Does the sum of digits of prime number written in base7 never divide by 3? The 2019 Stack Overflow Developer Survey Results Are InIs there a largest “nested” prime number?What prime numbers have the sum of their digits as a prime number?Whether any even number can be written as sum of odd no of primes?Can you make any number prime by adding some digits to the right?How often is a sum of $k$ consecutive primes also prime?Possible Prime Sum Pattern (Amateur)Proof: 1007 can not be written as the sum of two primes.Is this true for every prime $p>2$ , if$ m$ is even integer number then $m$ can't be written as :$m=prod _i=1^rp_i^a_i$?Weird Prime Number DiscoveryPrime Gap number runs
Does HR tell a hiring manager about salary negotiations?
Is it okay to consider publishing in my first year of PhD?
Keeping a retro style to sci-fi spaceships?
Why can't devices on different VLANs, but on the same subnet, communicate?
Geography at the pixel level
Accepted by European university, rejected by all American ones I applied to? Possible reasons?
For what reasons would an animal species NOT cross a *horizontal* land bridge?
Falsification in Math vs Science
Getting crown tickets for Statue of Liberty
Relationship between Gromov-Witten and Taubes' Gromov invariant
Straighten subgroup lattice
Output the Arecibo Message
Does adding complexity mean a more secure cipher?
How do you keep chess fun when your opponent constantly beats you?
Why doesn't shell automatically fix "useless use of cat"?
Can you cast a spell on someone in the Ethereal Plane, if you are on the Material Plane and have the True Seeing spell active?
Why couldn't they take pictures of a closer black hole?
Likelihood that a superbug or lethal virus could come from a landfill
APIPA and LAN Broadcast Domain
What is the meaning of Triage in Cybersec world?
How to charge AirPods to keep battery healthy?
Why “相同意思的词” is called “同义词” instead of "同意词"?
writing variables above the numbers in tikz picture
Why does the nucleus not repel itself?
Does the sum of digits of prime number written in base7 never divide by 3?
The 2019 Stack Overflow Developer Survey Results Are InIs there a largest “nested” prime number?What prime numbers have the sum of their digits as a prime number?Whether any even number can be written as sum of odd no of primes?Can you make any number prime by adding some digits to the right?How often is a sum of $k$ consecutive primes also prime?Possible Prime Sum Pattern (Amateur)Proof: 1007 can not be written as the sum of two primes.Is this true for every prime $p>2$ , if$ m$ is even integer number then $m$ can't be written as :$m=prod _i=1^rp_i^a_i$?Weird Prime Number DiscoveryPrime Gap number runs
$begingroup$
I have tested all the primes up to 50,000,000 and did not find a single prime which satisfies the condition "sum of digits of prime number written in base7 divides by 3". E.g.
- 13 (Base10) = 16 (Base7) --> 7 (sum of digits in base 7)
- 1021 (Base10) = 2656 (Base7) --> 19
- 823541 (Base10) = 6666665 (Base7) --> 41
- 46941953 (Base10) = 1110000002 (Base7) --> 5
Here you can see the distribution of sums in base 7:
http://s12.postimg.org/lcf3tntzx/prime_sum_in_base7_distribution.png
- COUNT(*) - the number of occurrences
- SUM7 - sum of digits in base7
- MIN(PRIME) - minimal prime in base10
- MAX(PRIME) - maximal prime in base10
As you can see sum7 of 9, 15, 21, 27, 33 are missing in the list, though other valid sums are widely represented. By 'valid sum' I mean that sum must be odd, because of "In an odd base, a number is odd if and only if it has an odd number of odd digits."
So what is the least prime whose sum of digits written in base7 divide by 3? Or is it possible to prove that all primes have such a feature?
prime-numbers
$endgroup$
add a comment |
$begingroup$
I have tested all the primes up to 50,000,000 and did not find a single prime which satisfies the condition "sum of digits of prime number written in base7 divides by 3". E.g.
- 13 (Base10) = 16 (Base7) --> 7 (sum of digits in base 7)
- 1021 (Base10) = 2656 (Base7) --> 19
- 823541 (Base10) = 6666665 (Base7) --> 41
- 46941953 (Base10) = 1110000002 (Base7) --> 5
Here you can see the distribution of sums in base 7:
http://s12.postimg.org/lcf3tntzx/prime_sum_in_base7_distribution.png
- COUNT(*) - the number of occurrences
- SUM7 - sum of digits in base7
- MIN(PRIME) - minimal prime in base10
- MAX(PRIME) - maximal prime in base10
As you can see sum7 of 9, 15, 21, 27, 33 are missing in the list, though other valid sums are widely represented. By 'valid sum' I mean that sum must be odd, because of "In an odd base, a number is odd if and only if it has an odd number of odd digits."
So what is the least prime whose sum of digits written in base7 divide by 3? Or is it possible to prove that all primes have such a feature?
prime-numbers
$endgroup$
add a comment |
$begingroup$
I have tested all the primes up to 50,000,000 and did not find a single prime which satisfies the condition "sum of digits of prime number written in base7 divides by 3". E.g.
- 13 (Base10) = 16 (Base7) --> 7 (sum of digits in base 7)
- 1021 (Base10) = 2656 (Base7) --> 19
- 823541 (Base10) = 6666665 (Base7) --> 41
- 46941953 (Base10) = 1110000002 (Base7) --> 5
Here you can see the distribution of sums in base 7:
http://s12.postimg.org/lcf3tntzx/prime_sum_in_base7_distribution.png
- COUNT(*) - the number of occurrences
- SUM7 - sum of digits in base7
- MIN(PRIME) - minimal prime in base10
- MAX(PRIME) - maximal prime in base10
As you can see sum7 of 9, 15, 21, 27, 33 are missing in the list, though other valid sums are widely represented. By 'valid sum' I mean that sum must be odd, because of "In an odd base, a number is odd if and only if it has an odd number of odd digits."
So what is the least prime whose sum of digits written in base7 divide by 3? Or is it possible to prove that all primes have such a feature?
prime-numbers
$endgroup$
I have tested all the primes up to 50,000,000 and did not find a single prime which satisfies the condition "sum of digits of prime number written in base7 divides by 3". E.g.
- 13 (Base10) = 16 (Base7) --> 7 (sum of digits in base 7)
- 1021 (Base10) = 2656 (Base7) --> 19
- 823541 (Base10) = 6666665 (Base7) --> 41
- 46941953 (Base10) = 1110000002 (Base7) --> 5
Here you can see the distribution of sums in base 7:
http://s12.postimg.org/lcf3tntzx/prime_sum_in_base7_distribution.png
- COUNT(*) - the number of occurrences
- SUM7 - sum of digits in base7
- MIN(PRIME) - minimal prime in base10
- MAX(PRIME) - maximal prime in base10
As you can see sum7 of 9, 15, 21, 27, 33 are missing in the list, though other valid sums are widely represented. By 'valid sum' I mean that sum must be odd, because of "In an odd base, a number is odd if and only if it has an odd number of odd digits."
So what is the least prime whose sum of digits written in base7 divide by 3? Or is it possible to prove that all primes have such a feature?
prime-numbers
prime-numbers
asked Nov 4 '15 at 8:12
BuchasBuchas
432
432
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Let $a = sum_i = 0^n a_i 7^i$ be a number written in base $7$, that is, $0 leq a_i leq 6$. Note that $7^i = (1 + 3 cdot 2)^i = 1 + 3 b_i$ for some $b_i geq 0$. Hence $a = sum_i = 0^n a_i + 3 sum_i = 0^n a_i b_i$ is divisible by $3$ if and only if its sum of digits is divisible by $3$. Hence the sum of digits in base $7$ of any prime larger than $3$ will not be divisible by $3$.
$endgroup$
add a comment |
$begingroup$
You know: If and only if a number can be devided by three than the sum of its digits to the base of then can be devided by three. Therefore each prime number larger than three has a sum of digits which could not be devided by three.
So what is about the base of $7$:
Now when switching from base $10$ to base $7$ you each increase one digit by a multiple of $3$. Therefore the sum of digits modulo three keeps the same. And the statement of above holds for the base $7$ as well.
Alternativly see the multiples of three in the base of $7$
$$3_7, 6_7, 12_7, 15_7, 21_7, 24_7, 30_7, 33_7, 36_7, 42_7,...$$
Each time you "overflow the digit" you have to increase the next digit by $1$ and decrease the current by $4$. So the sum of digits increases by $3$.
$endgroup$
add a comment |
$begingroup$
Do you see that in decimal the divisibility of 9 and 3 is checked by divisibility of sum of the digits for 9 or 3? Similarly you can check divisibility of 15, 5 and 3 in hexadecimal, 7 in octal, 3 in base 4 and 2, 3, 6 in base 7 because this is true for all factors of $b-1$ in base $b$.
Let's say we have a number $N_b$ in base b
$beginaligned
N_b &= overlinea_na_n-1 dots a_1a_0_b = a_nb^n + a_n-1b^n-1 + dots + a_1b^1 + a_0b^0 \
&= (a_n + a_n-1 + dots + a_0) + Big[ a_n(b^n-1) + a_n-1(b^n-1-1) + dots + a_2(b^2-1) + a_1(b-1) Big]
endaligned$
The latter part has $b^k-1$ in each term, thus is divisible by $b-1$. Hence it's also divisible by all factors of $b-1$. As a result we just need to check the first part which is the sum of all digits to see if it's divisible by that factor of $b-1$ or not
So in base 7 a number is divisible by 6/3/2 if and only if the sum of the digits is divisible by 6/3/2 respectively. Therefore a prime in base 7 can't have its digits accumulated to a multiple of 3.
$endgroup$
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%2f1512515%2fdoes-the-sum-of-digits-of-prime-number-written-in-base7-never-divide-by-3%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let $a = sum_i = 0^n a_i 7^i$ be a number written in base $7$, that is, $0 leq a_i leq 6$. Note that $7^i = (1 + 3 cdot 2)^i = 1 + 3 b_i$ for some $b_i geq 0$. Hence $a = sum_i = 0^n a_i + 3 sum_i = 0^n a_i b_i$ is divisible by $3$ if and only if its sum of digits is divisible by $3$. Hence the sum of digits in base $7$ of any prime larger than $3$ will not be divisible by $3$.
$endgroup$
add a comment |
$begingroup$
Let $a = sum_i = 0^n a_i 7^i$ be a number written in base $7$, that is, $0 leq a_i leq 6$. Note that $7^i = (1 + 3 cdot 2)^i = 1 + 3 b_i$ for some $b_i geq 0$. Hence $a = sum_i = 0^n a_i + 3 sum_i = 0^n a_i b_i$ is divisible by $3$ if and only if its sum of digits is divisible by $3$. Hence the sum of digits in base $7$ of any prime larger than $3$ will not be divisible by $3$.
$endgroup$
add a comment |
$begingroup$
Let $a = sum_i = 0^n a_i 7^i$ be a number written in base $7$, that is, $0 leq a_i leq 6$. Note that $7^i = (1 + 3 cdot 2)^i = 1 + 3 b_i$ for some $b_i geq 0$. Hence $a = sum_i = 0^n a_i + 3 sum_i = 0^n a_i b_i$ is divisible by $3$ if and only if its sum of digits is divisible by $3$. Hence the sum of digits in base $7$ of any prime larger than $3$ will not be divisible by $3$.
$endgroup$
Let $a = sum_i = 0^n a_i 7^i$ be a number written in base $7$, that is, $0 leq a_i leq 6$. Note that $7^i = (1 + 3 cdot 2)^i = 1 + 3 b_i$ for some $b_i geq 0$. Hence $a = sum_i = 0^n a_i + 3 sum_i = 0^n a_i b_i$ is divisible by $3$ if and only if its sum of digits is divisible by $3$. Hence the sum of digits in base $7$ of any prime larger than $3$ will not be divisible by $3$.
answered Nov 4 '15 at 8:32
Matthias KlupschMatthias Klupsch
6,2291227
6,2291227
add a comment |
add a comment |
$begingroup$
You know: If and only if a number can be devided by three than the sum of its digits to the base of then can be devided by three. Therefore each prime number larger than three has a sum of digits which could not be devided by three.
So what is about the base of $7$:
Now when switching from base $10$ to base $7$ you each increase one digit by a multiple of $3$. Therefore the sum of digits modulo three keeps the same. And the statement of above holds for the base $7$ as well.
Alternativly see the multiples of three in the base of $7$
$$3_7, 6_7, 12_7, 15_7, 21_7, 24_7, 30_7, 33_7, 36_7, 42_7,...$$
Each time you "overflow the digit" you have to increase the next digit by $1$ and decrease the current by $4$. So the sum of digits increases by $3$.
$endgroup$
add a comment |
$begingroup$
You know: If and only if a number can be devided by three than the sum of its digits to the base of then can be devided by three. Therefore each prime number larger than three has a sum of digits which could not be devided by three.
So what is about the base of $7$:
Now when switching from base $10$ to base $7$ you each increase one digit by a multiple of $3$. Therefore the sum of digits modulo three keeps the same. And the statement of above holds for the base $7$ as well.
Alternativly see the multiples of three in the base of $7$
$$3_7, 6_7, 12_7, 15_7, 21_7, 24_7, 30_7, 33_7, 36_7, 42_7,...$$
Each time you "overflow the digit" you have to increase the next digit by $1$ and decrease the current by $4$. So the sum of digits increases by $3$.
$endgroup$
add a comment |
$begingroup$
You know: If and only if a number can be devided by three than the sum of its digits to the base of then can be devided by three. Therefore each prime number larger than three has a sum of digits which could not be devided by three.
So what is about the base of $7$:
Now when switching from base $10$ to base $7$ you each increase one digit by a multiple of $3$. Therefore the sum of digits modulo three keeps the same. And the statement of above holds for the base $7$ as well.
Alternativly see the multiples of three in the base of $7$
$$3_7, 6_7, 12_7, 15_7, 21_7, 24_7, 30_7, 33_7, 36_7, 42_7,...$$
Each time you "overflow the digit" you have to increase the next digit by $1$ and decrease the current by $4$. So the sum of digits increases by $3$.
$endgroup$
You know: If and only if a number can be devided by three than the sum of its digits to the base of then can be devided by three. Therefore each prime number larger than three has a sum of digits which could not be devided by three.
So what is about the base of $7$:
Now when switching from base $10$ to base $7$ you each increase one digit by a multiple of $3$. Therefore the sum of digits modulo three keeps the same. And the statement of above holds for the base $7$ as well.
Alternativly see the multiples of three in the base of $7$
$$3_7, 6_7, 12_7, 15_7, 21_7, 24_7, 30_7, 33_7, 36_7, 42_7,...$$
Each time you "overflow the digit" you have to increase the next digit by $1$ and decrease the current by $4$. So the sum of digits increases by $3$.
edited Nov 4 '15 at 8:32
answered Nov 4 '15 at 8:26
MatthiasMatthias
2,023514
2,023514
add a comment |
add a comment |
$begingroup$
Do you see that in decimal the divisibility of 9 and 3 is checked by divisibility of sum of the digits for 9 or 3? Similarly you can check divisibility of 15, 5 and 3 in hexadecimal, 7 in octal, 3 in base 4 and 2, 3, 6 in base 7 because this is true for all factors of $b-1$ in base $b$.
Let's say we have a number $N_b$ in base b
$beginaligned
N_b &= overlinea_na_n-1 dots a_1a_0_b = a_nb^n + a_n-1b^n-1 + dots + a_1b^1 + a_0b^0 \
&= (a_n + a_n-1 + dots + a_0) + Big[ a_n(b^n-1) + a_n-1(b^n-1-1) + dots + a_2(b^2-1) + a_1(b-1) Big]
endaligned$
The latter part has $b^k-1$ in each term, thus is divisible by $b-1$. Hence it's also divisible by all factors of $b-1$. As a result we just need to check the first part which is the sum of all digits to see if it's divisible by that factor of $b-1$ or not
So in base 7 a number is divisible by 6/3/2 if and only if the sum of the digits is divisible by 6/3/2 respectively. Therefore a prime in base 7 can't have its digits accumulated to a multiple of 3.
$endgroup$
add a comment |
$begingroup$
Do you see that in decimal the divisibility of 9 and 3 is checked by divisibility of sum of the digits for 9 or 3? Similarly you can check divisibility of 15, 5 and 3 in hexadecimal, 7 in octal, 3 in base 4 and 2, 3, 6 in base 7 because this is true for all factors of $b-1$ in base $b$.
Let's say we have a number $N_b$ in base b
$beginaligned
N_b &= overlinea_na_n-1 dots a_1a_0_b = a_nb^n + a_n-1b^n-1 + dots + a_1b^1 + a_0b^0 \
&= (a_n + a_n-1 + dots + a_0) + Big[ a_n(b^n-1) + a_n-1(b^n-1-1) + dots + a_2(b^2-1) + a_1(b-1) Big]
endaligned$
The latter part has $b^k-1$ in each term, thus is divisible by $b-1$. Hence it's also divisible by all factors of $b-1$. As a result we just need to check the first part which is the sum of all digits to see if it's divisible by that factor of $b-1$ or not
So in base 7 a number is divisible by 6/3/2 if and only if the sum of the digits is divisible by 6/3/2 respectively. Therefore a prime in base 7 can't have its digits accumulated to a multiple of 3.
$endgroup$
add a comment |
$begingroup$
Do you see that in decimal the divisibility of 9 and 3 is checked by divisibility of sum of the digits for 9 or 3? Similarly you can check divisibility of 15, 5 and 3 in hexadecimal, 7 in octal, 3 in base 4 and 2, 3, 6 in base 7 because this is true for all factors of $b-1$ in base $b$.
Let's say we have a number $N_b$ in base b
$beginaligned
N_b &= overlinea_na_n-1 dots a_1a_0_b = a_nb^n + a_n-1b^n-1 + dots + a_1b^1 + a_0b^0 \
&= (a_n + a_n-1 + dots + a_0) + Big[ a_n(b^n-1) + a_n-1(b^n-1-1) + dots + a_2(b^2-1) + a_1(b-1) Big]
endaligned$
The latter part has $b^k-1$ in each term, thus is divisible by $b-1$. Hence it's also divisible by all factors of $b-1$. As a result we just need to check the first part which is the sum of all digits to see if it's divisible by that factor of $b-1$ or not
So in base 7 a number is divisible by 6/3/2 if and only if the sum of the digits is divisible by 6/3/2 respectively. Therefore a prime in base 7 can't have its digits accumulated to a multiple of 3.
$endgroup$
Do you see that in decimal the divisibility of 9 and 3 is checked by divisibility of sum of the digits for 9 or 3? Similarly you can check divisibility of 15, 5 and 3 in hexadecimal, 7 in octal, 3 in base 4 and 2, 3, 6 in base 7 because this is true for all factors of $b-1$ in base $b$.
Let's say we have a number $N_b$ in base b
$beginaligned
N_b &= overlinea_na_n-1 dots a_1a_0_b = a_nb^n + a_n-1b^n-1 + dots + a_1b^1 + a_0b^0 \
&= (a_n + a_n-1 + dots + a_0) + Big[ a_n(b^n-1) + a_n-1(b^n-1-1) + dots + a_2(b^2-1) + a_1(b-1) Big]
endaligned$
The latter part has $b^k-1$ in each term, thus is divisible by $b-1$. Hence it's also divisible by all factors of $b-1$. As a result we just need to check the first part which is the sum of all digits to see if it's divisible by that factor of $b-1$ or not
So in base 7 a number is divisible by 6/3/2 if and only if the sum of the digits is divisible by 6/3/2 respectively. Therefore a prime in base 7 can't have its digits accumulated to a multiple of 3.
edited Mar 24 at 3:43
answered Feb 29 '16 at 10:50
phuclvphuclv
1156
1156
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%2f1512515%2fdoes-the-sum-of-digits-of-prime-number-written-in-base7-never-divide-by-3%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