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How can I have $10$ to $99$ evaluate to $1$, $100$ to $999$ evaluate to $10$, and so on?

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0

$begingroup$

Sorry for the strange title, as I don't really know the proper terminology.

I need a formula that returns 1 if the supplied value is anything from 10 to 99, returns 10 if the value is anything from 100 to 999, returns 100 if the value is anything from 1000 to 9999, and so on.

I will be translating this to code and will ensure the value is never less than 1, in case that changes anything.

It's probably something really simple but I can't wrap my head around a nice way to do this so... thanks!

algebra-precalculus

share|cite|improve this question

edited May 9 at 8:34

YuiTo Cheng

3,26271345

asked May 9 at 7:38

calbarcalbar

132

$endgroup$

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  Try logarithms (and rounding)!
  $endgroup$
  – Babelfish
  May 9 at 7:41
  
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Unless time is absolutely of the essence for this operation, the best way to do it is to convert the number to a string and then take its length. This is robust and legible. Note that the mathematical functions below may suffer from floating point errors.
  $endgroup$
  – Mees de Vries
  May 9 at 12:59
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @MeesdeVries: It's not only robust and legible; it is way shorter and probably way faster (using floating-point operations here is slow and stupid)!
  $endgroup$
  – user21820
  May 11 at 14:32
  

  
  
  
  

add a comment | 

0

$begingroup$

Sorry for the strange title, as I don't really know the proper terminology.

I need a formula that returns 1 if the supplied value is anything from 10 to 99, returns 10 if the value is anything from 100 to 999, returns 100 if the value is anything from 1000 to 9999, and so on.

I will be translating this to code and will ensure the value is never less than 1, in case that changes anything.

It's probably something really simple but I can't wrap my head around a nice way to do this so... thanks!

algebra-precalculus

share|cite|improve this question

edited May 9 at 8:34

YuiTo Cheng

3,26271345

asked May 9 at 7:38

calbarcalbar

132

$endgroup$

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  Try logarithms (and rounding)!
  $endgroup$
  – Babelfish
  May 9 at 7:41
  
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Unless time is absolutely of the essence for this operation, the best way to do it is to convert the number to a string and then take its length. This is robust and legible. Note that the mathematical functions below may suffer from floating point errors.
  $endgroup$
  – Mees de Vries
  May 9 at 12:59
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @MeesdeVries: It's not only robust and legible; it is way shorter and probably way faster (using floating-point operations here is slow and stupid)!
  $endgroup$
  – user21820
  May 11 at 14:32
  

  
  
  
  

add a comment | 

$begingroup$

Sorry for the strange title, as I don't really know the proper terminology.

I need a formula that returns 1 if the supplied value is anything from 10 to 99, returns 10 if the value is anything from 100 to 999, returns 100 if the value is anything from 1000 to 9999, and so on.

I will be translating this to code and will ensure the value is never less than 1, in case that changes anything.

It's probably something really simple but I can't wrap my head around a nice way to do this so... thanks!

algebra-precalculus

share|cite|improve this question

edited May 9 at 8:34

YuiTo Cheng

3,26271345

asked May 9 at 7:38

calbarcalbar

132

$endgroup$

share|cite|improve this question

edited May 9 at 8:34

YuiTo Cheng

3,26271345

asked May 9 at 7:38

calbarcalbar

132

share|cite|improve this question

edited May 9 at 8:34

YuiTo Cheng

3,26271345

asked May 9 at 7:38

calbarcalbar

132

edited May 9 at 8:34

YuiTo Cheng

3,26271345

edited May 9 at 8:34

YuiTo Cheng

3,26271345

asked May 9 at 7:38

calbarcalbar

132

asked May 9 at 7:38

calbarcalbar

132

3 Answers
3

active

oldest

votes

4

$begingroup$

I think your function is
$$f(x) = 10^left lfloor log_10(x) right rfloor - 1$$
where $lfloor r rfloor$ denotes the largest integer less than or equal to $r$.

share|cite|improve this answer

answered May 9 at 7:43

Sharky KesaSharky Kesa

1,014413

$endgroup$

add a comment | 

2

$begingroup$

Is a mathematical formula really needed ? Or a function (in the sense of a programming language) that does this job ok? As you say you are going to translate this into code it is simpler to directly translate your description to code, without having to find a mathematical formula

Here is it in Python: (you may have to handle numbers less than 10 differently)

def mylog(x):

#assumes x is a positive integer between 10 and 10^25

tenpowers = [10**k for k in range(25)]

for k in range(25):

 if x-1 < tenpowers[k]:

 return tenpowers[k-1]

share|cite|improve this answer

answered May 9 at 8:26

P VanchinathanP Vanchinathan

15.8k12237

$endgroup$

add a comment | 

0

$begingroup$

$f(x) = 10^[log_10 x]-1$ where $[x]$ denotes floor

share|cite|improve this answer

edited May 9 at 7:45

answered May 9 at 7:42

Nilotpal Kanti SinhaNilotpal Kanti Sinha

4,96321641

$endgroup$

add a comment | 

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4

$begingroup$

I think your function is
$$f(x) = 10^left lfloor log_10(x) right rfloor - 1$$
where $lfloor r rfloor$ denotes the largest integer less than or equal to $r$.

share|cite|improve this answer

answered May 9 at 7:43

Sharky KesaSharky Kesa

1,014413

$endgroup$

add a comment | 

4

$begingroup$

I think your function is
$$f(x) = 10^left lfloor log_10(x) right rfloor - 1$$
where $lfloor r rfloor$ denotes the largest integer less than or equal to $r$.

share|cite|improve this answer

answered May 9 at 7:43

Sharky KesaSharky Kesa

1,014413

$endgroup$

add a comment | 

$begingroup$

I think your function is
$$f(x) = 10^left lfloor log_10(x) right rfloor - 1$$
where $lfloor r rfloor$ denotes the largest integer less than or equal to $r$.

share|cite|improve this answer

answered May 9 at 7:43

Sharky KesaSharky Kesa

1,014413

$endgroup$

share|cite|improve this answer

answered May 9 at 7:43

Sharky KesaSharky Kesa

1,014413

answered May 9 at 7:43

Sharky KesaSharky Kesa

1,014413

answered May 9 at 7:43

Sharky KesaSharky Kesa

1,014413

2

$begingroup$

Is a mathematical formula really needed ? Or a function (in the sense of a programming language) that does this job ok? As you say you are going to translate this into code it is simpler to directly translate your description to code, without having to find a mathematical formula

Here is it in Python: (you may have to handle numbers less than 10 differently)

def mylog(x):

#assumes x is a positive integer between 10 and 10^25

tenpowers = [10**k for k in range(25)]

for k in range(25):

 if x-1 < tenpowers[k]:

 return tenpowers[k-1]

share|cite|improve this answer

answered May 9 at 8:26

P VanchinathanP Vanchinathan

15.8k12237

$endgroup$

add a comment | 

2

$begingroup$

Is a mathematical formula really needed ? Or a function (in the sense of a programming language) that does this job ok? As you say you are going to translate this into code it is simpler to directly translate your description to code, without having to find a mathematical formula

Here is it in Python: (you may have to handle numbers less than 10 differently)

def mylog(x):

#assumes x is a positive integer between 10 and 10^25

tenpowers = [10**k for k in range(25)]

for k in range(25):

 if x-1 < tenpowers[k]:

 return tenpowers[k-1]

share|cite|improve this answer

answered May 9 at 8:26

P VanchinathanP Vanchinathan

15.8k12237

$endgroup$

add a comment | 

$begingroup$

Is a mathematical formula really needed ? Or a function (in the sense of a programming language) that does this job ok? As you say you are going to translate this into code it is simpler to directly translate your description to code, without having to find a mathematical formula

Here is it in Python: (you may have to handle numbers less than 10 differently)

def mylog(x):

#assumes x is a positive integer between 10 and 10^25

tenpowers = [10**k for k in range(25)]

for k in range(25):

 if x-1 < tenpowers[k]:

 return tenpowers[k-1]

share|cite|improve this answer

answered May 9 at 8:26

P VanchinathanP Vanchinathan

15.8k12237

$endgroup$

#assumes x is a positive integer between 10 and 10^25

tenpowers = [10**k for k in range(25)]

for k in range(25):

 if x-1 < tenpowers[k]:

 return tenpowers[k-1]

share|cite|improve this answer

answered May 9 at 8:26

P VanchinathanP Vanchinathan

15.8k12237

answered May 9 at 8:26

P VanchinathanP Vanchinathan

15.8k12237

answered May 9 at 8:26

P VanchinathanP Vanchinathan

15.8k12237

0

$begingroup$

$f(x) = 10^[log_10 x]-1$ where $[x]$ denotes floor

share|cite|improve this answer

edited May 9 at 7:45

answered May 9 at 7:42

Nilotpal Kanti SinhaNilotpal Kanti Sinha

4,96321641

$endgroup$

add a comment | 

0

$begingroup$

$f(x) = 10^[log_10 x]-1$ where $[x]$ denotes floor

share|cite|improve this answer

edited May 9 at 7:45

answered May 9 at 7:42

Nilotpal Kanti SinhaNilotpal Kanti Sinha

4,96321641

$endgroup$

add a comment | 

$begingroup$

$f(x) = 10^[log_10 x]-1$ where $[x]$ denotes floor

share|cite|improve this answer

edited May 9 at 7:45

answered May 9 at 7:42

Nilotpal Kanti SinhaNilotpal Kanti Sinha

4,96321641

$endgroup$

share|cite|improve this answer

edited May 9 at 7:45

answered May 9 at 7:42

Nilotpal Kanti SinhaNilotpal Kanti Sinha

4,96321641

answered May 9 at 7:42

Nilotpal Kanti SinhaNilotpal Kanti Sinha

4,96321641

answered May 9 at 7:42

Nilotpal Kanti SinhaNilotpal Kanti Sinha

4,96321641