Mixing
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Number theory involving sum of digits of a number:
$begingroup$
Let a fixed natural number m be given Call a positive integer n to be a GRT number iff:
$n equiv 1 pmod m$
Sum of digits in decimal representation of $n^2$ is greater than or equal to sum of digits in decimal representation of $n$.
How many GRT numbers are there ?
elementary-number-theory
edited Feb 1 at 17:56
Arturo Magidin
266k34591921
asked Feb 1 at 17:30
Mayank MishraMayank Mishra
605
$endgroup$
add a comment |
$begingroup$
Let a fixed natural number m be given Call a positive integer n to be a GRT number iff:
$n equiv 1 pmod m$
Sum of digits in decimal representation of $n^2$ is greater than or equal to sum of digits in decimal representation of $n$.
How many GRT numbers are there ?
elementary-number-theory
edited Feb 1 at 17:56
Arturo Magidin
266k34591921
asked Feb 1 at 17:30
Mayank MishraMayank Mishra
605
$endgroup$
4
$begingroup$
Please fix the question and please never type something like$2. Sum of digits in decimal representation of n^2 is greater than or equal to sum of digits in decimal representation of n.$again.
$endgroup$
– dan_fulea
Feb 1 at 17:34
add a comment |
$begingroup$
Let a fixed natural number m be given Call a positive integer n to be a GRT number iff:
$n equiv 1 pmod m$
Sum of digits in decimal representation of $n^2$ is greater than or equal to sum of digits in decimal representation of $n$.
How many GRT numbers are there ?
elementary-number-theory
edited Feb 1 at 17:56
Arturo Magidin
266k34591921
asked Feb 1 at 17:30
Mayank MishraMayank Mishra
605
$endgroup$
Let a fixed natural number m be given Call a positive integer n to be a GRT number iff:
$n equiv 1 pmod m$
Sum of digits in decimal representation of $n^2$ is greater than or equal to sum of digits in decimal representation of $n$.
How many GRT numbers are there ?
elementary-number-theory
elementary-number-theory
edited Feb 1 at 17:56
Arturo Magidin
266k34591921
asked Feb 1 at 17:30
Mayank MishraMayank Mishra
605
edited Feb 1 at 17:56
Arturo Magidin
266k34591921
asked Feb 1 at 17:30
Mayank MishraMayank Mishra
605
edited Feb 1 at 17:56
Arturo Magidin
266k34591921
edited Feb 1 at 17:56
Arturo Magidin
266k34591921
edited Feb 1 at 17:56
Arturo Magidin
266k34591921
266k34591921
asked Feb 1 at 17:30
Mayank MishraMayank Mishra
605
asked Feb 1 at 17:30
Mayank MishraMayank Mishra
605
asked Feb 1 at 17:30
Mayank MishraMayank Mishra
605
605
4
$begingroup$
Please fix the question and please never type something like$2. Sum of digits in decimal representation of n^2 is greater than or equal to sum of digits in decimal representation of n.$again.
$endgroup$
– dan_fulea
Feb 1 at 17:34
add a comment |
4
$begingroup$
Please fix the question and please never type something like$2. Sum of digits in decimal representation of n^2 is greater than or equal to sum of digits in decimal representation of n.$again.
$endgroup$
– dan_fulea
Feb 1 at 17:34
4
4
$begingroup$
Please fix the question and please never type something like
$2. Sum of digits in decimal representation of n^2 is greater than or equal to sum of digits in decimal representation of n.$ again.$endgroup$
– dan_fulea
Feb 1 at 17:34
$begingroup$
Please fix the question and please never type something like
$2. Sum of digits in decimal representation of n^2 is greater than or equal to sum of digits in decimal representation of n.$ again.$endgroup$
– dan_fulea
Feb 1 at 17:34
add a comment |
1 Answer
1
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$begingroup$
I can't prove it absolutely, but for any $m$ there must be infinitely many. Any large $n$ that is $equiv 1 bmod m$ will have a square with lots more digits than $n$ has, so the sum of digits of the square will be larger.
answered Feb 1 at 18:06
Ross MillikanRoss Millikan
301k24200375
$endgroup$
add a comment |
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$begingroup$
I can't prove it absolutely, but for any $m$ there must be infinitely many. Any large $n$ that is $equiv 1 bmod m$ will have a square with lots more digits than $n$ has, so the sum of digits of the square will be larger.
answered Feb 1 at 18:06
Ross MillikanRoss Millikan
301k24200375
$endgroup$
add a comment |
$begingroup$
I can't prove it absolutely, but for any $m$ there must be infinitely many. Any large $n$ that is $equiv 1 bmod m$ will have a square with lots more digits than $n$ has, so the sum of digits of the square will be larger.
answered Feb 1 at 18:06
Ross MillikanRoss Millikan
301k24200375
$endgroup$
add a comment |
$begingroup$
I can't prove it absolutely, but for any $m$ there must be infinitely many. Any large $n$ that is $equiv 1 bmod m$ will have a square with lots more digits than $n$ has, so the sum of digits of the square will be larger.
answered Feb 1 at 18:06
Ross MillikanRoss Millikan
301k24200375
$endgroup$
I can't prove it absolutely, but for any $m$ there must be infinitely many. Any large $n$ that is $equiv 1 bmod m$ will have a square with lots more digits than $n$ has, so the sum of digits of the square will be larger.
answered Feb 1 at 18:06
Ross MillikanRoss Millikan
301k24200375
answered Feb 1 at 18:06
Ross MillikanRoss Millikan
301k24200375
answered Feb 1 at 18:06
Ross MillikanRoss Millikan
301k24200375
answered Feb 1 at 18:06
Ross MillikanRoss Millikan
301k24200375
301k24200375
add a comment |
add a comment |
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$begingroup$
Please fix the question and please never type something like
$2. Sum of digits in decimal representation of n^2 is greater than or equal to sum of digits in decimal representation of n.$again.$endgroup$
– dan_fulea
Feb 1 at 17:34