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How many digits are there in the numbers from 0 to 10^9?
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1
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A dedicated person is going to write down all numbers in the range $[0, 10^9)$. It takes 1 second to write down 1 digit.
The numbers in the range $[0, 100)$ takes 190 seconds to write: $10$ seconds for $[0, 9]$ + $2*90$ seconds for $[10, 99]$.
How long would it take to write down all numbers in the range $[0, 10^9)$?
elementary-number-theory
asked Aug 2 at 10:05
Alexander
197311
 |Â
show 6 more comments
up vote
1
down vote
favorite
A dedicated person is going to write down all numbers in the range $[0, 10^9)$. It takes 1 second to write down 1 digit.
The numbers in the range $[0, 100)$ takes 190 seconds to write: $10$ seconds for $[0, 9]$ + $2*90$ seconds for $[10, 99]$.
How long would it take to write down all numbers in the range $[0, 10^9)$?
elementary-number-theory
asked Aug 2 at 10:05
Alexander
197311
Hint : There are $9cdot 10^n-1$ $n$-digit-numbers for $nge 2$
– Peter
Aug 2 at 10:10
1
How many numbers are there with each number of digits? Write down those counts. Multiply each by the number of digits. Add them all. (Even with pencil and paper this is probably just as fast as getting details of some more clever trick for summing the series right).
– Henning Makholm
Aug 2 at 10:10
2
Also, your dedicated person writes curiously slowly :-)
– Henning Makholm
Aug 2 at 10:12
1
@HenningMakholm One digit per second slow ?
– Peter
Aug 2 at 10:13
1
@Servaes: Not quite that fast -- I just timed myself for 6 rows of 0123456789 and got 30 seconds. But then again I don't have the decades and decades of practice our hypothetical person will get.
– Henning Makholm
Aug 2 at 10:24
 |Â
show 6 more comments
up vote
1
down vote
favorite
up vote
1
down vote
favorite
A dedicated person is going to write down all numbers in the range $[0, 10^9)$. It takes 1 second to write down 1 digit.
The numbers in the range $[0, 100)$ takes 190 seconds to write: $10$ seconds for $[0, 9]$ + $2*90$ seconds for $[10, 99]$.
How long would it take to write down all numbers in the range $[0, 10^9)$?
elementary-number-theory
asked Aug 2 at 10:05
Alexander
197311
A dedicated person is going to write down all numbers in the range $[0, 10^9)$. It takes 1 second to write down 1 digit.
The numbers in the range $[0, 100)$ takes 190 seconds to write: $10$ seconds for $[0, 9]$ + $2*90$ seconds for $[10, 99]$.
How long would it take to write down all numbers in the range $[0, 10^9)$?
elementary-number-theory
asked Aug 2 at 10:05
Alexander
197311
asked Aug 2 at 10:05
Alexander
197311
asked Aug 2 at 10:05
Alexander
197311
asked Aug 2 at 10:05
Alexander
197311
197311
Hint : There are $9cdot 10^n-1$ $n$-digit-numbers for $nge 2$
– Peter
Aug 2 at 10:10
1
How many numbers are there with each number of digits? Write down those counts. Multiply each by the number of digits. Add them all. (Even with pencil and paper this is probably just as fast as getting details of some more clever trick for summing the series right).
– Henning Makholm
Aug 2 at 10:10
2
Also, your dedicated person writes curiously slowly :-)
– Henning Makholm
Aug 2 at 10:12
1
@HenningMakholm One digit per second slow ?
– Peter
Aug 2 at 10:13
1
@Servaes: Not quite that fast -- I just timed myself for 6 rows of 0123456789 and got 30 seconds. But then again I don't have the decades and decades of practice our hypothetical person will get.
– Henning Makholm
Aug 2 at 10:24
 |Â
show 6 more comments
Hint : There are $9cdot 10^n-1$ $n$-digit-numbers for $nge 2$
– Peter
Aug 2 at 10:10
1
How many numbers are there with each number of digits? Write down those counts. Multiply each by the number of digits. Add them all. (Even with pencil and paper this is probably just as fast as getting details of some more clever trick for summing the series right).
– Henning Makholm
Aug 2 at 10:10
2
Also, your dedicated person writes curiously slowly :-)
– Henning Makholm
Aug 2 at 10:12
1
@HenningMakholm One digit per second slow ?
– Peter
Aug 2 at 10:13
1
@Servaes: Not quite that fast -- I just timed myself for 6 rows of 0123456789 and got 30 seconds. But then again I don't have the decades and decades of practice our hypothetical person will get.
– Henning Makholm
Aug 2 at 10:24
Hint : There are $9cdot 10^n-1$ $n$-digit-numbers for $nge 2$
– Peter
Aug 2 at 10:10
Hint : There are $9cdot 10^n-1$ $n$-digit-numbers for $nge 2$
– Peter
Aug 2 at 10:10
1
1
How many numbers are there with each number of digits? Write down those counts. Multiply each by the number of digits. Add them all. (Even with pencil and paper this is probably just as fast as getting details of some more clever trick for summing the series right).
– Henning Makholm
Aug 2 at 10:10
How many numbers are there with each number of digits? Write down those counts. Multiply each by the number of digits. Add them all. (Even with pencil and paper this is probably just as fast as getting details of some more clever trick for summing the series right).
– Henning Makholm
Aug 2 at 10:10
2
2
Also, your dedicated person writes curiously slowly :-)
– Henning Makholm
Aug 2 at 10:12
Also, your dedicated person writes curiously slowly :-)
– Henning Makholm
Aug 2 at 10:12
1
1
@HenningMakholm One digit per second slow ?
– Peter
Aug 2 at 10:13
@HenningMakholm One digit per second slow ?
– Peter
Aug 2 at 10:13
1
1
@Servaes: Not quite that fast -- I just timed myself for 6 rows of 0123456789 and got 30 seconds. But then again I don't have the decades and decades of practice our hypothetical person will get.
– Henning Makholm
Aug 2 at 10:24
@Servaes: Not quite that fast -- I just timed myself for 6 rows of 0123456789 and got 30 seconds. But then again I don't have the decades and decades of practice our hypothetical person will get.
– Henning Makholm
Aug 2 at 10:24
 |Â
show 6 more comments
3 Answers
3
active
oldest
votes
up vote
3
down vote
accepted
As others have already pointed out, it is easy to count the number of digits of all numbers between $10^m-1$ (included) and $10^m$ (excluded) - the $m$-digits-numbers:
$$
underbracetextdigits per number_mcdotunderbracetexthow many numbers_10^m - 10^m-1
$$
$$
= 9cdot 10^m-1 cdot m
$$
Note that for two different values of $m$ there is no number that contributes more than once.
Be $mathrmd(n)$ the number of digits of all numbers between $0$ (included) and $10^n$ (excluded), it is a matter of considering the target quantity until $10^n-1$ (represented by $mathrmd(n-1)$) and the number of $n$-digits-numbers:
$$
mathrmd(n) =
begincases
1 & n = 0 \
mathrmd(n-1) + 9cdot 10^n-1 cdot n & n > 0
endcases
$$
Giving:
$$
beginalign
mathrmd(n) &= 1 + sum_k=1^n 9cdot 10^n-1 cdot n
endalign
$$
We’ll use Wolfram Alpha to obtain the closed formula of that series:
$$
beginalign
sum_k=1^n 9cdot 10^n-1 cdot n&=10^ncdot n-frac10^n9+frac19\
&=10^ncdot n + frac1-10^n9
endalign
$$
Finally:
$$
mathrmd(n) = 1+10^ncdot n + frac1-10^n9
$$
Let’s verify that we have indeed found the right formula:
$$
beginarrayc
n & mathrmd(n) \
hline
1 & 10\
2 & 190\
vdots & vdots \
9 & 8888888890
endarray
$$
Epilogue - Be $mathrmt(n)$ how many seconds it would take to write down all numbers one next to the other between $0$ and $10^n$ given a speed of $v$ digits per second then:
$$mathrmt(n) = fracmathrmd(n)v$$
The number of years $mathrmy(n)$ needed at least (given a $365$ days long year) is:
[mathrmy(n) = leftlfloorfracmathrmt(n)3600cdot24cdot365rightrfloor]
answered Aug 2 at 10:55
Giulio Scattolin
15619
1
Your simplification of the closed form of the sum seems wrong; it isn't even an integer for $n=1,2,3,ldots,8$. And in the epilogue, shouldn't that be $mathrmt(n) = fracmathrmd(n)v$?
– Servaes
Aug 3 at 9:10
@Servaes Thank you very much, I corrected them.
– Giulio Scattolin
Aug 3 at 9:17
add a comment |Â
up vote
3
down vote
As another way to calculate it, first imagine all $10^9$ numbers written out padded with zeroes on the left to make each of them 9 digits long. That makes $9,000,000,000$ digits.
Then erase $10^8$ leading zeroes in the first column.
Then erase $10^7$ leading zeroes in the second column.
Then erase $10^6$ leading zeroes in the third column.
And so forth. At the end you don't erase the single remaining zero in the number $0$, so you end up with
$$ 9,000,000,000 - 111,111,110 $$
digits.
answered Aug 2 at 10:54
Henning Makholm
225k16290516
add a comment |Â
up vote
1
down vote
$8888888890$ seconds
Which can also be expressed as:
Roughly 282 years of writing digits non-stop 24/7
The answer was found using this Python expression:
1 + sum([(10**(x+1) - 10**x) * (x+1) for x in range(9)])
By replacing range(9) with other ranges, it's easy to check that the lower values are correct:
- With
range(0): $[0,10^0)$ has 1 digit. - With
range(1): $[0,10^1)$ has 10 digits. - With
range(2): $[0,10^2)$ has 190 digits. - With
range(3): $[0,10^3)$ has 2890 digits.
This is not a proper proof, though.
answered Aug 2 at 10:32
Alexander
197311
1
Good enough to answer the question, I think.
– David R.
Aug 2 at 21:27
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
As others have already pointed out, it is easy to count the number of digits of all numbers between $10^m-1$ (included) and $10^m$ (excluded) - the $m$-digits-numbers:
$$
underbracetextdigits per number_mcdotunderbracetexthow many numbers_10^m - 10^m-1
$$
$$
= 9cdot 10^m-1 cdot m
$$
Note that for two different values of $m$ there is no number that contributes more than once.
Be $mathrmd(n)$ the number of digits of all numbers between $0$ (included) and $10^n$ (excluded), it is a matter of considering the target quantity until $10^n-1$ (represented by $mathrmd(n-1)$) and the number of $n$-digits-numbers:
$$
mathrmd(n) =
begincases
1 & n = 0 \
mathrmd(n-1) + 9cdot 10^n-1 cdot n & n > 0
endcases
$$
Giving:
$$
beginalign
mathrmd(n) &= 1 + sum_k=1^n 9cdot 10^n-1 cdot n
endalign
$$
We’ll use Wolfram Alpha to obtain the closed formula of that series:
$$
beginalign
sum_k=1^n 9cdot 10^n-1 cdot n&=10^ncdot n-frac10^n9+frac19\
&=10^ncdot n + frac1-10^n9
endalign
$$
Finally:
$$
mathrmd(n) = 1+10^ncdot n + frac1-10^n9
$$
Let’s verify that we have indeed found the right formula:
$$
beginarrayc
n & mathrmd(n) \
hline
1 & 10\
2 & 190\
vdots & vdots \
9 & 8888888890
endarray
$$
Epilogue - Be $mathrmt(n)$ how many seconds it would take to write down all numbers one next to the other between $0$ and $10^n$ given a speed of $v$ digits per second then:
$$mathrmt(n) = fracmathrmd(n)v$$
The number of years $mathrmy(n)$ needed at least (given a $365$ days long year) is:
[mathrmy(n) = leftlfloorfracmathrmt(n)3600cdot24cdot365rightrfloor]
answered Aug 2 at 10:55
Giulio Scattolin
15619
1
Your simplification of the closed form of the sum seems wrong; it isn't even an integer for $n=1,2,3,ldots,8$. And in the epilogue, shouldn't that be $mathrmt(n) = fracmathrmd(n)v$?
– Servaes
Aug 3 at 9:10
@Servaes Thank you very much, I corrected them.
– Giulio Scattolin
Aug 3 at 9:17
add a comment |Â
up vote
3
down vote
accepted
As others have already pointed out, it is easy to count the number of digits of all numbers between $10^m-1$ (included) and $10^m$ (excluded) - the $m$-digits-numbers:
$$
underbracetextdigits per number_mcdotunderbracetexthow many numbers_10^m - 10^m-1
$$
$$
= 9cdot 10^m-1 cdot m
$$
Note that for two different values of $m$ there is no number that contributes more than once.
Be $mathrmd(n)$ the number of digits of all numbers between $0$ (included) and $10^n$ (excluded), it is a matter of considering the target quantity until $10^n-1$ (represented by $mathrmd(n-1)$) and the number of $n$-digits-numbers:
$$
mathrmd(n) =
begincases
1 & n = 0 \
mathrmd(n-1) + 9cdot 10^n-1 cdot n & n > 0
endcases
$$
Giving:
$$
beginalign
mathrmd(n) &= 1 + sum_k=1^n 9cdot 10^n-1 cdot n
endalign
$$
We’ll use Wolfram Alpha to obtain the closed formula of that series:
$$
beginalign
sum_k=1^n 9cdot 10^n-1 cdot n&=10^ncdot n-frac10^n9+frac19\
&=10^ncdot n + frac1-10^n9
endalign
$$
Finally:
$$
mathrmd(n) = 1+10^ncdot n + frac1-10^n9
$$
Let’s verify that we have indeed found the right formula:
$$
beginarrayc
n & mathrmd(n) \
hline
1 & 10\
2 & 190\
vdots & vdots \
9 & 8888888890
endarray
$$
Epilogue - Be $mathrmt(n)$ how many seconds it would take to write down all numbers one next to the other between $0$ and $10^n$ given a speed of $v$ digits per second then:
$$mathrmt(n) = fracmathrmd(n)v$$
The number of years $mathrmy(n)$ needed at least (given a $365$ days long year) is:
[mathrmy(n) = leftlfloorfracmathrmt(n)3600cdot24cdot365rightrfloor]
answered Aug 2 at 10:55
Giulio Scattolin
15619
1
Your simplification of the closed form of the sum seems wrong; it isn't even an integer for $n=1,2,3,ldots,8$. And in the epilogue, shouldn't that be $mathrmt(n) = fracmathrmd(n)v$?
– Servaes
Aug 3 at 9:10
@Servaes Thank you very much, I corrected them.
– Giulio Scattolin
Aug 3 at 9:17
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
As others have already pointed out, it is easy to count the number of digits of all numbers between $10^m-1$ (included) and $10^m$ (excluded) - the $m$-digits-numbers:
$$
underbracetextdigits per number_mcdotunderbracetexthow many numbers_10^m - 10^m-1
$$
$$
= 9cdot 10^m-1 cdot m
$$
Note that for two different values of $m$ there is no number that contributes more than once.
Be $mathrmd(n)$ the number of digits of all numbers between $0$ (included) and $10^n$ (excluded), it is a matter of considering the target quantity until $10^n-1$ (represented by $mathrmd(n-1)$) and the number of $n$-digits-numbers:
$$
mathrmd(n) =
begincases
1 & n = 0 \
mathrmd(n-1) + 9cdot 10^n-1 cdot n & n > 0
endcases
$$
Giving:
$$
beginalign
mathrmd(n) &= 1 + sum_k=1^n 9cdot 10^n-1 cdot n
endalign
$$
We’ll use Wolfram Alpha to obtain the closed formula of that series:
$$
beginalign
sum_k=1^n 9cdot 10^n-1 cdot n&=10^ncdot n-frac10^n9+frac19\
&=10^ncdot n + frac1-10^n9
endalign
$$
Finally:
$$
mathrmd(n) = 1+10^ncdot n + frac1-10^n9
$$
Let’s verify that we have indeed found the right formula:
$$
beginarrayc
n & mathrmd(n) \
hline
1 & 10\
2 & 190\
vdots & vdots \
9 & 8888888890
endarray
$$
Epilogue - Be $mathrmt(n)$ how many seconds it would take to write down all numbers one next to the other between $0$ and $10^n$ given a speed of $v$ digits per second then:
$$mathrmt(n) = fracmathrmd(n)v$$
The number of years $mathrmy(n)$ needed at least (given a $365$ days long year) is:
[mathrmy(n) = leftlfloorfracmathrmt(n)3600cdot24cdot365rightrfloor]
answered Aug 2 at 10:55
Giulio Scattolin
15619
As others have already pointed out, it is easy to count the number of digits of all numbers between $10^m-1$ (included) and $10^m$ (excluded) - the $m$-digits-numbers:
$$
underbracetextdigits per number_mcdotunderbracetexthow many numbers_10^m - 10^m-1
$$
$$
= 9cdot 10^m-1 cdot m
$$
Note that for two different values of $m$ there is no number that contributes more than once.
Be $mathrmd(n)$ the number of digits of all numbers between $0$ (included) and $10^n$ (excluded), it is a matter of considering the target quantity until $10^n-1$ (represented by $mathrmd(n-1)$) and the number of $n$-digits-numbers:
$$
mathrmd(n) =
begincases
1 & n = 0 \
mathrmd(n-1) + 9cdot 10^n-1 cdot n & n > 0
endcases
$$
Giving:
$$
beginalign
mathrmd(n) &= 1 + sum_k=1^n 9cdot 10^n-1 cdot n
endalign
$$
We’ll use Wolfram Alpha to obtain the closed formula of that series:
$$
beginalign
sum_k=1^n 9cdot 10^n-1 cdot n&=10^ncdot n-frac10^n9+frac19\
&=10^ncdot n + frac1-10^n9
endalign
$$
Finally:
$$
mathrmd(n) = 1+10^ncdot n + frac1-10^n9
$$
Let’s verify that we have indeed found the right formula:
$$
beginarrayc
n & mathrmd(n) \
hline
1 & 10\
2 & 190\
vdots & vdots \
9 & 8888888890
endarray
$$
Epilogue - Be $mathrmt(n)$ how many seconds it would take to write down all numbers one next to the other between $0$ and $10^n$ given a speed of $v$ digits per second then:
$$mathrmt(n) = fracmathrmd(n)v$$
The number of years $mathrmy(n)$ needed at least (given a $365$ days long year) is:
[mathrmy(n) = leftlfloorfracmathrmt(n)3600cdot24cdot365rightrfloor]
answered Aug 2 at 10:55
Giulio Scattolin
15619
edited Aug 3 at 9:16
answered Aug 2 at 10:55
Giulio Scattolin
15619
answered Aug 2 at 10:55
Giulio Scattolin
15619
answered Aug 2 at 10:55
Giulio Scattolin
15619
15619
1
Your simplification of the closed form of the sum seems wrong; it isn't even an integer for $n=1,2,3,ldots,8$. And in the epilogue, shouldn't that be $mathrmt(n) = fracmathrmd(n)v$?
– Servaes
Aug 3 at 9:10
@Servaes Thank you very much, I corrected them.
– Giulio Scattolin
Aug 3 at 9:17
add a comment |Â
1
Your simplification of the closed form of the sum seems wrong; it isn't even an integer for $n=1,2,3,ldots,8$. And in the epilogue, shouldn't that be $mathrmt(n) = fracmathrmd(n)v$?
– Servaes
Aug 3 at 9:10
@Servaes Thank you very much, I corrected them.
– Giulio Scattolin
Aug 3 at 9:17
1
1
Your simplification of the closed form of the sum seems wrong; it isn't even an integer for $n=1,2,3,ldots,8$. And in the epilogue, shouldn't that be $mathrmt(n) = fracmathrmd(n)v$?
– Servaes
Aug 3 at 9:10
Your simplification of the closed form of the sum seems wrong; it isn't even an integer for $n=1,2,3,ldots,8$. And in the epilogue, shouldn't that be $mathrmt(n) = fracmathrmd(n)v$?
– Servaes
Aug 3 at 9:10
@Servaes Thank you very much, I corrected them.
– Giulio Scattolin
Aug 3 at 9:17
@Servaes Thank you very much, I corrected them.
– Giulio Scattolin
Aug 3 at 9:17
add a comment |Â
up vote
3
down vote
As another way to calculate it, first imagine all $10^9$ numbers written out padded with zeroes on the left to make each of them 9 digits long. That makes $9,000,000,000$ digits.
Then erase $10^8$ leading zeroes in the first column.
Then erase $10^7$ leading zeroes in the second column.
Then erase $10^6$ leading zeroes in the third column.
And so forth. At the end you don't erase the single remaining zero in the number $0$, so you end up with
$$ 9,000,000,000 - 111,111,110 $$
digits.
answered Aug 2 at 10:54
Henning Makholm
225k16290516
add a comment |Â
up vote
3
down vote
As another way to calculate it, first imagine all $10^9$ numbers written out padded with zeroes on the left to make each of them 9 digits long. That makes $9,000,000,000$ digits.
Then erase $10^8$ leading zeroes in the first column.
Then erase $10^7$ leading zeroes in the second column.
Then erase $10^6$ leading zeroes in the third column.
And so forth. At the end you don't erase the single remaining zero in the number $0$, so you end up with
$$ 9,000,000,000 - 111,111,110 $$
digits.
answered Aug 2 at 10:54
Henning Makholm
225k16290516
add a comment |Â
up vote
3
down vote
up vote
3
down vote
As another way to calculate it, first imagine all $10^9$ numbers written out padded with zeroes on the left to make each of them 9 digits long. That makes $9,000,000,000$ digits.
Then erase $10^8$ leading zeroes in the first column.
Then erase $10^7$ leading zeroes in the second column.
Then erase $10^6$ leading zeroes in the third column.
And so forth. At the end you don't erase the single remaining zero in the number $0$, so you end up with
$$ 9,000,000,000 - 111,111,110 $$
digits.
answered Aug 2 at 10:54
Henning Makholm
225k16290516
As another way to calculate it, first imagine all $10^9$ numbers written out padded with zeroes on the left to make each of them 9 digits long. That makes $9,000,000,000$ digits.
Then erase $10^8$ leading zeroes in the first column.
Then erase $10^7$ leading zeroes in the second column.
Then erase $10^6$ leading zeroes in the third column.
And so forth. At the end you don't erase the single remaining zero in the number $0$, so you end up with
$$ 9,000,000,000 - 111,111,110 $$
digits.
answered Aug 2 at 10:54
Henning Makholm
225k16290516
edited Aug 2 at 11:05
answered Aug 2 at 10:54
Henning Makholm
225k16290516
answered Aug 2 at 10:54
Henning Makholm
225k16290516
answered Aug 2 at 10:54
Henning Makholm
225k16290516
225k16290516
add a comment |Â
add a comment |Â
up vote
1
down vote
$8888888890$ seconds
Which can also be expressed as:
Roughly 282 years of writing digits non-stop 24/7
The answer was found using this Python expression:
1 + sum([(10**(x+1) - 10**x) * (x+1) for x in range(9)])
By replacing range(9) with other ranges, it's easy to check that the lower values are correct:
- With
range(0): $[0,10^0)$ has 1 digit. - With
range(1): $[0,10^1)$ has 10 digits. - With
range(2): $[0,10^2)$ has 190 digits. - With
range(3): $[0,10^3)$ has 2890 digits.
This is not a proper proof, though.
answered Aug 2 at 10:32
Alexander
197311
1
Good enough to answer the question, I think.
– David R.
Aug 2 at 21:27
add a comment |Â
up vote
1
down vote
$8888888890$ seconds
Which can also be expressed as:
Roughly 282 years of writing digits non-stop 24/7
The answer was found using this Python expression:
1 + sum([(10**(x+1) - 10**x) * (x+1) for x in range(9)])
By replacing range(9) with other ranges, it's easy to check that the lower values are correct:
- With
range(0): $[0,10^0)$ has 1 digit. - With
range(1): $[0,10^1)$ has 10 digits. - With
range(2): $[0,10^2)$ has 190 digits. - With
range(3): $[0,10^3)$ has 2890 digits.
This is not a proper proof, though.
answered Aug 2 at 10:32
Alexander
197311
1
Good enough to answer the question, I think.
– David R.
Aug 2 at 21:27
add a comment |Â
up vote
1
down vote
up vote
1
down vote
$8888888890$ seconds
Which can also be expressed as:
Roughly 282 years of writing digits non-stop 24/7
The answer was found using this Python expression:
1 + sum([(10**(x+1) - 10**x) * (x+1) for x in range(9)])
By replacing range(9) with other ranges, it's easy to check that the lower values are correct:
- With
range(0): $[0,10^0)$ has 1 digit. - With
range(1): $[0,10^1)$ has 10 digits. - With
range(2): $[0,10^2)$ has 190 digits. - With
range(3): $[0,10^3)$ has 2890 digits.
This is not a proper proof, though.
answered Aug 2 at 10:32
Alexander
197311
$8888888890$ seconds
Which can also be expressed as:
Roughly 282 years of writing digits non-stop 24/7
The answer was found using this Python expression:
1 + sum([(10**(x+1) - 10**x) * (x+1) for x in range(9)])
By replacing range(9) with other ranges, it's easy to check that the lower values are correct:
- With
range(0): $[0,10^0)$ has 1 digit. - With
range(1): $[0,10^1)$ has 10 digits. - With
range(2): $[0,10^2)$ has 190 digits. - With
range(3): $[0,10^3)$ has 2890 digits.
This is not a proper proof, though.
answered Aug 2 at 10:32
Alexander
197311
edited Aug 2 at 22:02
answered Aug 2 at 10:32
Alexander
197311
answered Aug 2 at 10:32
Alexander
197311
answered Aug 2 at 10:32
Alexander
197311
197311
1
Good enough to answer the question, I think.
– David R.
Aug 2 at 21:27
add a comment |Â
1
Good enough to answer the question, I think.
– David R.
Aug 2 at 21:27
1
1
Good enough to answer the question, I think.
– David R.
Aug 2 at 21:27
Good enough to answer the question, I think.
– David R.
Aug 2 at 21:27
add a comment |Â
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Hint : There are $9cdot 10^n-1$ $n$-digit-numbers for $nge 2$
– Peter
Aug 2 at 10:10
1
How many numbers are there with each number of digits? Write down those counts. Multiply each by the number of digits. Add them all. (Even with pencil and paper this is probably just as fast as getting details of some more clever trick for summing the series right).
– Henning Makholm
Aug 2 at 10:10
2
Also, your dedicated person writes curiously slowly :-)
– Henning Makholm
Aug 2 at 10:12
1
@HenningMakholm One digit per second slow ?
– Peter
Aug 2 at 10:13
1
@Servaes: Not quite that fast -- I just timed myself for 6 rows of 0123456789 and got 30 seconds. But then again I don't have the decades and decades of practice our hypothetical person will get.
– Henning Makholm
Aug 2 at 10:24