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Merge numbers to a new number and find digit on position (n)
Brackets in output make unable to use output to identify matrix elementWhat does the slash-colon symbol do?Splicing a list of arguments into a function with SequenceBincounting a listimport a list of replacement rulesFinding a number which has the same remainder when dividing certain other numbersNegative accuracy numerics ( 0``-128 notation )Please explain the following code containing Partition and FlattenDifference between f[x] and f as arguments of functionsEvaluating an expression on one line
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty
margin-bottom:0;
.everyonelovesstackoverflowposition:absolute;height:1px;width:1px;opacity:0;top:0;left:0;pointer-events:none;
$begingroup$
I'm trying to construct a 'Mathematica solution' for the following 'puzzle'
The natural numbers are placed in a row;
12345678910111213...
What digit is the 10,000th digit?
I don't know the syntax how to form this number, or to extract digit $n$ (in the example $n=10000$) in Mathematica. Anybody that can help? TIA.
syntax
asked Oct 16 at 18:14
mf67mf67
6091 silver badge6 bronze badges
$endgroup$
add a comment
|
$begingroup$
I'm trying to construct a 'Mathematica solution' for the following 'puzzle'
The natural numbers are placed in a row;
12345678910111213...
What digit is the 10,000th digit?
I don't know the syntax how to form this number, or to extract digit $n$ (in the example $n=10000$) in Mathematica. Anybody that can help? TIA.
syntax
asked Oct 16 at 18:14
mf67mf67
6091 silver badge6 bronze badges
$endgroup$
$begingroup$
Take a look at the documentation ofIntegerDigits, that should solve your problem. To get enough digits, you can either try to work out exactly how many numbers you need to concatenate, or you could also just take enough (i.e. 10000) to make sure
$endgroup$
– Lukas Lang
Oct 16 at 18:16
$begingroup$
Flatten[IntegerDigits /@ Range[5000]][[10000]]
$endgroup$
– OkkesDulgerci
Oct 16 at 19:47
add a comment
|
$begingroup$
I'm trying to construct a 'Mathematica solution' for the following 'puzzle'
The natural numbers are placed in a row;
12345678910111213...
What digit is the 10,000th digit?
I don't know the syntax how to form this number, or to extract digit $n$ (in the example $n=10000$) in Mathematica. Anybody that can help? TIA.
syntax
asked Oct 16 at 18:14
mf67mf67
6091 silver badge6 bronze badges
$endgroup$
I'm trying to construct a 'Mathematica solution' for the following 'puzzle'
The natural numbers are placed in a row;
12345678910111213...
What digit is the 10,000th digit?
I don't know the syntax how to form this number, or to extract digit $n$ (in the example $n=10000$) in Mathematica. Anybody that can help? TIA.
syntax
syntax
asked Oct 16 at 18:14
mf67mf67
6091 silver badge6 bronze badges
asked Oct 16 at 18:14
mf67mf67
6091 silver badge6 bronze badges
asked Oct 16 at 18:14
mf67mf67
6091 silver badge6 bronze badges
asked Oct 16 at 18:14
mf67mf67
6091 silver badge6 bronze badges
asked Oct 16 at 18:14
mf67mf67
6091 silver badge6 bronze badges
6091 silver badge6 bronze badges
$begingroup$
Take a look at the documentation ofIntegerDigits, that should solve your problem. To get enough digits, you can either try to work out exactly how many numbers you need to concatenate, or you could also just take enough (i.e. 10000) to make sure
$endgroup$
– Lukas Lang
Oct 16 at 18:16
$begingroup$
Flatten[IntegerDigits /@ Range[5000]][[10000]]
$endgroup$
– OkkesDulgerci
Oct 16 at 19:47
add a comment
|
$begingroup$
Take a look at the documentation ofIntegerDigits, that should solve your problem. To get enough digits, you can either try to work out exactly how many numbers you need to concatenate, or you could also just take enough (i.e. 10000) to make sure
$endgroup$
– Lukas Lang
Oct 16 at 18:16
$begingroup$
Flatten[IntegerDigits /@ Range[5000]][[10000]]
$endgroup$
– OkkesDulgerci
Oct 16 at 19:47
$begingroup$
Take a look at the documentation of
IntegerDigits, that should solve your problem. To get enough digits, you can either try to work out exactly how many numbers you need to concatenate, or you could also just take enough (i.e. 10000) to make sure$endgroup$
– Lukas Lang
Oct 16 at 18:16
$begingroup$
Take a look at the documentation of
IntegerDigits, that should solve your problem. To get enough digits, you can either try to work out exactly how many numbers you need to concatenate, or you could also just take enough (i.e. 10000) to make sure$endgroup$
– Lukas Lang
Oct 16 at 18:16
$begingroup$
Flatten[IntegerDigits /@ Range[5000]][[10000]]$endgroup$
– OkkesDulgerci
Oct 16 at 19:47
$begingroup$
Flatten[IntegerDigits /@ Range[5000]][[10000]]$endgroup$
– OkkesDulgerci
Oct 16 at 19:47
add a comment
|
2 Answers
2
active
oldest
votes
$begingroup$
The number you are constructing is related to the base 10 Champernowne constant, except that the Champernowne is a real number that starts with 0.123... while your "number" is an integer with an infinite number of digits.
In Mathematica, the Champernowne constant in base 10 is given by ChampernowneNumber[]. To get the 10,000 digit, just use RealDigits:
RealDigits[ChampernowneNumber[], 10, 1, -10000][[1, 1]]
7
answered Oct 16 at 19:03
Carl WollCarl Woll
98.4k4 gold badges133 silver badges247 bronze badges
$endgroup$
add a comment
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$begingroup$
A (much slower) procedural method.
First we find the number of digits of 123...(n-1)n.
numlength[n_] := With[
pow = Floor@N@Log10[n],
Range[pow].(9*10^Range[0, pow - 1]) + (pow + 1) (n - 10^pow + 1)
]
Here pow = Floor@N@Log10[n] gives the order of magnitude (OoM) of n, Range[pow] and (pow + 1) gives the number of digits in each OoM and the last OoM, respectively, 9*10^Range[0, pow - 1] and (n - 10^pow + 1) gives the number of numbers in each OoM and the last OoM. So the final output is the number of digits in the number 123...(n-1)n.
Then, because I couldn't think of a better way,
proc[limit_] := Block[
n = 1,
While[numlength[n] <= limit, n++];
If[numlength[n - 1] == limit, Mod[n - 1, 10], IntegerDigits[n][[limit - numlength[n - 1]]]]
]
While it's much slower than @CarlWoll's method, and definitely not the best method, it still works.
proc[10^4] // RepeatedTiming
RealDigits[ChampernowneNumber[], 10, 1, -10^4][[1, 1]] // RepeatedTiming
0.032, 7
0.000086, 7
answered Oct 17 at 2:02
That Gravity GuyThat Gravity Guy
4,1161 gold badge7 silver badges21 bronze badges
$endgroup$
add a comment
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The number you are constructing is related to the base 10 Champernowne constant, except that the Champernowne is a real number that starts with 0.123... while your "number" is an integer with an infinite number of digits.
In Mathematica, the Champernowne constant in base 10 is given by ChampernowneNumber[]. To get the 10,000 digit, just use RealDigits:
RealDigits[ChampernowneNumber[], 10, 1, -10000][[1, 1]]
7
answered Oct 16 at 19:03
Carl WollCarl Woll
98.4k4 gold badges133 silver badges247 bronze badges
$endgroup$
add a comment
|
$begingroup$
The number you are constructing is related to the base 10 Champernowne constant, except that the Champernowne is a real number that starts with 0.123... while your "number" is an integer with an infinite number of digits.
In Mathematica, the Champernowne constant in base 10 is given by ChampernowneNumber[]. To get the 10,000 digit, just use RealDigits:
RealDigits[ChampernowneNumber[], 10, 1, -10000][[1, 1]]
7
answered Oct 16 at 19:03
Carl WollCarl Woll
98.4k4 gold badges133 silver badges247 bronze badges
$endgroup$
add a comment
|
$begingroup$
The number you are constructing is related to the base 10 Champernowne constant, except that the Champernowne is a real number that starts with 0.123... while your "number" is an integer with an infinite number of digits.
In Mathematica, the Champernowne constant in base 10 is given by ChampernowneNumber[]. To get the 10,000 digit, just use RealDigits:
RealDigits[ChampernowneNumber[], 10, 1, -10000][[1, 1]]
7
answered Oct 16 at 19:03
Carl WollCarl Woll
98.4k4 gold badges133 silver badges247 bronze badges
$endgroup$
The number you are constructing is related to the base 10 Champernowne constant, except that the Champernowne is a real number that starts with 0.123... while your "number" is an integer with an infinite number of digits.
In Mathematica, the Champernowne constant in base 10 is given by ChampernowneNumber[]. To get the 10,000 digit, just use RealDigits:
RealDigits[ChampernowneNumber[], 10, 1, -10000][[1, 1]]
7
answered Oct 16 at 19:03
Carl WollCarl Woll
98.4k4 gold badges133 silver badges247 bronze badges
answered Oct 16 at 19:03
Carl WollCarl Woll
98.4k4 gold badges133 silver badges247 bronze badges
answered Oct 16 at 19:03
Carl WollCarl Woll
98.4k4 gold badges133 silver badges247 bronze badges
answered Oct 16 at 19:03
Carl WollCarl Woll
98.4k4 gold badges133 silver badges247 bronze badges
98.4k4 gold badges133 silver badges247 bronze badges
add a comment
|
add a comment
|
$begingroup$
A (much slower) procedural method.
First we find the number of digits of 123...(n-1)n.
numlength[n_] := With[
pow = Floor@N@Log10[n],
Range[pow].(9*10^Range[0, pow - 1]) + (pow + 1) (n - 10^pow + 1)
]
Here pow = Floor@N@Log10[n] gives the order of magnitude (OoM) of n, Range[pow] and (pow + 1) gives the number of digits in each OoM and the last OoM, respectively, 9*10^Range[0, pow - 1] and (n - 10^pow + 1) gives the number of numbers in each OoM and the last OoM. So the final output is the number of digits in the number 123...(n-1)n.
Then, because I couldn't think of a better way,
proc[limit_] := Block[
n = 1,
While[numlength[n] <= limit, n++];
If[numlength[n - 1] == limit, Mod[n - 1, 10], IntegerDigits[n][[limit - numlength[n - 1]]]]
]
While it's much slower than @CarlWoll's method, and definitely not the best method, it still works.
proc[10^4] // RepeatedTiming
RealDigits[ChampernowneNumber[], 10, 1, -10^4][[1, 1]] // RepeatedTiming
0.032, 7
0.000086, 7
answered Oct 17 at 2:02
That Gravity GuyThat Gravity Guy
4,1161 gold badge7 silver badges21 bronze badges
$endgroup$
add a comment
|
$begingroup$
A (much slower) procedural method.
First we find the number of digits of 123...(n-1)n.
numlength[n_] := With[
pow = Floor@N@Log10[n],
Range[pow].(9*10^Range[0, pow - 1]) + (pow + 1) (n - 10^pow + 1)
]
Here pow = Floor@N@Log10[n] gives the order of magnitude (OoM) of n, Range[pow] and (pow + 1) gives the number of digits in each OoM and the last OoM, respectively, 9*10^Range[0, pow - 1] and (n - 10^pow + 1) gives the number of numbers in each OoM and the last OoM. So the final output is the number of digits in the number 123...(n-1)n.
Then, because I couldn't think of a better way,
proc[limit_] := Block[
n = 1,
While[numlength[n] <= limit, n++];
If[numlength[n - 1] == limit, Mod[n - 1, 10], IntegerDigits[n][[limit - numlength[n - 1]]]]
]
While it's much slower than @CarlWoll's method, and definitely not the best method, it still works.
proc[10^4] // RepeatedTiming
RealDigits[ChampernowneNumber[], 10, 1, -10^4][[1, 1]] // RepeatedTiming
0.032, 7
0.000086, 7
answered Oct 17 at 2:02
That Gravity GuyThat Gravity Guy
4,1161 gold badge7 silver badges21 bronze badges
$endgroup$
add a comment
|
$begingroup$
A (much slower) procedural method.
First we find the number of digits of 123...(n-1)n.
numlength[n_] := With[
pow = Floor@N@Log10[n],
Range[pow].(9*10^Range[0, pow - 1]) + (pow + 1) (n - 10^pow + 1)
]
Here pow = Floor@N@Log10[n] gives the order of magnitude (OoM) of n, Range[pow] and (pow + 1) gives the number of digits in each OoM and the last OoM, respectively, 9*10^Range[0, pow - 1] and (n - 10^pow + 1) gives the number of numbers in each OoM and the last OoM. So the final output is the number of digits in the number 123...(n-1)n.
Then, because I couldn't think of a better way,
proc[limit_] := Block[
n = 1,
While[numlength[n] <= limit, n++];
If[numlength[n - 1] == limit, Mod[n - 1, 10], IntegerDigits[n][[limit - numlength[n - 1]]]]
]
While it's much slower than @CarlWoll's method, and definitely not the best method, it still works.
proc[10^4] // RepeatedTiming
RealDigits[ChampernowneNumber[], 10, 1, -10^4][[1, 1]] // RepeatedTiming
0.032, 7
0.000086, 7
answered Oct 17 at 2:02
That Gravity GuyThat Gravity Guy
4,1161 gold badge7 silver badges21 bronze badges
$endgroup$
A (much slower) procedural method.
First we find the number of digits of 123...(n-1)n.
numlength[n_] := With[
pow = Floor@N@Log10[n],
Range[pow].(9*10^Range[0, pow - 1]) + (pow + 1) (n - 10^pow + 1)
]
Here pow = Floor@N@Log10[n] gives the order of magnitude (OoM) of n, Range[pow] and (pow + 1) gives the number of digits in each OoM and the last OoM, respectively, 9*10^Range[0, pow - 1] and (n - 10^pow + 1) gives the number of numbers in each OoM and the last OoM. So the final output is the number of digits in the number 123...(n-1)n.
Then, because I couldn't think of a better way,
proc[limit_] := Block[
n = 1,
While[numlength[n] <= limit, n++];
If[numlength[n - 1] == limit, Mod[n - 1, 10], IntegerDigits[n][[limit - numlength[n - 1]]]]
]
While it's much slower than @CarlWoll's method, and definitely not the best method, it still works.
proc[10^4] // RepeatedTiming
RealDigits[ChampernowneNumber[], 10, 1, -10^4][[1, 1]] // RepeatedTiming
0.032, 7
0.000086, 7
answered Oct 17 at 2:02
That Gravity GuyThat Gravity Guy
4,1161 gold badge7 silver badges21 bronze badges
answered Oct 17 at 2:02
That Gravity GuyThat Gravity Guy
4,1161 gold badge7 silver badges21 bronze badges
answered Oct 17 at 2:02
That Gravity GuyThat Gravity Guy
4,1161 gold badge7 silver badges21 bronze badges
answered Oct 17 at 2:02
That Gravity GuyThat Gravity Guy
4,1161 gold badge7 silver badges21 bronze badges
4,1161 gold badge7 silver badges21 bronze badges
add a comment
|
add a comment
|
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$begingroup$
Take a look at the documentation of
IntegerDigits, that should solve your problem. To get enough digits, you can either try to work out exactly how many numbers you need to concatenate, or you could also just take enough (i.e. 10000) to make sure$endgroup$
– Lukas Lang
Oct 16 at 18:16
$begingroup$
Flatten[IntegerDigits /@ Range[5000]][[10000]]$endgroup$
– OkkesDulgerci
Oct 16 at 19:47