Multiply a number so that it's always in the thousand billions

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1
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Excuse my lack of mathematical notation, hopefully this makes sense.

Say I have a number which is x

10,000,000,000,000 > x > 999,999,999

I always want to multiply this number by some other number y, so that the result z in the thousand billions, so for example:

x = 1,230,000,000 
x * y = z
z = 1,230,000,000,000
y = 1000

x = 55,123,000,000 
x * y = z
z = 5,512,300,000,000
y = 100

x = 456,123,000,000 
x * y = z
z = 4,561,230,000,000
y = 10

x = 7,894,560,000,000 
x * y = z
z = 7,894,560,000,000
y = 1

All I have is x and the I know that x is be between 1,000,000,000 and 9,999,999,999,999.

How can I find the value of y so that I can calculate z? I'm currently doing this manually (in code) with if statements,

if x < 10 => x * 1000
if x < 100 => x * 100
if x < 1000 => x * 10
if x < 1000 => x

But I'm looking for a single mathematical function that can do this for me.

Solved with Arnaud Mortier's answer

Using the marked answer I created the function (in Javascript)

const fn = x => 
 let k = 12 - Math.floor(Math.log10(x));
 let y = Math.pow(10, k);
 let z = x * y;

 console.log(`x = $x, k = $k, y = $y, z = $z`);

 return z;
;

Even more than I hoped for, this works for any positive integer, not just those in my range.

edited Jul 19 at 6:29

asked Jul 19 at 5:52

James Hay

1084

add a commentÂ |Â

up vote
1
down vote

favorite

Excuse my lack of mathematical notation, hopefully this makes sense.

Say I have a number which is x

10,000,000,000,000 > x > 999,999,999

I always want to multiply this number by some other number y, so that the result z in the thousand billions, so for example:

x = 1,230,000,000 
x * y = z
z = 1,230,000,000,000
y = 1000

x = 55,123,000,000 
x * y = z
z = 5,512,300,000,000
y = 100

x = 456,123,000,000 
x * y = z
z = 4,561,230,000,000
y = 10

x = 7,894,560,000,000 
x * y = z
z = 7,894,560,000,000
y = 1

All I have is x and the I know that x is be between 1,000,000,000 and 9,999,999,999,999.

How can I find the value of y so that I can calculate z? I'm currently doing this manually (in code) with if statements,

if x < 10 => x * 1000
if x < 100 => x * 100
if x < 1000 => x * 10
if x < 1000 => x

But I'm looking for a single mathematical function that can do this for me.

Solved with Arnaud Mortier's answer

Using the marked answer I created the function (in Javascript)

const fn = x => 
 let k = 12 - Math.floor(Math.log10(x));
 let y = Math.pow(10, k);
 let z = x * y;

 console.log(`x = $x, k = $k, y = $y, z = $z`);

 return z;
;

Even more than I hoped for, this works for any positive integer, not just those in my range.

edited Jul 19 at 6:29

asked Jul 19 at 5:52

James Hay

1084

add a commentÂ |Â

up vote
1
down vote

favorite

Excuse my lack of mathematical notation, hopefully this makes sense.

Say I have a number which is x

10,000,000,000,000 > x > 999,999,999

I always want to multiply this number by some other number y, so that the result z in the thousand billions, so for example:

x = 1,230,000,000 
x * y = z
z = 1,230,000,000,000
y = 1000

x = 55,123,000,000 
x * y = z
z = 5,512,300,000,000
y = 100

x = 456,123,000,000 
x * y = z
z = 4,561,230,000,000
y = 10

x = 7,894,560,000,000 
x * y = z
z = 7,894,560,000,000
y = 1

All I have is x and the I know that x is be between 1,000,000,000 and 9,999,999,999,999.

How can I find the value of y so that I can calculate z? I'm currently doing this manually (in code) with if statements,

if x < 10 => x * 1000
if x < 100 => x * 100
if x < 1000 => x * 10
if x < 1000 => x

But I'm looking for a single mathematical function that can do this for me.

Solved with Arnaud Mortier's answer

Using the marked answer I created the function (in Javascript)

const fn = x => 
 let k = 12 - Math.floor(Math.log10(x));
 let y = Math.pow(10, k);
 let z = x * y;

 console.log(`x = $x, k = $k, y = $y, z = $z`);

 return z;
;

Even more than I hoped for, this works for any positive integer, not just those in my range.

edited Jul 19 at 6:29

asked Jul 19 at 5:52

James Hay

1084

Excuse my lack of mathematical notation, hopefully this makes sense.

Say I have a number which is x

10,000,000,000,000 > x > 999,999,999

I always want to multiply this number by some other number y, so that the result z in the thousand billions, so for example:

x = 1,230,000,000 
x * y = z
z = 1,230,000,000,000
y = 1000

x = 55,123,000,000 
x * y = z
z = 5,512,300,000,000
y = 100

x = 456,123,000,000 
x * y = z
z = 4,561,230,000,000
y = 10

x = 7,894,560,000,000 
x * y = z
z = 7,894,560,000,000
y = 1

All I have is x and the I know that x is be between 1,000,000,000 and 9,999,999,999,999.

How can I find the value of y so that I can calculate z? I'm currently doing this manually (in code) with if statements,

if x < 10 => x * 1000
if x < 100 => x * 100
if x < 1000 => x * 10
if x < 1000 => x

But I'm looking for a single mathematical function that can do this for me.

Solved with Arnaud Mortier's answer

Using the marked answer I created the function (in Javascript)

const fn = x => 
 let k = 12 - Math.floor(Math.log10(x));
 let y = Math.pow(10, k);
 let z = x * y;

 console.log(`x = $x, k = $k, y = $y, z = $z`);

 return z;
;

Even more than I hoped for, this works for any positive integer, not just those in my range.

edited Jul 19 at 6:29

asked Jul 19 at 5:52

James Hay

1084

edited Jul 19 at 6:29

asked Jul 19 at 5:52

James Hay

1084

asked Jul 19 at 5:52

James Hay

1084

asked Jul 19 at 5:52

James Hay

1084

add a commentÂ |Â

3 Answers
3

active

oldest

votes

up vote
1
down vote

accepted

Set $y=10^k$. beginalign10^12leq z<10^13&Longleftrightarrow 12leq log_10(xcdot 10^k)<13\&Longleftrightarrow leftlfloor log_10(xcdot 10^k)rightrfloor=12\&Longleftrightarrow leftlfloor log_10(x)+krightrfloor=12\&Longleftrightarrow k=12-leftlfloor log_10(x)rightrfloor\endalign

If you don't have $log_10$ implemented in your computer, use $log_10(x)=fracln(x)ln(10)$.

The function $leftlfloor xrightrfloor$ is usually implemented as $operatornamefloor(x)$.

answered Jul 19 at 6:06

Arnaud Mortier

19k22159

Thanks, going to give this one a go. All x are integers so I won't be needing the floor
â€“Â James Hay
Jul 19 at 6:09

@JamesHay You will most likely need the floor function after you apply log.
â€“Â Arnaud Mortier
Jul 19 at 6:09

Oh my god, that completely worked! Thanks heaps.
â€“Â James Hay
Jul 19 at 6:18

@JamesHay You're welcome.
â€“Â Arnaud Mortier
Jul 19 at 6:19

add a commentÂ |Â

up vote
0
down vote

Pick a number $k$ between one thousand billion and ten thousand billion and take $y = frackx$ ?

answered Jul 19 at 6:01

Daniel Pietrobon

3,26421836

If you look at the examples, the OP wants $y$ to be a power of $10$.
â€“Â Arnaud Mortier
Jul 19 at 6:09

add a commentÂ |Â

up vote
0
down vote

Writing $x$ in scientific notation we get something like $ktimes10^n$ where $1<k<9$. For example $x = 1,230,000,000 = 1.23times10^9$ Now you want the $z = 1,230,000,000,000=1.23times10^12$. So you need to multiply $1000$ or $10^3$. Just find the value $n$ when $x$ is written in scientific notation and if see how much you are short of $12$. And multiply that much.

answered Jul 19 at 6:01

Piyush Divyanakar

3,258122

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3 Answers
3

active

oldest

votes

3 Answers
3

active

oldest

votes

up vote
1
down vote

accepted

If you don't have $log_10$ implemented in your computer, use $log_10(x)=fracln(x)ln(10)$.

The function $leftlfloor xrightrfloor$ is usually implemented as $operatornamefloor(x)$.

answered Jul 19 at 6:06

Arnaud Mortier

19k22159

Thanks, going to give this one a go. All x are integers so I won't be needing the floor
â€“Â James Hay
Jul 19 at 6:09

@JamesHay You will most likely need the floor function after you apply log.
â€“Â Arnaud Mortier
Jul 19 at 6:09

Oh my god, that completely worked! Thanks heaps.
â€“Â James Hay
Jul 19 at 6:18

@JamesHay You're welcome.
â€“Â Arnaud Mortier
Jul 19 at 6:19

add a commentÂ |Â

up vote
1
down vote

accepted

If you don't have $log_10$ implemented in your computer, use $log_10(x)=fracln(x)ln(10)$.

The function $leftlfloor xrightrfloor$ is usually implemented as $operatornamefloor(x)$.

answered Jul 19 at 6:06

Arnaud Mortier

19k22159

Thanks, going to give this one a go. All x are integers so I won't be needing the floor
â€“Â James Hay
Jul 19 at 6:09

@JamesHay You will most likely need the floor function after you apply log.
â€“Â Arnaud Mortier
Jul 19 at 6:09

Oh my god, that completely worked! Thanks heaps.
â€“Â James Hay
Jul 19 at 6:18

@JamesHay You're welcome.
â€“Â Arnaud Mortier
Jul 19 at 6:19

add a commentÂ |Â

up vote
1
down vote

accepted

If you don't have $log_10$ implemented in your computer, use $log_10(x)=fracln(x)ln(10)$.

The function $leftlfloor xrightrfloor$ is usually implemented as $operatornamefloor(x)$.

answered Jul 19 at 6:06

Arnaud Mortier

19k22159

If you don't have $log_10$ implemented in your computer, use $log_10(x)=fracln(x)ln(10)$.

The function $leftlfloor xrightrfloor$ is usually implemented as $operatornamefloor(x)$.

answered Jul 19 at 6:06

Arnaud Mortier

19k22159

answered Jul 19 at 6:06

Arnaud Mortier

19k22159

answered Jul 19 at 6:06

Arnaud Mortier

19k22159

answered Jul 19 at 6:06

Arnaud Mortier

19k22159

Thanks, going to give this one a go. All x are integers so I won't be needing the floor
â€“Â James Hay
Jul 19 at 6:09

@JamesHay You will most likely need the floor function after you apply log.
â€“Â Arnaud Mortier
Jul 19 at 6:09

Oh my god, that completely worked! Thanks heaps.
â€“Â James Hay
Jul 19 at 6:18

@JamesHay You're welcome.
â€“Â Arnaud Mortier
Jul 19 at 6:19

add a commentÂ |Â

Thanks, going to give this one a go. All x are integers so I won't be needing the floor
â€“Â James Hay
Jul 19 at 6:09

@JamesHay You will most likely need the floor function after you apply log.
â€“Â Arnaud Mortier
Jul 19 at 6:09

Oh my god, that completely worked! Thanks heaps.
â€“Â James Hay
Jul 19 at 6:18

@JamesHay You're welcome.
â€“Â Arnaud Mortier
Jul 19 at 6:19

Thanks, going to give this one a go. All x are integers so I won't be needing the floor
â€“Â James Hay
Jul 19 at 6:09

@JamesHay You will most likely need the floor function after you apply log.
â€“Â Arnaud Mortier
Jul 19 at 6:09

Oh my god, that completely worked! Thanks heaps.
â€“Â James Hay
Jul 19 at 6:18

@JamesHay You're welcome.
â€“Â Arnaud Mortier
Jul 19 at 6:19

add a commentÂ |Â

up vote
0
down vote

Pick a number $k$ between one thousand billion and ten thousand billion and take $y = frackx$ ?

answered Jul 19 at 6:01

Daniel Pietrobon

3,26421836

If you look at the examples, the OP wants $y$ to be a power of $10$.
â€“Â Arnaud Mortier
Jul 19 at 6:09

add a commentÂ |Â

up vote
0
down vote

Pick a number $k$ between one thousand billion and ten thousand billion and take $y = frackx$ ?

answered Jul 19 at 6:01

Daniel Pietrobon

3,26421836

If you look at the examples, the OP wants $y$ to be a power of $10$.
â€“Â Arnaud Mortier
Jul 19 at 6:09

add a commentÂ |Â

up vote
0
down vote

Pick a number $k$ between one thousand billion and ten thousand billion and take $y = frackx$ ?

answered Jul 19 at 6:01

Daniel Pietrobon

3,26421836

Pick a number $k$ between one thousand billion and ten thousand billion and take $y = frackx$ ?

answered Jul 19 at 6:01

Daniel Pietrobon

3,26421836

answered Jul 19 at 6:01

Daniel Pietrobon

3,26421836

answered Jul 19 at 6:01

Daniel Pietrobon

3,26421836

answered Jul 19 at 6:01

Daniel Pietrobon

3,26421836

If you look at the examples, the OP wants $y$ to be a power of $10$.
â€“Â Arnaud Mortier
Jul 19 at 6:09

add a commentÂ |Â

If you look at the examples, the OP wants $y$ to be a power of $10$.
â€“Â Arnaud Mortier
Jul 19 at 6:09

If you look at the examples, the OP wants $y$ to be a power of $10$.
â€“Â Arnaud Mortier
Jul 19 at 6:09

add a commentÂ |Â

up vote
0
down vote

answered Jul 19 at 6:01

Piyush Divyanakar

3,258122

add a commentÂ |Â

up vote
0
down vote

answered Jul 19 at 6:01

Piyush Divyanakar

3,258122

add a commentÂ |Â

up vote
0
down vote

answered Jul 19 at 6:01

Piyush Divyanakar

3,258122

answered Jul 19 at 6:01

Piyush Divyanakar

3,258122

answered Jul 19 at 6:01

Piyush Divyanakar

3,258122

answered Jul 19 at 6:01

Piyush Divyanakar

3,258122

answered Jul 19 at 6:01

Piyush Divyanakar

3,258122

add a commentÂ |Â

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