 1. Then you count the number of levels. This creates a sort of magnitude of the number. In this case, that magnitude is 3, because it required 3 levels to reach 1's and 0's.
Now the question is, is there any mathematical way of doing this? Any formula that could produce the magnitude of an integer?
mathematics
asked 3 hours ago
RigidityRigidity
483
New contributor
Rigidity is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
So I recently thought of something. What if you take an integer (like 17 in this case), and find the set of numbers that add together in the closest form, of length 2, but increasing each level. So you would divide 17 into 8 and 9. But then those two numbers would be divided 3 times, producing 2, 3, 3 and 3, 3, 3 respectively. You go any number of levels until the resulting numbers are 0 or 1. Then you count the number of levels. This creates a sort of magnitude of the number. In this case, that magnitude is 3, because it required 3 levels to reach 1's and 0's.
Now the question is, is there any mathematical way of doing this? Any formula that could produce the magnitude of an integer?
mathematics
mathematics
asked 3 hours ago
RigidityRigidity
483
New contributor
Rigidity is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 3 hours ago
RigidityRigidity
483
New contributor
Rigidity is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 3 hours ago
RigidityRigidity
483
New contributor
Rigidity is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 3 hours ago
RigidityRigidity
483
asked 3 hours ago
RigidityRigidity
483
483
New contributor
Rigidity is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Rigidity is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Rigidity is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
$begingroup$
Probably more suited for math.se.
$endgroup$
– Zimonze
3 hours ago
$begingroup$
Can $17$ be reduced to any two numbers e.g. $6$ and $11$?
$endgroup$
– athin
2 hours ago
$begingroup$
Inverse factorial, rounded down
$endgroup$
– Dr Xorile
1 hour ago
$begingroup$
Actually rounded up and minus 1
$endgroup$
– Dr Xorile
1 hour ago
$begingroup$
@athin It says 'that add together in the closest form'. Bit obscure, but I think it means the set of numbers on the lower level can differ by at most 1.
$endgroup$
– ZanyG
1 hour ago
add a comment |
1
$begingroup$
Probably more suited for math.se.
$endgroup$
– Zimonze
3 hours ago
$begingroup$
Can $17$ be reduced to any two numbers e.g. $6$ and $11$?
$endgroup$
– athin
2 hours ago
$begingroup$
Inverse factorial, rounded down
$endgroup$
– Dr Xorile
1 hour ago
$begingroup$
Actually rounded up and minus 1
$endgroup$
– Dr Xorile
1 hour ago
$begingroup$
@athin It says 'that add together in the closest form'. Bit obscure, but I think it means the set of numbers on the lower level can differ by at most 1.
$endgroup$
– ZanyG
1 hour ago
1
1
$begingroup$
Probably more suited for math.se.
$endgroup$
– Zimonze
3 hours ago
$begingroup$
Probably more suited for math.se.
$endgroup$
– Zimonze
3 hours ago
$begingroup$
Can $17$ be reduced to any two numbers e.g. $6$ and $11$?
$endgroup$
– athin
2 hours ago
$begingroup$
Can $17$ be reduced to any two numbers e.g. $6$ and $11$?
$endgroup$
– athin
2 hours ago
$begingroup$
Inverse factorial, rounded down
$endgroup$
– Dr Xorile
1 hour ago
$begingroup$
Inverse factorial, rounded down
$endgroup$
– Dr Xorile
1 hour ago
$begingroup$
Actually rounded up and minus 1
$endgroup$
– Dr Xorile
1 hour ago
$begingroup$
Actually rounded up and minus 1
$endgroup$
– Dr Xorile
1 hour ago
$begingroup$
@athin It says 'that add together in the closest form'. Bit obscure, but I think it means the set of numbers on the lower level can differ by at most 1.
$endgroup$
– ZanyG
1 hour ago
$begingroup$
@athin It says 'that add together in the closest form'. Bit obscure, but I think it means the set of numbers on the lower level can differ by at most 1.
$endgroup$
– ZanyG
1 hour ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
As Dr Xorile noted in the comments, all numbers, $k$, which are of magnitude $N$ satisfy
$N! <k leq (N+1)!$
Reasoning
Consider the largest number represented by a given magnitude $N$.
For this number, the bottom layer would consist entirely of groups of $N+1$ $1$s. The number of groups in this bottom layer is just $N!$ (since the $n$th layer has $n$ times more factors than the $(n-1)$th layer) and the sum of the numbers in the bottom layer is equal to the original number.
Hence, this largest number of magnitude $N$ is $k = (N+1)N! = (N+1)!$
Increasing this number by $1$ guarantees that we are forced to have at least one $2$ in the $(N+1)$th layer and so the magnitude must increase by $1$.
answered 1 hour ago
hexominohexomino
40.8k3122191
$endgroup$
add a comment |
$begingroup$
SPOILER ALERT
I have no idea how to put code in spoiler notation
Well, I put together a little java program:
public class NumberMagnitude {
public static void main(String args) {
int value = 17;
int k = value;
int magnitude = 2;
while(true) {
k=(k/magnitude);
(k<=1) { break; }
magnitude++;
}
System.out.println("The magnitude of " + k + " equals " + magnitude);
}
}
I'm not sure if this is exactly what you where looking for when you asked for a formula, but I figured it couldn't hurt.
You can modify the variable named "value" to test a different number; you could also create a fancier application that asks the user for input or used GUI, but I didn't have time for that.
answered 28 mins ago
Brandon_JBrandon_J
1,635228
$endgroup$
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: "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
});
}
});
Rigidity is a new contributor. Be nice, and check out our Code of Conduct.
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%2f79739%2fexpanding-factors%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
$begingroup$
As Dr Xorile noted in the comments, all numbers, $k$, which are of magnitude $N$ satisfy
$N! <k leq (N+1)!$
Reasoning
Consider the largest number represented by a given magnitude $N$.
For this number, the bottom layer would consist entirely of groups of $N+1$ $1$s. The number of groups in this bottom layer is just $N!$ (since the $n$th layer has $n$ times more factors than the $(n-1)$th layer) and the sum of the numbers in the bottom layer is equal to the original number.
Hence, this largest number of magnitude $N$ is $k = (N+1)N! = (N+1)!$
Increasing this number by $1$ guarantees that we are forced to have at least one $2$ in the $(N+1)$th layer and so the magnitude must increase by $1$.
answered 1 hour ago
hexominohexomino
40.8k3122191
$endgroup$
add a comment |
$begingroup$
As Dr Xorile noted in the comments, all numbers, $k$, which are of magnitude $N$ satisfy
$N! <k leq (N+1)!$
Reasoning
Consider the largest number represented by a given magnitude $N$.
For this number, the bottom layer would consist entirely of groups of $N+1$ $1$s. The number of groups in this bottom layer is just $N!$ (since the $n$th layer has $n$ times more factors than the $(n-1)$th layer) and the sum of the numbers in the bottom layer is equal to the original number.
Hence, this largest number of magnitude $N$ is $k = (N+1)N! = (N+1)!$
Increasing this number by $1$ guarantees that we are forced to have at least one $2$ in the $(N+1)$th layer and so the magnitude must increase by $1$.
answered 1 hour ago
hexominohexomino
40.8k3122191
$endgroup$
add a comment |
$begingroup$
As Dr Xorile noted in the comments, all numbers, $k$, which are of magnitude $N$ satisfy
$N! <k leq (N+1)!$
Reasoning
Consider the largest number represented by a given magnitude $N$.
For this number, the bottom layer would consist entirely of groups of $N+1$ $1$s. The number of groups in this bottom layer is just $N!$ (since the $n$th layer has $n$ times more factors than the $(n-1)$th layer) and the sum of the numbers in the bottom layer is equal to the original number.
Hence, this largest number of magnitude $N$ is $k = (N+1)N! = (N+1)!$
Increasing this number by $1$ guarantees that we are forced to have at least one $2$ in the $(N+1)$th layer and so the magnitude must increase by $1$.
answered 1 hour ago
hexominohexomino
40.8k3122191
$endgroup$
As Dr Xorile noted in the comments, all numbers, $k$, which are of magnitude $N$ satisfy
$N! <k leq (N+1)!$
Reasoning
Consider the largest number represented by a given magnitude $N$.
For this number, the bottom layer would consist entirely of groups of $N+1$ $1$s. The number of groups in this bottom layer is just $N!$ (since the $n$th layer has $n$ times more factors than the $(n-1)$th layer) and the sum of the numbers in the bottom layer is equal to the original number.
Hence, this largest number of magnitude $N$ is $k = (N+1)N! = (N+1)!$
Increasing this number by $1$ guarantees that we are forced to have at least one $2$ in the $(N+1)$th layer and so the magnitude must increase by $1$.
answered 1 hour ago
hexominohexomino
40.8k3122191
answered 1 hour ago
hexominohexomino
40.8k3122191
answered 1 hour ago
hexominohexomino
40.8k3122191
answered 1 hour ago
hexominohexomino
40.8k3122191
40.8k3122191
add a comment |
add a comment |
$begingroup$
SPOILER ALERT
I have no idea how to put code in spoiler notation
Well, I put together a little java program:
public class NumberMagnitude {
public static void main(String args) {
int value = 17;
int k = value;
int magnitude = 2;
while(true) {
k=(k/magnitude);
(k<=1) { break; }
magnitude++;
}
System.out.println("The magnitude of " + k + " equals " + magnitude);
}
}
I'm not sure if this is exactly what you where looking for when you asked for a formula, but I figured it couldn't hurt.
You can modify the variable named "value" to test a different number; you could also create a fancier application that asks the user for input or used GUI, but I didn't have time for that.
answered 28 mins ago
Brandon_JBrandon_J
1,635228
$endgroup$
add a comment |
$begingroup$
SPOILER ALERT
I have no idea how to put code in spoiler notation
Well, I put together a little java program:
public class NumberMagnitude {
public static void main(String args) {
int value = 17;
int k = value;
int magnitude = 2;
while(true) {
k=(k/magnitude);
(k<=1) { break; }
magnitude++;
}
System.out.println("The magnitude of " + k + " equals " + magnitude);
}
}
I'm not sure if this is exactly what you where looking for when you asked for a formula, but I figured it couldn't hurt.
You can modify the variable named "value" to test a different number; you could also create a fancier application that asks the user for input or used GUI, but I didn't have time for that.
answered 28 mins ago
Brandon_JBrandon_J
1,635228
$endgroup$
add a comment |
$begingroup$
SPOILER ALERT
I have no idea how to put code in spoiler notation
Well, I put together a little java program:
public class NumberMagnitude {
public static void main(String args) {
int value = 17;
int k = value;
int magnitude = 2;
while(true) {
k=(k/magnitude);
(k<=1) { break; }
magnitude++;
}
System.out.println("The magnitude of " + k + " equals " + magnitude);
}
}
I'm not sure if this is exactly what you where looking for when you asked for a formula, but I figured it couldn't hurt.
You can modify the variable named "value" to test a different number; you could also create a fancier application that asks the user for input or used GUI, but I didn't have time for that.
answered 28 mins ago
Brandon_JBrandon_J
1,635228
$endgroup$
SPOILER ALERT
I have no idea how to put code in spoiler notation
Well, I put together a little java program:
public class NumberMagnitude {
public static void main(String args) {
int value = 17;
int k = value;
int magnitude = 2;
while(true) {
k=(k/magnitude);
(k<=1) { break; }
magnitude++;
}
System.out.println("The magnitude of " + k + " equals " + magnitude);
}
}
I'm not sure if this is exactly what you where looking for when you asked for a formula, but I figured it couldn't hurt.
You can modify the variable named "value" to test a different number; you could also create a fancier application that asks the user for input or used GUI, but I didn't have time for that.
answered 28 mins ago
Brandon_JBrandon_J
1,635228
edited 17 mins ago
answered 28 mins ago
Brandon_JBrandon_J
1,635228
answered 28 mins ago
Brandon_JBrandon_J
1,635228
answered 28 mins ago
Brandon_JBrandon_J
1,635228
1,635228
add a comment |
add a comment |
Rigidity is a new contributor. Be nice, and check out our Code of Conduct.
Rigidity is a new contributor. Be nice, and check out our Code of Conduct.
Rigidity is a new contributor. Be nice, and check out our Code of Conduct.
Rigidity is a new contributor. Be nice, and check out our Code of Conduct.
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%2f79739%2fexpanding-factors%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
$begingroup$
Probably more suited for math.se.
$endgroup$
– Zimonze
3 hours ago
$begingroup$
Can $17$ be reduced to any two numbers e.g. $6$ and $11$?
$endgroup$
– athin
2 hours ago
$begingroup$
Inverse factorial, rounded down
$endgroup$
– Dr Xorile
1 hour ago
$begingroup$
Actually rounded up and minus 1
$endgroup$
– Dr Xorile
1 hour ago
$begingroup$
@athin It says 'that add together in the closest form'. Bit obscure, but I think it means the set of numbers on the lower level can differ by at most 1.
$endgroup$
– ZanyG
1 hour ago